# CBSE Class 12 Maths Revision Notes Chapter 11

## Class 12 Mathematics Chapter 11 Notes

Class 12 is a very important stage in a student’s academic life. The marks scored in the board exam are considered for  competitive exams and admission to prestigious colleges for  higher studies. Mathematics is regarded as  one of the most challenging subjects. A strong conceptual understanding of all the fundamental concepts and consistent practice are necessary. Using the best reference material, such as the Class 12 Mathematics Chapter 11 notes, helps students to gain a clear understanding of three-dimensional geometry.

Extramarks, an online learning platform,  offers  professional assistance  and the best study material for students to  prepare each and every topic thoroughly  to ace their exams.

## NCERT Class 12 Mathematics Chapter 11 Notes: Key Topics

### Introduction:

This Chapter provides a detailed description of 3-D geometry and the coordinate system. Students will learn to use vector algebra, direction cosines and direction ratios of a line joining two points. The  Class 12 Mathematics Chapter 11 notes include all equations of planes and lines in Space. The three-dimensional coordinate planes divide Space into a total of eight parts called the Octants.

### Representation of a point in Cartesian and Vector form

The point P (1, 2, 3) in the Cartesian form will be represented as (x,y,z)=(1, 2, 3).

In the vector form, the point P is represented as i+2j+3k.

Represent line in 2-D where the endpoints are given as (2,4) and (3,5):-

In the Cartesian form, we write it as

y-4x-3=5-43-2=1

Therefore, (y-3) = 1 (x-3)

In the vector form, we write it as

AB= (ba) = (3i + 2j) – (5i + 4j)

AB= – 2i – 2j

Represent line in 3-D where the endpoints are (2, 4, 5) and (3, 6, 2)

In the vector form, we write it as

AB= (ba) = (3i + 6j + 2k) – (2i + 4j + 5k)

AB= – i – 2j +3k

### Direction Cosines of a Line

Consider any line passing through the origin (0, 0) and making angles α, β and γ with the x-axis, y-axis and z-axis, respectively; then cosines of the angles, i.e., cosα, cosβ and cosγ are called the direction cosines of the line.

Suppose, cos α =l, cos β =m and cos = n, then l2+ m2+ n2= 1 i.e. cos2α + cos2β+ cos2 =1.

Refer to the Class 12 Mathematics Chapter 11 Notes to get more detailed information on this topic.

### Direction Ratios of a Line

Direction ratios are defined as the three numbers which are proportional to the direction cosines of the line. They are denoted by (a, b and c).

For a vector OP= aî +bĵ +ck̂ =r (cos α î + cosβ ĵ + cos γ k̂) where r= a2+b2+c2 then cos α, cos β and cos are the direction cosines, whereas r cos α=a, r cos β =b, r cos =c are termed as direction ratios.

#### Relationship between the Direction Ratios and Direction Cosines

Suppose a line whose one point is at origin (0, 0) and another point is at P. For a vector OP= aî +bĵ +ck̂ =r (cos α î + cosβ ĵ + cos γ k̂) where r= a2+b2+c2

we get a=(r cos α) ,b =(r cos β) and c=(r cosγ)

Therefore, we get cos α =(ar), cos β=(br) and cos =(cr).

Hence, cos α = (ar)= (±) aa2+b2+c2

Similarly, cos β=(br)= (±) ba2+b2+c2 and cos = (cr)= (±) ca2+b2+c2.

#### Relation between the direction ratios and cosines when a line is passing through two points:

Only one line passes through the given two points. For a line PQ, and the coordinates are given as P(x1, y1, z1) and Q(x2, y2, z2). Consider l, m and n as the direction cosines of line PQ that will make an angle α, β and with the x-axis, y-axis and z-axis, respectively.

Hence, cos α= x2x1PQ , cosβ = y2y1PQ and cosγ = z2z1PQ, where PQ=(x2x1)2+(y2y1)2+(z2z1)2

The direction ratios of a line segment can also be considered as (x2x1, y2y1, z2z1 or x1x2, y1y2, z1z2).

### Equation of a line in Space

1. A line passing through a point and is parallel to a given vector is given as r = a + b, where

a = position vector at a point A w.r.t. the origin and is parallel to vector b

l = line passing through the point A and vector b and

r = position vector at P

= real number.

In Cartesian form:

x-x1a=y-y1b=z-z1c, where coordinates of point A is (x1, y1, z1) and direction ratios of the given line be a, b, and c. Coordinates of point P be (x, y, z).

Then r = xî +yĵ +zk̂, a = x1î +y1ĵ +z1

2. A line passing through points A(x1, y1, z1) and B(x2, y2, z2) will be r = a + (ba), where

a and b = position vectors

r = position vector at P(x, y, z)

= real number.

In Cartesian form:

x-x1x2x1=y-y1y2y1=z-z1z2z1, where x = x1 + (x2x1), y = y1 + (y2y1) and z = z1 + (z2z1)

Solve unlimited problems included in the Class 12 Mathematics Chapter 11 Notes at Extramarks website.

### Angle between the two lines with respect to their direction ratios:

Consider L1 and L2 as two lines which pass through the origin (0, 0). L1 and L2 has direction ratios a1,b1, c1 and a2, b2, c2, respectively. If P is a point on the line L1 and Q is a point on the line L2, then the directed lines will be OP and OQ. Suppose, q is the acute angle between OP and OQ, then the angle q is given by

Cos = a1a2+b1b2+c1c2a12+b12+c12a22+b22+c22 and Sin = (a1b2a2b1)2+(b1c2b2c1)2+(c1a2c2a1)2a12+b12+c12a22+b22+c22

The direction ratios of the two lines a1,b1, c1 and a2, b2, c2 are

(i) perpendicular i.e. if q = 90° and

(ii) parallel i.e. if q = 0

#### Distance between two lines:

1. For two skew lines, r1 = a1 + b1 and r2 = a2 + b2

Let PQ be the shortest distance between lines l1 and l2.  If n is the unit vector along PQ, then

n = b1 X b2b1 X b2. If PQ = d. n, where d is the magnitude of distance.

d= (b1 X b2) . (a1 X a2)b1 X b2

2. For two parallel lines, r1 = a1 + b1 and r2 = a2 + b2

l1 and l2 are coplanar lines as they are parallel, d = PT= b . (a1a2)b

#### PLANE:

A plane can be determined uniquely if:

(i) the normal to the plane and the distance from the origin (0, 0) is given, that is, the equation of the plane in the normal form.

(ii) it passes through any point and is perpendicular to the direction.

(iii) it passes through the given non-collinear points.

#### Equation in normal form:

If d is the distance from the origin, ON is the normal and n= unit normal. Suppose NP is perpendicular to ON and r is the position vector at P(x, y, z), then r.n = d is the vector equation and lx + my + nz = d is the Cartesian equation of the plane

Refer to the Class 12 Chapter 11 Mathematics notes to access unlimited problems for extra practice.

#### Equation of a given plane which is perpendicular to a vector and passes through any point:

Consider a plane passing through point A with a position vector a perpendicular to the vector N. If  r is the position vector of the point P(x, y, z), then (ra). N = 0 is the vector equation and A (x-x1) + B (y-y1) + C (z-z1) = 0 is the cartesian equation of the plane.

#### Equation of a plane through three non-collinear points

Let points R, S and T be non-collinear on the plane having position vectors a, b, and c, respectively.

The equation of the plane through point R and perpendicular to the RS and RT in the vector equation is given as (ra) [(ba) x (ca)] = 0. The Cartesian equation is in the form of determinants, we have |D| = 0, where D =

 x-x1 y-y1 z-z1 x-x2 y-y2 z-z2 x-x3 y-y3 z-z3

### Intercept form of an equation:

Consider Ax + By + Cz + D = 0 as the equation of the plane. If a, b, c are the intercepts on x, y and z-axes at (a, 0, 0), (0, b, 0) and also (0, 0, c) respectively. Then the required plane equation in the form of intercepts is given as xa+yb+zc=0

The Class 12 Mathematics Chapter 11 Notes includes several questions based on this concept. Students must consider referring to these notes for extra practice.

#### Equation of a given plane passing through the intersection of two different planes:

If P1 and P2 are two planes having equations r.n1 = d1 and r.n2 = d2, where r is the position vector, then we express any plane which passes through the intersection of two planes as r.(n1+n2) = d1+ d2 in the vector equation.

In Cartesian equation, we have

(A1x + B1y + C1z – d1) + (A2x + B2y + C2z – d2) = 0

### Coplanarity of Two Lines:

Consider lines r = a1 + b1 and r = a2 + b2, if AB is perpendicular to b1 X b2. Then in vector form, we have AB. (b1 X b2)= 0 or (a2a1).(b1 X b2) = 0. In cartesian form, the equation is expressed as |D|=0, where D =

 x2–x1 y2–y1 z2–z1 a1 b1 c1 a2 b2 c2

#### Angle between Two Planes

Cos = n1.n2n1n2, where is the angle between the two planes drawn from a common point.

In Cartesian form, Cos = a1a2+b1b2+c1c2a12+b12+c12a22+b22+c22 where a1,b1, c1 and a2, b2, c2 are the direction ratios of the equation of planes.

#### Angle between line and plane:

The angle between a plane r.n1 = d1 and a line r = a + b is given as:

In vector form: Cos = Sin = b . nbn

Visit the links provided below and get the key points and summary in the Class 12 Mathematics Chapter 11 notes associated with the CBSE syllabus.

## Class 12 Mathematics Chapter 11 Notes:Exercise &  Solutions

The expert faculty at Extramarks has prepared the Class 12 Mathematics Chapter 11 notes by analysing several CBSE sample papers and CBSE previous year question papers. The notes include solutions to all Exercise and Answer problems in the NCERT books. Students can clarify their doubts and seek help if they face any difficulties. The CBSE revision notes include all theorems, formulas, and derivations in the CBSE syllabus in a very descriptive way. With the help of Class 12 Mathematics notes Chapter 11, students can efficiently prepare for the boards and national level competitive exams. Our notes are provided with detailed information and apt knowledge for students’ overall development.

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### NCERT Class 12 Mathematics Chapter 11 Notes: Key Features

• The notes are prepared by experienced  through Extramarks Research & Development Wing.
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Q.1 Find the vector equation for the line passing through the points (–1, 0, 2) and (3, 4, 6).

Ans

$\begin{array}{l}\text{We know}\stackrel{\to }{\mathrm{r}}=\stackrel{\to }{\mathrm{a}}+\mathrm{\lambda }\left(\stackrel{\to }{\mathrm{b}}-\stackrel{\to }{\mathrm{a}}\right)\\ \therefore \text{}\stackrel{\to }{\mathrm{r}}=-\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{k}}+\mathrm{\lambda }\left[3\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}+6\stackrel{^}{\mathrm{k}}-\left(-\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{k}}\right)\right]\\ \stackrel{\to }{\mathrm{r}}=-\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{k}}+\mathrm{\lambda }\left[4\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}\right]\end{array}$

Q.2 Find the coordinates of the foot of the perpendicular drawn from the point A(1, 8, 4) to the line joining the points B(0, –1, 3) and C(2, –3, –1).

Ans

The equation of line joining the points B and C $\begin{array}{l}\frac{\mathrm{x}-0}{2-0}=\frac{\mathrm{y}+1}{-3+1}=\frac{\mathrm{z}-3}{-1-3}\\ \frac{\mathrm{x}}{2}=\frac{\mathrm{y}+1}{-2}=\frac{\mathrm{z}-3}{-4}\\ \frac{\mathrm{x}}{1}=\frac{\mathrm{y}+1}{-1}=\frac{\mathrm{z}-3}{-2}=\mathrm{\lambda }\left(\text{say}\right)\\ \mathrm{x}=\mathrm{\lambda },\mathrm{y}=-\mathrm{\lambda }-1,\mathrm{z}=-2\mathrm{\lambda }+3\\ \text{Let the coordinates of point M be}\\ \text{M}\left(\mathrm{\lambda },-\mathrm{\lambda }-1,-2\mathrm{\lambda }+3\right)\\ \text{Now d.r’s of AM are}\\ \mathrm{\lambda }-1,-\mathrm{\lambda }-1-8,-2\mathrm{\lambda }+3-4\\ \mathrm{\lambda }-1,-\mathrm{\lambda }-9,-2\mathrm{\lambda }-1\\ \text{AM}\perp \text{BC}\\ \left(\mathrm{\lambda }-1\right)1+\left(-\mathrm{\lambda }-9\right).\left(-1\right)+\left(-2\mathrm{\lambda }-1\right).\left(-2\right)=0\\ 6\mathrm{\lambda }+10=0\\ \mathrm{\lambda }=-\frac{5}{3}\\ \text{Thus, M}\left(-\frac{5}{3},\frac{2}{3},\frac{19}{3}\right)\end{array}$

Q.3 If a line has direction ratios 2, – 1, – 2, determine its direction cosines.

Ans

Direction cosines are

$\begin{array}{l}\mathrm{}\mathrm{l}\mathrm{}=\mathrm{}\frac{2}{\sqrt{{2}^{2}+{\left(-1\right)}^{2}+{\left(-2\right)}^{2}}}\mathrm{}=\mathrm{}\frac{2}{3}\\ \mathrm{}\mathrm{m}\mathrm{}=\mathrm{}\frac{-1}{\sqrt{{2}^{2}+{\left(-1\right)}^{2}+{\left(-2\right)}^{2}}}\mathrm{}=\mathrm{}\frac{-1}{3}\\ \mathrm{n}\mathrm{}=\mathrm{}\frac{-2}{\sqrt{{2}^{2}+{\left(-1\right)}^{2}+{\left(-2\right)}^{2}}}\mathrm{}=\mathrm{}\frac{-2}{3}\end{array}$

Q.4 Find the direction cosines of the line passing through the two points
(– 2, 4, – 5) and (1, 2, 3).

Ans

$\begin{array}{l}\mathrm{Direction}\mathrm{cosines}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{passing}\mathrm{through}\mathrm{two}\mathrm{points}\\ \mathrm{P}\left({\mathrm{x}}_{1},\mathrm{}{\mathrm{y}}_{1},\mathrm{}{\mathrm{z}}_{1}\right)\mathrm{}\mathrm{and}\mathrm{}\mathrm{Q}\left({\mathrm{x}}_{2},\mathrm{}{\mathrm{y}}_{2},\mathrm{}{\mathrm{z}}_{2}\right)\\ \mathrm{}\frac{{\mathrm{x}}_{2}-{\mathrm{x}}_{1}}{\mathrm{PQ}},\frac{{\mathrm{y}}_{2}-{\mathrm{y}}_{1}}{\mathrm{PQ}},\frac{{\mathrm{z}}_{2}-{\mathrm{z}}_{1}}{\mathrm{PQ}},\mathrm{Where},\mathrm{PQ}\mathrm{}\sqrt{{\left({\mathrm{x}}_{2}-{\mathrm{x}}_{1}\right)}^{2}+{\left({\mathrm{y}}_{2}-{\mathrm{y}}_{1}\right)}^{2}+{\left({\mathrm{z}}_{2}-{\mathrm{z}}_{1}\right)}^{2}}\\ \therefore \mathrm{}\mathrm{PQ}\mathrm{}=\mathrm{}\sqrt{{\left(1-2\right)}^{2}+{\left(2-4\right)}^{2}+{\left(3+5\right)}^{2}}\mathrm{}=\mathrm{}\sqrt{77}\\ \mathrm{Thus},\mathrm{the}\mathrm{direction}\mathrm{cosines}\\ \therefore \mathrm{}\frac{3}{\sqrt{77}},\mathrm{}\frac{-2}{\sqrt{77}},\mathrm{}\frac{8}{\sqrt{77}}\end{array}$

Q.5 Show that the points A (2, 3, – 4), B (1, – 2, 3) and C (3, 8, – 1) are collinear.

AnsDirection ratios of line joining A and B are

1 – 2, –2 – 3, 3 + 4 i.e., –1, –5, 7.

The direction ratios of line joining B and C are

3 – 1, 8 + 2, –11 – 3, i.e., 2, 10, –14.

It is clear that direction ratios of AB and BC are proportional, hence, AB is parallel to BC. But point B is common to both AB and BC. Therefore, A, B, C are collinear points.

Q.6

$\text{Write the equation of a line that passes through a given point A and parallel to a given vector}\stackrel{\to }{\text{b}}\text{.}$

Ans

$\stackrel{\to }{\mathrm{r}}=\stackrel{^}{\mathrm{a}}+\mathrm{\lambda }\stackrel{^}{\mathrm{b}},\mathrm{where}\stackrel{^}{\mathrm{a}}\mathrm{}\mathrm{is}\mathrm{the}\mathrm{position}\mathrm{vector}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{point}\mathrm{A}.$

Q.7 Write the Cartesian equation of the line, if the coordinates of the given point A are (x1, y1, z1) and the direction ratios of the line are a, b, c.

Ans

$\mathrm{The}\mathrm{cartesian}\mathrm{equationof}\mathrm{the}\mathrm{line}\mathrm{is}\frac{\mathrm{x}-{\mathrm{x}}_{1}}{\mathrm{a}},\frac{\mathrm{y}-{\mathrm{y}}_{1}}{\mathrm{b}},\mathrm{}\frac{\mathrm{z}-{\mathrm{z}}_{1}}{\mathrm{c}}$

Q.8 Find the distance of the point (3, -5, 12) from x-axis.

Ans

$\begin{array}{l}\text{The distance of the point P}\left(\text{x, y, z}\right)\text{from x-axis =}\sqrt{{\mathrm{y}}^{2}+{\mathrm{z}}^{2}}\\ \text{The distance of the point P}\left(\text{3, -5, 12}\right)\text{from x-axis =}\sqrt{{\left(-5\right)}^{2}+{\left(12\right)}^{2}}\\ \text{=}13\text{units.}\end{array}$

Q.9 The Cartesian equation of a line is

$\frac{\mathrm{x}+3}{2}=\frac{\mathrm{y}-5}{4}=\frac{\mathrm{z}+6}{2}$

. Find the vector equation for the line.

Ans

Comparing the given equation with the standard form

$\frac{\mathrm{x}-{\mathrm{x}}_{1}}{\mathrm{a}},\frac{\mathrm{y}-{\mathrm{y}}_{1}}{\mathrm{b}},\frac{\mathrm{z}-{\mathrm{z}}_{1}}{\mathrm{c}}$

, we get

x1= – 3, y1= 5, z1 = – 6; a = 2, b = 4, c = 2.
Thus, the line passes through the point (–3, 5, –6) and is parallel to
the vector

$2\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}$

Then, vector equation of the line

$\stackrel{\to }{\mathrm{r}}=-3\stackrel{^}{\mathrm{i}}+5\stackrel{^}{\mathrm{j}}-6\stackrel{^}{\mathrm{k}}+\mathrm{\lambda }\left(2\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right)$

Q.10 Find the angle between the pair of lines given by .

$\stackrel{\to }{\mathrm{r}}=3\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-4\stackrel{^}{\mathrm{k}}+\mathrm{\lambda }\left(\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right)\text{and}\stackrel{\to }{\mathrm{r}}=5\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}+\mathrm{\mu }\left(3\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+6\stackrel{^}{\mathrm{k}}\right)$

Ans

$\begin{array}{l}\text{Here,}\stackrel{\to }{{\mathrm{b}}_{1}}=\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\text{and}\stackrel{\to }{{\mathrm{b}}_{2}}=3\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+6\stackrel{^}{\mathrm{k}}\\ \text{Therefore,}\\ \mathrm{cos}\mathrm{\theta }=\frac{{\stackrel{\to }{\mathrm{b}}}_{1}.{\stackrel{\to }{\mathrm{b}}}_{2}}{\left|{\stackrel{\to }{\mathrm{b}}}_{1}\right|\left|{\stackrel{\to }{\mathrm{b}}}_{2}\right|}\\ \text{}=\frac{\left(\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right).\left(3\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+6\stackrel{^}{\mathrm{k}}\right)}{\sqrt{1+4+4}\sqrt{9+4+36}}\\ \text{}=\frac{3+4+12}{\sqrt{9}\sqrt{49}}=\frac{19}{21}\\ ⇒\mathrm{\theta }={\mathrm{cos}}^{-1}\left(\frac{19}{21}\right)\end{array}$

Q.11

$\begin{array}{l}\text{If the \hspace{0.17em}}\mathrm{line}\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{r}}\text{\hspace{0.17em}\hspace{0.17em}}=\text{\hspace{0.17em}\hspace{0.17em}}\left(\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}\hspace{0.17em}}-\text{\hspace{0.17em}\hspace{0.17em}}2\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\right)\text{\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{\lambda }\text{\hspace{0.17em}\hspace{0.17em}}\left(2\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}}2\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\right)\text{\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{plane}\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}}\stackrel{\to }{\mathrm{r}}.\text{\hspace{0.17em}\hspace{0.17em}}\left(3\text{\hspace{0.17em}\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}\hspace{0.17em}}-2\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}\hspace{0.17em}}+\mathrm{m}\text{\hspace{0.17em}\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{\hspace{0.17em}\hspace{0.17em}}14.\mathrm{Find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{m}.\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\text{given line is parallel to the vector}\stackrel{\to }{\mathrm{b}}\text{=}2\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}}2\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ \mathrm{and}\text{the given vector is normal to the vector}\stackrel{\to }{\mathrm{n}}\text{=}3\text{\hspace{0.17em}\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}\hspace{0.17em}}-2\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}\hspace{0.17em}}+\mathrm{m}\text{\hspace{0.17em}\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\\ \mathrm{If}\text{the line is parallel to th the plane then normal is perpendicular}\\ \mathrm{to}\text{}\mathrm{the}\text{}\mathrm{line}.\\ \therefore \stackrel{\to }{\mathrm{b}}\perp \stackrel{\to }{\mathrm{n}}\\ ⇒\stackrel{\to }{\mathrm{b}}\text{.}\stackrel{\to }{\mathrm{n}}\text{=0}\\ ⇒\left(2\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}}2\text{\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\text{}\right).\left(3\text{\hspace{0.17em}\hspace{0.17em}}\stackrel{^}{\mathrm{i}}\text{\hspace{0.17em}\hspace{0.17em}}-2\text{\hspace{0.17em}}\stackrel{^}{\mathrm{j}}\text{\hspace{0.17em}\hspace{0.17em}}+\mathrm{m}\text{\hspace{0.17em}\hspace{0.17em}}\stackrel{^}{\mathrm{k}}\right)=0\\ ⇒\text{}6-2+2\mathrm{m}=0\\ ⇒\text{}\mathrm{m}=-2\end{array}$

Q.12 Find the distance of the plane 2x-3y+4z-6=0 from the origin.

Ans

Direction ratios of the normal to the plane are 2, –3, 4
Direction cosines are

$\frac{2}{\sqrt{4+9+16}},\frac{-3}{\sqrt{4+9+16}},\frac{4}{\sqrt{4+9+16}}\text{or}\frac{2}{\sqrt{29}},\frac{-3}{\sqrt{29}},\frac{4}{\sqrt{29}}$

Hence, dividing the equation 2x – 3y + 4z – 6 = 0 i.e., 2x – 3y + 4z = 6 throughout by

$\sqrt{29}$

, we get

$\begin{array}{l}\frac{2}{\sqrt{29}}\mathrm{x}-\frac{-3}{\sqrt{29}}\mathrm{y}+\frac{4}{\sqrt{29}}\mathrm{z}=\frac{6}{\sqrt{29}}\\ \therefore \text{the distance of the plane from the origin}=\frac{6}{\sqrt{29}}\end{array}$

Q.13 Find the equation of the plane through the point (1, 4, -2) and parallel to the plane -2x + y – 3z = 7.

Ans

Let the equation of a plane parallel to the plane -2x + y – 3z = 7 be -2x + y – 3z + k = 0.
This passes through (1, 4, -2). Therefore,
(-2)(1) + 4 – 3(-2) + k = 0
⇒ -2 + 4 + 6 + k = 0
⇒ k = -8

Q.14 Find the coordinates of the point where the line through the points
A (3, 4, 1) and
B(5, 1, 6) crosses the XY-plane.

Ans

The vector equation of the line through the points A and B is

$\begin{array}{l}\stackrel{\to }{\mathrm{r}}=\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}+\mathrm{\lambda }\left[\left(5-3\right)\stackrel{^}{\mathrm{i}}+\left(1-4\right)\stackrel{^}{\mathrm{j}}+\left(6-1\right)\stackrel{^}{\mathrm{k}}\right]\\ \mathrm{or}\stackrel{\to }{\mathrm{r}}=\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}+\mathrm{\lambda }\left(2\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+5\stackrel{^}{\mathrm{k}}\right).\dots ..\left(\mathrm{i}\right)\\ \mathrm{Let}\mathrm{P}\mathrm{be}\mathrm{the}\mathrm{point}\mathrm{where}\mathrm{the}\mathrm{line}\mathrm{AB}\mathrm{crosses}\mathrm{the}\mathrm{XY}–\mathrm{plane}.\\ \mathrm{The}\mathrm{the}\mathrm{position}\mathrm{vector}\mathrm{of}\mathrm{thepoint}\mathrm{P}\mathrm{is}\mathrm{of}\mathrm{the}\mathrm{form}\mathrm{x}\stackrel{^}{\mathrm{i}}+\mathrm{y}\stackrel{^}{\mathrm{j}}\\ \mathrm{Since}\mathrm{point}\mathrm{P}\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{plane}\left(\mathrm{i}\right),\mathrm{then}\\ \mathrm{x}\stackrel{^}{\mathrm{i}}+\mathrm{y}\stackrel{^}{\mathrm{j}}=\mathrm{}\left(3+2\mathrm{\lambda }\right)\mathrm{}\stackrel{^}{\mathrm{i}}+\left(4-3\mathrm{\lambda }\right)\stackrel{^}{\mathrm{j}}+\left(1+5\mathrm{\lambda }\right)\stackrel{^}{\mathrm{k}}\\ \mathrm{Comparing}\mathrm{the}\mathrm{like}\mathrm{coefficients}\mathrm{of}\stackrel{^}{\mathrm{i}},\mathrm{}\stackrel{^}{\mathrm{j}},\mathrm{}\stackrel{^}{\mathrm{k}},\mathrm{we}\mathrm{have}\\ \mathrm{x}= 3+2\mathrm{\lambda },\mathrm{y}= 4-3\mathrm{\lambda }\mathrm{and}0 = 1+5\mathrm{\lambda }\\ \mathrm{Now}, 0 = 1+5\mathrm{\lambda }\mathrm{}⇒\mathrm{}\mathrm{\lambda }= –\frac{1}{5}\\ \therefore \mathrm{x}\mathrm{}=\mathrm{}3-\frac{2}{5}\mathrm{}=\mathrm{}\frac{13}{5}\mathrm{}\mathrm{and}\\ \mathrm{y}\mathrm{}=\mathrm{}4+\frac{3}{5}=\frac{23}{5}\\ \mathrm{Hence},\mathrm{coordinates}\mathrm{of}\mathrm{the}\mathrm{point}\mathrm{P}\mathrm{are}\left(\frac{13}{5},\mathrm{}\frac{23}{5},\mathrm{}0\right)\end{array}$

Q.15 If the points (1, 1, p) and (– 3, 0, 1) be equidistant from the plane

$\stackrel{\to }{\mathrm{r}}.\left(3\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}-12\stackrel{^}{\mathrm{k}}\right)+13\mathrm{}=0$

, then find the value of p.

Ans

$\begin{array}{l}\mathrm{Perpendicular}\mathrm{distance}\mathrm{from}\mathrm{a}\mathrm{point}\mathrm{whose}\mathrm{position}\\ \mathrm{vector}\mathrm{is}\stackrel{\to }{\mathrm{a}}\mathrm{to}\mathrm{the}\mathrm{plane}\stackrel{\to }{\mathrm{r}}.\stackrel{\to }{\mathrm{N}}\mathrm{}=\mathrm{d}\mathrm{is}\mathrm{given}\mathrm{by}\\ \left|\frac{\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{N}}-\mathrm{d}}{\stackrel{\to }{\mathrm{N}}}\right|\\ \mathrm{Position}\mathrm{vector}\mathrm{of}\left(1,\mathrm{}1,\mathrm{}\mathrm{p}\right)\mathrm{}=\mathrm{}\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}-\mathrm{p}\stackrel{^}{\mathrm{k}}\\ \mathrm{Position}\mathrm{vector}\mathrm{of}\left(-3,\mathrm{}0,\mathrm{}1\right)\mathrm{}=\mathrm{}-3\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{k}}\\ \therefore \mathrm{Distance}\mathrm{of}\mathrm{plane}\stackrel{\to }{\mathrm{r}}.\left(3\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}-12\stackrel{^}{\mathrm{k}}\right)+13=\mathrm{}0\mathrm{}\left(1,\mathrm{ }1,\mathrm{ }\mathrm{p}\right)\\ =\mathrm{}\left|\frac{3+4-12\mathrm{p}+13}{\sqrt{{3}^{2}+{4}^{2}+{12}^{2}}}\right|\\ =\mathrm{}\left|\frac{20-12\mathrm{p}}{\sqrt{169}}\right|\\ =\mathrm{}\left|\frac{20-12\mathrm{p}}{\sqrt{169}}\right|\mathrm{}=\mathrm{}\left|\frac{20-12\mathrm{p}}{13}\right|.\dots ..\mathrm{}\left(\mathrm{i}\right)\\ \mathrm{Now}\mathrm{distance}\mathrm{of}\mathrm{plane}\stackrel{\to }{\mathrm{r}}.\left(3\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}-12\stackrel{^}{\mathrm{k}}\right)+13\mathrm{}=0\mathrm{}\mathrm{from}\mathrm{}\left(-3,\mathrm{}0,\mathrm{}1\right)\mathrm{}\\ =\mathrm{}\left|\frac{-9+0-12+13}{\sqrt{{3}^{2}+{4}^{2}+{12}^{2}}}\right|\\ =\mathrm{}\left|\frac{-8}{\sqrt{169}}\right|\\ =\mathrm{}\left|\frac{-8}{13}\right|\\ =\frac{8}{13}.\dots \dots \left(\mathrm{ii}\right)\\ \mathrm{Since}\mathrm{point}\mathrm{is}\mathrm{equidistant}\mathrm{from}\mathrm{the}\mathrm{plane},\\ \mathrm{therefore},\mathrm{from}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right)\mathrm{}\mathrm{we}\mathrm{}\mathrm{get}\\ =\mathrm{}\left|\frac{20-12\mathrm{p}}{\sqrt{169}}\right|=\frac{8}{13}\\ ⇒20-12\mathrm{p}\mathrm{}=\mathrm{}8\mathrm{or}–\left(20-12\mathrm{p}\right)\mathrm{ }=\mathrm{}8\\ ⇒\mathrm{p}=\mathrm{}1\mathrm{or}\mathrm{p}\mathrm{}=\mathrm{}\frac{7}{3}\end{array}$

Q.16 If a line makes angle 90°, 60° and 30° with the positive direction of x, y and z-axis respectively, find its direction cosines.

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{the}\mathrm{d}.\mathrm{c}‘\mathrm{s}\mathrm{of}\mathrm{the}\mathrm{lines}\mathrm{be}\mathrm{l},\mathrm{m},\mathrm{n}.\mathrm{Then}\\ \mathrm{l}=\mathrm{cos}90°= 0,\\ \mathrm{m}=\mathrm{cos}60°= \frac{1}{2}\\ \mathrm{n}=\mathrm{cos}30°= \frac{\sqrt{3}}{2}\end{array}$

Q.17 Show that the planes 2x + 6y + 6z = 7 and 3x + 4y – 5z = 8 are at ight angles.

AnsWe kknow that the planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 are at right angles, if a1a2 + b1b2 + c1c2 = 0
Now, (2)(3) + (6)(4) + (6)(-5) = 0
Therefore, planes 2x + 6y + 6z = 7 and 3x + 4y – 5z = 8 are at right angles.

Q.18 Write the vector equation of a plane passing through the intersection of two given planes.

Ans

If

$\mathrm{If}\stackrel{\to }{\mathrm{r}}.{\stackrel{\to }{\mathrm{n}}}_{{}_{1}}\mathrm{}=\mathrm{}{\mathrm{d}}_{1}\mathrm{}\mathrm{and}\mathrm{}\stackrel{\to }{\mathrm{r}}.{\stackrel{\to }{\mathrm{n}}}_{{}_{2}}={\mathrm{d}}_{2}\mathrm{}$

are two planes, then equation of a plane passing through the intersection of two given planes is given by

$\stackrel{\to }{\mathrm{r}}.\left({\stackrel{\to }{\mathrm{n}}}_{{}_{1}}+\mathrm{\lambda }{\stackrel{\to }{\mathrm{n}}}_{2}\mathrm{}\right)=\mathrm{}{\mathrm{d}}_{1}+{\mathrm{\lambda d}}_{2}$

Q.19

$\begin{array}{l}\mathrm{If}\mathrm{the}\mathrm{plane}4\mathrm{x}+4\mathrm{y}–\mathrm{\lambda z}=0\mathrm{contains}\mathrm{the}\mathrm{line}\left(\mathrm{x}-1\right)/2=\\ \left(\mathrm{y}+1\right)/3=\mathrm{z}/4,\mathrm{then}\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{\lambda }.\end{array}$

Ans

$\begin{array}{l}\mathrm{If}\mathrm{}\mathrm{the}\mathrm{}\mathrm{plane}\mathrm{}4\mathrm{x}\mathrm{}+4\mathrm{y}\mathrm{}-\mathrm{\lambda z}\mathrm{}=0\mathrm{}\mathrm{contains}\mathrm{}\mathrm{the}\mathrm{}\mathrm{line}\mathrm{}\\ \left(\mathrm{x}-1\right)/2=\left(\mathrm{y}+1\right)/3=\mathrm{z}/4,\mathrm{}\mathrm{then}\\ 2×4\mathrm{}+3×4-4×\mathrm{\lambda }=0\\ ⇒20-4×\mathrm{\lambda }=0\\ ⇒\mathrm{\lambda }=5\end{array}$

Q.20 Find the equation of the plane passing through the points (1, 2, 3) and (0, –1, 0) and parallel to the line .

$\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}+2}{3}=\frac{\mathrm{z}}{-3}$

Ans

$\begin{array}{l}\text{The equation of the plane passing through}\left(\text{1},\text{2},\text{3}\right)\text{is}\\ \mathrm{a}\left(\mathrm{x}-1\right)+\mathrm{b}\left(\mathrm{y}-2\right)+\mathrm{c}\left(\mathrm{z}-3\right)=0\\ \text{If the plane passes through}\left(0,-\text{1},0\right),\text{then}\\ \mathrm{a}\left(0-1\right)+\mathrm{b}\left(-1-2\right)+\mathrm{c}\left(0-3\right)=0\\ ⇒\mathrm{a}+3\mathrm{b}+3\mathrm{c}=0\dots \left(\text{ii}\right)\\ \text{Now},\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}+2}{3}=\frac{\mathrm{z}}{-3}\text{is parallel to plane}\\ \text{Therefore},2\mathrm{a}+3\mathrm{b}-3\mathrm{c}=0...\left(\text{iii}\right)\\ \text{By}\left(\text{i}\right)\text{}\left(\text{ii}\right)\text{and}\left(\text{iii}\right)\\ \left|\begin{array}{ccc}\mathrm{x}-1& \mathrm{y}-2& \mathrm{z}-3\\ 1& 3& 3\\ 2& 3& -3\end{array}\right|=0\\ 6\left(\mathrm{x}-1\right)-3\left(\mathrm{y}-2\right)+\left(\mathrm{z}-3\right)=0\\ 6\mathrm{x}-3\mathrm{y}+\mathrm{z}-3=0\end{array}$

Q.21 Find the angle between the planes x + y + 2z = 9 and 2x – y + z = 15.

Ans

$\begin{array}{l}\mathrm{The}\mathrm{angle}\mathrm{between}\mathrm{two}\mathrm{planes}{\mathrm{A}}_{1}\mathrm{x}+{\mathrm{B}}_{1}\mathrm{y}+{\mathrm{C}}_{1}\mathrm{z}+{\mathrm{D}}_{1}=0\text{and}\\ {\mathrm{A}}_{2}\mathrm{x}+{\mathrm{B}}_{2}\mathrm{y}+{\mathrm{C}}_{2}\mathrm{z}+{\mathrm{D}}_{2}=0\mathrm{is}\mathrm{given}\mathrm{by}\\ \mathrm{\theta }={\mathrm{cos}}^{–1}\left|\frac{{\mathrm{A}}_{1}{\mathrm{A}}_{2}+{\mathrm{B}}_{1}{\mathrm{B}}_{2}+{\mathrm{C}}_{1}{\mathrm{C}}_{2}}{\sqrt{{{\mathrm{A}}_{1}}^{2}+{{\mathrm{B}}_{1}}^{2}+{{\mathrm{C}}_{1}}^{2}}\sqrt{{{\mathrm{A}}_{2}}^{2}+{{\mathrm{B}}_{2}}^{2}+{{\mathrm{C}}_{2}}^{2}}}\right|\\ \\ \text{Hence, the angle between the planes x+y+2z = 9}\\ \text{and 2x-y+ z=15 is given by}\\ \mathrm{\theta }={\mathrm{cos}}^{–1}\left|\frac{1.2+1.\left(–1\right)+2.1}{\sqrt{{1}^{2}+{1}^{2}+{2}^{2}}\sqrt{{2}^{2}+{\left(–1\right)}^{2}+{1}^{2}}}\right|\\ \mathrm{\theta }={\mathrm{cos}}^{–1}\left|\frac{2–1+2}{6}\right|\\ \mathrm{\theta }={\mathrm{cos}}^{–1}\left|\frac{1}{2}\right|\\ \mathrm{\theta }=\frac{\mathrm{\pi }}{3}\end{array}$

Q.22 Find the angle between the line (x+1)/2 = y/3 = (z-3)/6
and the plane 10x +2y-11z = 3 .

Ans

$\begin{array}{l}\text{Converting the equations in vector form we have}\\ \stackrel{\to }{\mathrm{r}}=\left(-\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{k}}\right)+\mathrm{\lambda }\left(2\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+6\stackrel{^}{\mathrm{k}}\right)\text{and}\stackrel{\to }{\mathrm{r}}.\left(10\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-11\stackrel{^}{\mathrm{k}}\right)=3\\ \text{Here}\stackrel{\to }{\mathrm{b}}=2\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+6\stackrel{^}{\mathrm{k}}\text{and}\stackrel{\to }{\mathrm{n}}=10\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-11\stackrel{^}{\mathrm{k}}\\ \mathrm{sin}\mathrm{Ï•}=\left|\frac{\left(2\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+6\stackrel{^}{\mathrm{k}}\right).\left(10\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-11\stackrel{^}{\mathrm{k}}\right)}{\sqrt{{2}^{2}+{3}^{2}+{6}^{2}}\sqrt{{10}^{2}+{2}^{2}+{11}^{2}}}\right|\\ \text{}=\left|\frac{20+6-66}{\sqrt{49}\sqrt{225}}\right|\\ \text{}=\left|\frac{-40}{7×15}\right|=\left|\frac{-8}{21}\right|=\frac{8}{21}\\ ⇒\mathrm{Ï•}={\mathrm{sin}}^{-1}\left(\frac{8}{21}\right)\end{array}$

Q.23 The vector equations of two lines are

$\stackrel{\to }{\mathrm{r}}\mathrm{}=\mathrm{}\stackrel{^}{\mathrm{i}}\mathrm{}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}+\mathrm{\lambda }\mathrm{}\left(\stackrel{^}{\mathrm{i}}\mathrm{}-3\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right)\mathrm{and}\stackrel{\to }{\mathrm{r}}= 4\stackrel{^}{\mathrm{i}}\mathrm{}+5\stackrel{^}{\mathrm{j}}+6\stackrel{^}{\mathrm{k}}\mathrm{}+\mathrm{}\mathrm{\mu }\left(\stackrel{^}{2\mathrm{i}}\mathrm{}+3\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)$

.
Find the shortest distance between the above lines.

Ans

$\begin{array}{l}\mathrm{Comparing}\mathrm{given}\mathrm{lines}\mathrm{with}\mathrm{standard}\mathrm{vector}\mathrm{form}\mathrm{of}\mathrm{lines}\\ \stackrel{\to }{\mathrm{r}}\mathrm{}=\stackrel{\to }{\mathrm{a}}\mathrm{}+\mathrm{\lambda }\left(\stackrel{\to }{{\mathrm{b}}_{1}}\right)\mathrm{and}\stackrel{\to }{\mathrm{r}}\mathrm{}=\mathrm{}\stackrel{\to }{{\mathrm{a}}_{2}}+\mathrm{}\mathrm{\mu }\mathrm{}\left(\stackrel{\to }{{\mathrm{b}}_{2}}\right)\mathrm{we}\mathrm{get}\\ {\stackrel{\to }{\mathrm{a}}}_{1}\mathrm{}=\mathrm{}\stackrel{^}{\mathrm{i}}\mathrm{}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}},\mathrm{}\stackrel{\to }{{\mathrm{b}}_{1}}\mathrm{}=\mathrm{}\stackrel{^}{\mathrm{i}}\mathrm{}-3\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}},\mathrm{}{\stackrel{\to }{\mathrm{a}}}_{2}\mathrm{}=4\stackrel{^}{\mathrm{i}}\mathrm{}+5\stackrel{^}{\mathrm{j}}+6\stackrel{^}{\mathrm{k}}\\ \mathrm{and}{\stackrel{\to }{\mathrm{b}}}_{2}\mathrm{}=2\stackrel{^}{\mathrm{i}}\mathrm{}+3\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\\ \therefore \stackrel{\to }{{\mathrm{a}}_{2}}-{\stackrel{\to }{\mathrm{a}}}_{1}\mathrm{}=4\stackrel{^}{\mathrm{i}}\mathrm{}+5\stackrel{^}{\mathrm{j}}+6\stackrel{^}{\mathrm{k}}-\stackrel{^}{\mathrm{i}}\mathrm{}-2\stackrel{^}{\mathrm{j}}-3\stackrel{^}{\mathrm{k}}\\ =\mathrm{}3\stackrel{^}{\mathrm{i}}\mathrm{}+3\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\\ \mathrm{Now},\stackrel{\to }{{\mathrm{b}}_{1}}×{\stackrel{\to }{\mathrm{b}}}_{2}\mathrm{}=\mathrm{}\left|\begin{array}{ccc}\begin{array}{l}\stackrel{^}{\mathrm{i}}\\ 1\mathrm{}\\ 2\end{array}& \begin{array}{l}\stackrel{^}{\mathrm{j}}\\ -3\\ 3\mathrm{}\end{array}& \stackrel{^}{\begin{array}{l}\mathrm{k}\\ 2\\ 1\end{array}}\mathrm{}\end{array}\right|\\ =\mathrm{}\stackrel{^}{\mathrm{i}}\mathrm{}\left(-3-6\right)\mathrm{}-\stackrel{^}{\mathrm{j}}\left(1-4\right)+\stackrel{^}{\mathrm{k}}\left(3+6\right)\\ =\mathrm{}-9\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+9\stackrel{^}{\mathrm{k}}\\ \mathrm{Therefore},\\ \left|\stackrel{\to }{{\mathrm{b}}_{1}}×{\stackrel{\to }{\mathrm{b}}}_{2}\right|\mathrm{}=\mathrm{}\sqrt{81+9+81}\\ =\mathrm{}\sqrt{171}\\ =\mathrm{}3\sqrt{19}\\ \mathrm{Also},\left(\stackrel{\to }{{\mathrm{a}}_{2}}-{\stackrel{\to }{\mathrm{a}}}_{1}\right).\left(\stackrel{\to }{{\mathrm{b}}_{1}}×{\stackrel{\to }{\mathrm{b}}}_{2}\right)\mathrm{}=\mathrm{}\left(3\stackrel{^}{\mathrm{i}}\mathrm{}+3\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\right).\left(-9\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+9\stackrel{^}{\mathrm{k}}\right)\\ =\mathrm{}-27+9+27\\ =\mathrm{}9\\ \therefore \mathrm{Required}\mathrm{shortest}\mathrm{distance}\\ =\mathrm{}\left|\frac{\left(\stackrel{\to }{{\mathrm{a}}_{2}}-{\stackrel{\to }{\mathrm{a}}}_{1}\right).\left(\stackrel{\to }{{\mathrm{b}}_{1}}×{\stackrel{\to }{\mathrm{b}}}_{2}\right)}{\left|\stackrel{\to }{{\mathrm{b}}_{1}}×{\stackrel{\to }{\mathrm{b}}}_{2}\right|}\right|\\ =\mathrm{}\left|\frac{9}{3\sqrt{19}}\right|\\ =\mathrm{}\frac{3}{\sqrt{19}}\end{array}$

Q.24 Find the equation of the plane through the intersection of the plane x+2y+3z-4 = 0 and 2x+y-z+5 = 0 and perpendicular to the plane 5x+3y+6z+1 = 0.

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{required}\mathrm{plane}\mathrm{be}\\ \mathrm{x}+2\mathrm{y}+3\mathrm{z}-4+\mathrm{\lambda }\left(2\mathrm{x}+\mathrm{y}-\mathrm{z}+5\right)=0\\ \left(1+2\mathrm{\lambda }\right)+\left(2+\mathrm{\lambda }\right)\mathrm{y}+\left(3-\mathrm{\lambda }\right)\mathrm{z}-4+5\mathrm{\lambda }=\mathrm{}0\\ \mathrm{This}\mathrm{plane}\mathrm{is}\mathrm{perpendicular}\mathrm{to}5\mathrm{x}+3\mathrm{y}+6\mathrm{z}+1 = 0\\ ⇒\mathrm{}\left(1+2\mathrm{\lambda }\right)5+\left(2+\mathrm{\lambda }\right)3+\left(3-\mathrm{\lambda }\right)6\mathrm{}=\mathrm{}0\\ ⇒7\mathrm{\lambda }+29\mathrm{}=\mathrm{}0\\ ⇒\mathrm{\lambda }\mathrm{}=\mathrm{}-\frac{29}{7}\\ \therefore \mathrm{x}+2\mathrm{y}+3\mathrm{z}-4-\frac{29}{7}\mathrm{}\left(2\mathrm{x}+\mathrm{y}-\mathrm{z}+5\right)\mathrm{}=\mathrm{}0\\ ⇒7\mathrm{x}+14\mathrm{y}+21\mathrm{z}-28-58\mathrm{x}-29\mathrm{y}+29\mathrm{z}-145\mathrm{}=\mathrm{}0\\ ⇒-51\mathrm{x}-15\mathrm{y}+50\mathrm{z}-173\mathrm{}=\mathrm{}0\\ ⇒51\mathrm{x}+15\mathrm{y}-50\mathrm{z}+173\mathrm{}=\mathrm{}0\\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{plane}.\end{array}$

Q.25

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{distance}\mathrm{of}\mathrm{a}\mathrm{point}\left(1,0,1\right)\mathrm{from}\mathrm{the}\mathrm{plane}\\ \stackrel{\to }{\mathrm{r}}.\left(6\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right)=4.\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{plane}\mathrm{in}\mathrm{vector}\mathrm{form}.\\ \stackrel{\to }{\mathrm{r}}.\left(6\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right)=4,\mathrm{where}\mathrm{}\stackrel{\to }{\mathrm{r}}=\mathrm{x}\stackrel{^}{\mathrm{i}}+\mathrm{y}\stackrel{^}{\mathrm{j}}+\mathrm{z}\stackrel{^}{\mathrm{k}}\\ \mathrm{Equation}\mathrm{of}\mathrm{the}\mathrm{plane}\mathrm{in}\mathrm{cartesian}\mathrm{form}\mathrm{is}\\ \left(\mathrm{x}\stackrel{^}{\mathrm{i}}+\mathrm{y}\stackrel{^}{\mathrm{j}}+\mathrm{z}\stackrel{^}{\mathrm{k}}\right).\left(6\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right)=4\\ ⇒6\mathrm{x}-3\mathrm{y}+2\mathrm{z}=4\\ ⇒6\mathrm{x}-3\mathrm{y}+2\mathrm{z}-4=0\\ \mathrm{Now},\mathrm{the}\mathrm{distance}\mathrm{of}\mathrm{plane}\mathrm{from}\mathrm{point}\left(1,0,1\right)–\\ \mathrm{}=\left|\frac{6\left(1\right)-3\left(0\right)+2\left(1\right)-4}{\sqrt{{6}^{2}+{\left(-3\right)}^{2}+{2}^{2}}}\right|\\ \mathrm{}=\left|\frac{6-0+2-4}{\sqrt{36+9+4}}\right|\\ \mathrm{}=\left|\frac{4}{\sqrt{49}}\right|\\ \mathrm{}=\frac{4}{7}\mathrm{units}\\ \end{array}$

Q.26 Find the equation of the plane through the intersection of the planes 3x+y+2z-8 = 0
and x+2y+5z-1 = 0 and the point (0,0,1).

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{plane}\mathrm{passing}\mathrm{through}\\ \mathrm{the}\mathrm{intersection}\mathrm{of}\mathrm{the}\mathrm{planes}3\mathrm{x}+\mathrm{y}+2\mathrm{z}-8=0\mathrm{and}\\ \mathrm{x}+2\mathrm{y}+5\mathrm{z}–\mathrm{\lambda }=0\mathrm{is}\\ \mathrm{x}+2\mathrm{y}+5\mathrm{z}-1+\mathrm{\lambda }\left(3\mathrm{x}+\mathrm{y}+2\mathrm{z}-8\right)=0\\ \left(1+3\mathrm{\lambda }\right)\mathrm{x}+\left(2+\mathrm{\lambda }\right)\mathrm{y}+ \left(5+2\mathrm{\lambda }\right)\mathrm{z}=8\mathrm{\lambda }+1\dots \left(\mathrm{1}\right)\\ \mathrm{Since}\mathrm{this}\mathrm{plane}\mathrm{passes}\mathrm{through}\mathrm{point}\left(0, 0, 1\right)\mathrm{,}\\ \therefore 0+0+5+2\mathrm{\lambda }=8\mathrm{\lambda }+1\\ ⇒6\mathrm{l}=4\\ ⇒\mathrm{l}=4/6\\ ⇒\mathrm{l}=2/3\\ \mathrm{On}\mathrm{putting}\mathrm{the}\mathrm{value}\mathrm{oflin}\mathrm{equation}\left(\mathrm{1}\right),\mathrm{we}\mathrm{get}\\ \left(1+2\right)\mathrm{x}+\left(2+2/3\right)\mathrm{y}+\left(5+4/3\right)\mathrm{z}=16/3+1 \dots \left(\mathrm{1}\right)\\ ⇒3\mathrm{x}+\left(8/3\right)\mathrm{y}+\left(19/3\right)\mathrm{z}=19/3\\ ⇒9\mathrm{x}+8\mathrm{y}+19\mathrm{z}= 19\\ ⇒9\mathrm{x}+8\mathrm{y}+19\mathrm{z}-19 = 0\end{array}$

Q.27

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{distance}\mathrm{between}\mathrm{the}\mathrm{lines}\\ \stackrel{\to }{\mathrm{r}}=\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)+\mathrm{\lambda }\left(3\stackrel{^}{\mathrm{i}}+5\stackrel{^}{\mathrm{j}}-2\stackrel{^}{\mathrm{k}}\right)\mathrm{and}\\ \stackrel{\to }{\mathrm{r}}=\left(3\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}\right)+\mathrm{\mu }\left(3\stackrel{^}{\mathrm{i}}+5\stackrel{^}{\mathrm{j}}-2\stackrel{^}{\mathrm{k}}\right)\end{array}$

Ans

$\begin{array}{l}\stackrel{\to }{\mathrm{r}}=\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)+\mathrm{\lambda }\left(3\stackrel{^}{\mathrm{i}}+5\stackrel{^}{\mathrm{j}}-2\stackrel{^}{\mathrm{k}}\right)\mathrm{and}\\ \stackrel{\to }{\mathrm{r}}=\left(3\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}\right)+\mathrm{\mu }\left(3\stackrel{^}{\mathrm{i}}+5\stackrel{^}{\mathrm{j}}-2\stackrel{^}{\mathrm{k}}\right)\\ \mathrm{Since}\mathrm{coefficient}\mathrm{of}\mathrm{\lambda }\mathrm{and}\mathrm{\mu }\mathrm{are}\mathrm{same},\mathrm{therefore},\mathrm{lines}\\ \mathrm{are}\mathrm{parallel}.\\ \mathrm{Here},\stackrel{\to }{{\mathrm{a}}_{1}}=\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}},{\stackrel{\to }{\mathrm{a}}}_{2}=3\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}\\ \mathrm{and}\mathrm{}\stackrel{\to }{\mathrm{b}}=3\stackrel{^}{\mathrm{i}}+5\stackrel{^}{\mathrm{j}}-2\stackrel{^}{\mathrm{k}}\\ {\stackrel{\to }{\mathrm{a}}}_{2}-\stackrel{\to }{{\mathrm{a}}_{1}}=2\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\\ \therefore \stackrel{\to }{\mathrm{b}}×\left({\stackrel{\to }{\mathrm{a}}}_{2}-\stackrel{\to }{{\mathrm{a}}_{1}}\right)=\left|\begin{array}{ccc}\stackrel{^}{\mathrm{i}}& \stackrel{^}{\mathrm{j}}& \stackrel{^}{\mathrm{k}}\\ 3& 5& -2\\ 2& 1& 3\end{array}\right|\\ =\stackrel{^}{\mathrm{i}}\left(15+2\right)-\stackrel{^}{\mathrm{j}}\left(9+4\right)+\stackrel{^}{\mathrm{k}}\left(3-10\right)\\ =\stackrel{^}{\mathrm{i}}\left(17\right)-\stackrel{^}{\mathrm{j}}\left(13\right)+\stackrel{^}{\mathrm{k}}\left(-7\right)\\ \left|\stackrel{\to }{\mathrm{b}}×\left({\stackrel{\to }{\mathrm{a}}}_{2}-\stackrel{\to }{{\mathrm{a}}_{1}}\right)\right|=\sqrt{289+169+49}\\ =\sqrt{507}\\ \stackrel{\to }{\mathrm{b}}=3\stackrel{^}{\mathrm{i}}+5\stackrel{^}{\mathrm{j}}-2\stackrel{^}{\mathrm{k}}\\ \therefore \left|\stackrel{\to }{\mathrm{b}}\right|=\sqrt{9+25+4}\\ =\sqrt{38}\end{array}$ $\begin{array}{l}\mathrm{Distance}\mathrm{between}\mathrm{lines},\mathrm{d}=\frac{\left|\stackrel{\to }{\mathrm{b}}×\left({\stackrel{\to }{\mathrm{a}}}_{2}-{\stackrel{\to }{\mathrm{a}}}_{1}\right)\right|}{\left|\stackrel{\to }{\mathrm{b}}\right|}\\ =\frac{\sqrt{507}}{\sqrt{38}}\\ =\sqrt{\frac{507}{38}}\\ \end{array}$

Q.28 Prove that the sum of square of direction cosines of a line is unity.

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{P}\left(\mathrm{x},\mathrm{}\mathrm{y},\mathrm{}\mathrm{z}\right)\mathrm{be}\mathrm{any}\mathrm{point}\mathrm{such}\mathrm{that}\mathrm{OP}=\mathrm{r},\\ \mathrm{Where}\mathrm{O}\mathrm{is}\mathrm{origin}.\mathrm{Let}\mathrm{\alpha },\mathrm{\beta }\mathrm{and}\mathrm{\gamma }\mathrm{be}\mathrm{the}\mathrm{angle}\mathrm{made}\\ \mathrm{by}\mathrm{a}\mathrm{line}\mathrm{OP}\mathrm{with}\mathrm{coordinate}\mathrm{axes}.\\ \mathrm{OP}=\mathrm{r}\\ \mathrm{OR}=\mathrm{x}\\ \mathrm{QR}=\mathrm{y}\\ ⇒\mathrm{OQ}=\sqrt{{\mathrm{x}}^{2}+{\mathrm{y}}^{2}}\\ \mathrm{PR}=\mathrm{z}\\ \therefore \mathrm{OP}=\mathrm{r}=\sqrt{{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}}\end{array}$ $\begin{array}{l}\therefore \mathrm{}\mathrm{x}\mathrm{}=\mathrm{}\mathrm{rcos\alpha }\\ \mathrm{y}\mathrm{}=\mathrm{}\mathrm{rcos\beta }\\ \mathrm{z}\mathrm{}=\mathrm{}\mathrm{rcos\gamma }\\ \mathrm{We}\mathrm{have},\mathrm{OP}=\mathrm{r}\\ ⇒\mathrm{}\sqrt{{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}}=\mathrm{r}\\ ⇒\mathrm{}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}\mathrm{}=\mathrm{}{\mathrm{r}}^{2}\\ ⇒\mathrm{}{\left(\mathrm{rcos\alpha }\right)}^{2}+{\left(\mathrm{rcos\beta }\right)}^{2}+{\left(\mathrm{rcos\gamma }\right)}^{2}=\mathrm{}{\mathrm{r}}^{2}\\ ⇒\mathrm{}{\mathrm{r}}^{2}\mathrm{}\left({\mathrm{cos}}^{2}\mathrm{\alpha }\mathrm{}+{\mathrm{cos}}^{2}\mathrm{\beta }\mathrm{}+{\mathrm{cos}}^{2}\mathrm{\gamma }\right)\mathrm{}{\mathrm{r}}^{2}\\ ⇒{\mathrm{cos}}^{2}\mathrm{\alpha }\mathrm{}+{\mathrm{cos}}^{2}\mathrm{\beta }\mathrm{}+{\mathrm{cos}}^{2}\mathrm{\gamma }\mathrm{}=\mathrm{}1\\ ⇒\mathrm{}{\mathrm{l}}^{2}+{\mathrm{m}}^{2}+{\mathrm{n}}^{2}=\mathrm{ }1\end{array}$

Q.29

$\begin{array}{l}\mathrm{If}\mathrm{\alpha },\mathrm{\beta }\mathrm{and}\mathrm{\gamma }\mathrm{are}\mathrm{the}\mathrm{angles}\mathrm{made}\mathrm{by}\mathrm{a}\mathrm{line}\mathrm{OP}\mathrm{with}\\ \mathrm{coordinate}\mathrm{axes},\mathrm{then}\mathrm{prove}\mathrm{that}{\mathrm{sin}}^{2}\mathrm{\alpha }+{\mathrm{sin}}^{2}\mathrm{\beta }+{\mathrm{sin}}^{2}\mathrm{\gamma }=2.\end{array}$

Ans

$\begin{array}{l}\mathrm{Since}\mathrm{\alpha },\mathrm{\beta }\mathrm{and}\mathrm{\gamma }\mathrm{are}\mathrm{the}\mathrm{angles}\mathrm{made}\mathrm{by}\mathrm{a}\mathrm{line}\mathrm{OP}\mathrm{with}\\ \mathrm{coordinate}\mathrm{axes},\mathrm{let}\mathrm{P}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)\mathrm{be}\mathrm{any}\mathrm{point}\mathrm{such}\mathrm{that}\mathrm{OP}=\mathrm{r}.\\ \mathrm{OP}=\mathrm{r}\\ \mathrm{OR}=\mathrm{x}\\ \mathrm{QR}=\mathrm{y}\\ ⇒\mathrm{OQ}=\sqrt{{\mathrm{x}}^{2}+{\mathrm{y}}^{2}}\\ \mathrm{PQ}=\mathrm{z}\\ \therefore \mathrm{OP}=\mathrm{r}=\sqrt{{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}}\\ \therefore \mathrm{cos\alpha }=\frac{\mathrm{x}}{\mathrm{r}}\\ ⇒\mathrm{x}=\mathrm{rcos\alpha }\\ ⇒\mathrm{y}=\mathrm{rcos\beta }\\ ⇒\mathrm{z}=\mathrm{rcos\gamma }\end{array}$ $\begin{array}{l}\mathrm{We}\mathrm{have},\\ ⇒\sqrt{{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}}=\mathrm{r}\\ ⇒{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}={\mathrm{r}}^{2}\\ ⇒{\left(\mathrm{rcos\alpha }\right)}^{2}+{\left(\mathrm{rcos\beta }\right)}^{2}+{\left(\mathrm{rcos\gamma }\right)}^{2}={\mathrm{r}}^{2}\\ ⇒{\mathrm{r}}^{2}\left({\mathrm{cos}}^{2}\mathrm{\alpha }+{\mathrm{cos}}^{2}\mathrm{\beta }+{\mathrm{cos}}^{2}\mathrm{\gamma }\right)={\mathrm{r}}^{2}\\ ⇒{\mathrm{cos}}^{2}\mathrm{\alpha }+{\mathrm{cos}}^{2}\mathrm{\beta }+{\mathrm{cos}}^{2}\mathrm{\gamma }=1\\ ⇒1-{\mathrm{sin}}^{2}\mathrm{\alpha }+1-{\mathrm{sin}}^{2}\mathrm{\beta }+1-{\mathrm{sin}}^{2}\mathrm{\gamma }=1\\ ⇒3-\left({\mathrm{sin}}^{2}\mathrm{\alpha }+{\mathrm{sin}}^{2}\mathrm{\beta }+{\mathrm{sin}}^{2}\mathrm{\gamma }\right)=1\\ ⇒{\mathrm{sin}}^{2}\mathrm{\alpha }+{\mathrm{sin}}^{2}\mathrm{\beta }+{\mathrm{sin}}^{2}\mathrm{\gamma }=2\end{array}$

Q.30 Prove that sum of the square of direction cosines of a line is unity.

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{P}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)\mathrm{be}\mathrm{any}\mathrm{point}\mathrm{such}\mathrm{that}\mathrm{OP}=\mathrm{r},\mathrm{}\\ \mathrm{where}\mathrm{O}\mathrm{is}\mathrm{origin}.\mathrm{Let}\mathrm{\alpha },\mathrm{\beta }\mathrm{and}\mathrm{\gamma }\mathrm{be}\mathrm{the}\mathrm{angle}\mathrm{made}\\ \mathrm{by}\mathrm{a}\mathrm{line}\mathrm{OP}\mathrm{with}\mathrm{coordinate}\mathrm{axes}.\\ \mathrm{OP}=\mathrm{r}\\ \mathrm{OR}=\mathrm{x}\\ \mathrm{QR}=\mathrm{y}\\ ⇒\mathrm{OQ}=\sqrt{{\mathrm{x}}^{2}+{\mathrm{y}}^{2}}\\ \mathrm{PQ}=\mathrm{z}\\ \therefore \mathrm{OP}=\mathrm{r}=\sqrt{{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}}\end{array}$ $\begin{array}{l}\therefore \mathrm{x}=\mathrm{rcos\alpha }\\ \mathrm{y}=\mathrm{rcos\beta }\\ \mathrm{z}=\mathrm{rcos\gamma }\\ \mathrm{We}\mathrm{have},\mathrm{OP}=\mathrm{r}\\ ⇒\sqrt{{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}}=\mathrm{r}\\ ⇒{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}={\mathrm{r}}^{2}\\ ⇒{\left(\mathrm{rcos\alpha }\right)}^{2}+{\left(\mathrm{rcos\beta }\right)}^{2}+{\left(\mathrm{rcos\gamma }\right)}^{2}={\mathrm{r}}^{2}\\ ⇒{\mathrm{r}}^{2}\left({\mathrm{cos}}^{2}\mathrm{\alpha }+{\mathrm{cos}}^{2}\mathrm{\beta }+{\mathrm{cos}}^{2}\mathrm{\gamma }\right)={\mathrm{r}}^{2}\\ ⇒{\mathrm{cos}}^{2}\mathrm{\alpha }+{\mathrm{cos}}^{2}\mathrm{\beta }+{\mathrm{cos}}^{2}\mathrm{\gamma }=1\\ ⇒{\mathrm{l}}^{2}+{\mathrm{m}}^{2}+{\mathrm{n}}^{2}=1\end{array}$

Q.31

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{in}\mathrm{vector}\mathrm{form}\mathrm{that}\mathrm{pases}\mathrm{through}\\ \mathrm{the}\mathrm{point}\left(1,\mathrm{ }2,\mathrm{ }3\right)\mathrm{and}\mathrm{perpendicular}\mathrm{to}\mathrm{the}\mathrm{plane}\mathrm{x}+2\mathrm{y}-5\mathrm{z}+9 = 0\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{a}\mathrm{plane}\mathrm{in}\mathrm{cartesian}\mathrm{form}\mathrm{is}\\ \mathrm{x}+2\mathrm{y}-5\mathrm{z}+9 = 0\\ \mathrm{The}\mathrm{direction}\mathrm{ratio}\mathrm{of}\mathrm{this}\mathrm{plane}\mathrm{is}\left(1,\mathrm{ }2,\mathrm{ }-5\right),\mathrm{which}\mathrm{is}\mathrm{directed}\\ \mathrm{along}\mathrm{a}\mathrm{line}\mathrm{perpendicular}\mathrm{to}\mathrm{the}\mathrm{plane}.\\ \therefore \mathrm{The}\mathrm{line}\mathrm{perpendicular}\mathrm{to}\mathrm{the}\mathrm{plane}\mathrm{is}\mathrm{along}\stackrel{\to }{\mathrm{b}}\mathrm{}=\mathrm{}\stackrel{^}{\mathrm{i}}\mathrm{}+2\stackrel{^}{\mathrm{j}}-5\stackrel{^}{\mathrm{k}}.\\ \mathrm{Since}\mathrm{the}\mathrm{line}\mathrm{is}\mathrm{passing}\mathrm{through}\mathrm{the}\mathrm{point}\left(1,\mathrm{}2,\mathrm{}3\right),\mathrm{therefore}\mathrm{its}\\ \mathrm{position}\mathrm{vetor}\mathrm{is}\stackrel{\to }{\mathrm{a}}\mathrm{}=\mathrm{}\stackrel{^}{\mathrm{i}}\mathrm{}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}.\\ \mathrm{Now},\mathrm{we}\mathrm{have}–\\ \mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{line}\mathrm{in}\mathrm{vector}\mathrm{form}\mathrm{is}\\ \stackrel{\to }{\mathrm{r}}\mathrm{}=\stackrel{\to }{\mathrm{a}}+\mathrm{\lambda }\stackrel{\to }{\mathrm{b}},\mathrm{where}\stackrel{\to }{\mathrm{a}}\mathrm{represents}\mathrm{point}\mathrm{and}\stackrel{\to }{\mathrm{b}}\mathrm{}\mathrm{direction}.\\ ⇒\stackrel{\to }{\mathrm{r}}\mathrm{}=\mathrm{}\left(\stackrel{^}{\mathrm{i}}\mathrm{}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\right)+\mathrm{\lambda }\mathrm{ }\left(\stackrel{^}{\mathrm{i}}\mathrm{}+2\stackrel{^}{\mathrm{j}}-5\stackrel{^}{\mathrm{k}}\right)\end{array}$

Q.32

$\begin{array}{l}\mathrm{If}\mathrm{\alpha },\mathrm{\beta }\mathrm{and}\mathrm{\gamma }\mathrm{are}\mathrm{the}\mathrm{angles}\mathrm{made}\mathrm{by}\mathrm{a}\mathrm{line}\mathrm{OP}\mathrm{with}\\ \mathrm{coordinate}\mathrm{axes},\mathrm{then}\mathrm{prove}\mathrm{that}{\mathrm{cos}}^{2}\mathrm{\alpha }+{\mathrm{cos}}^{2}\mathrm{\beta }+{\mathrm{cos}}^{2}\mathrm{\gamma }=1.\end{array}$

Ans

$\begin{array}{l}\mathrm{Since}\mathrm{\alpha },\mathrm{\beta }\mathrm{and}\mathrm{\gamma }\mathrm{are}\mathrm{the}\mathrm{angles}\mathrm{made}\mathrm{by}\mathrm{the}\mathrm{line}\mathrm{OP}\mathrm{with}\\ \mathrm{coordinate}\mathrm{axes},\mathrm{let}\mathrm{P}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)\mathrm{be}\mathrm{any}\mathrm{point}\mathrm{such}\mathrm{that}\\ \mathrm{OP}=\mathrm{r}\\ \mathrm{OR}=\mathrm{x}\\ \mathrm{QR}=\mathrm{y}\\ ⇒\mathrm{OQ}=\sqrt{{\mathrm{x}}^{2}+{\mathrm{y}}^{2}}\\ \mathrm{PQ}=\mathrm{z}\\ \therefore \mathrm{OP}=\mathrm{r}=\sqrt{{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}}\end{array}$ $\begin{array}{l}\therefore \mathrm{cos\alpha }=\frac{\mathrm{x}}{\mathrm{r}}\\ ⇒\mathrm{x}=\mathrm{rcos\alpha }\\ ⇒\mathrm{y}=\mathrm{rcos\beta }\\ ⇒\mathrm{z}=\mathrm{rcos\gamma }\\ \mathrm{We}\mathrm{have}\mathrm{OP}=\mathrm{r}\\ ⇒\sqrt{{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}}=\mathrm{r}\\ ⇒{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}={\mathrm{r}}^{2}\\ ⇒{\left(\mathrm{rcos\alpha }\right)}^{2}+{\left(\mathrm{rcos\beta }\right)}^{2}+{\left(\mathrm{rcos\gamma }\right)}^{2}={\mathrm{r}}^{2}\\ ⇒{\mathrm{r}}^{2}\left({\mathrm{cos}}^{2}\mathrm{\alpha }+{\mathrm{cos}}^{2}\mathrm{\beta }+{\mathrm{cos}}^{2}\mathrm{\gamma }\right)={\mathrm{r}}^{2}\\ ⇒{\mathrm{cos}}^{2}\mathrm{\alpha }+{\mathrm{cos}}^{2}\mathrm{\beta }+{\mathrm{cos}}^{2}\mathrm{\gamma }=1\end{array}$

Q.33

$\text{Find the vector equation of the line}\frac{\mathrm{x}+3}{5}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}+4}{3}.$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have},\\ \mathrm{}\frac{\mathrm{x}+3}{5}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}+4}{3}\\ \mathrm{The}\mathrm{standard}\mathrm{equation}\mathrm{of}\mathrm{line}\mathrm{in}\mathrm{cartesian}\mathrm{form}\mathrm{is}\\ \frac{\mathrm{x}-{\mathrm{x}}_{1}}{\mathrm{a}}=\frac{\mathrm{y}-{\mathrm{y}}_{1}}{\mathrm{b}}=\frac{\mathrm{z}-{\mathrm{z}}_{1}}{\mathrm{c}}\\ \mathrm{On}\mathrm{comparing}\mathrm{the}\mathrm{equations},\mathrm{we}\mathrm{get}\\ {\mathrm{x}}_{1}=-3\\ {\mathrm{y}}_{1}=1\\ {\mathrm{z}}_{1}=-4\\ \mathrm{a}=5\\ \mathrm{b}=2\\ \mathrm{c}=3\\ \therefore \mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{line}\mathrm{in}\mathrm{vector}\mathrm{form}\mathrm{is}\\ \stackrel{\to }{\mathrm{r}}=\stackrel{\to }{\mathrm{a}}+\mathrm{\lambda }\stackrel{\to }{\mathrm{b}},\mathrm{where}\mathrm{a}\mathrm{represents}\mathrm{point}\mathrm{and}\mathrm{b}\mathrm{direction}.\\ ⇒\stackrel{\to }{\mathrm{r}}=\left(-3\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}-4\stackrel{^}{\mathrm{k}}\right)+\mathrm{\lambda }\left(5\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\right)\end{array}$

Q.34

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{perpedicular}\mathrm{distance}\mathrm{of}\mathrm{a}\mathrm{plane}\mathrm{from}\mathrm{the}\\ \mathrm{origin}\mathrm{that}\mathrm{makes}\mathrm{intercepts}\mathrm{a},\mathrm{b},\mathrm{c}\mathrm{with}\mathrm{X},\mathrm{Y},\mathrm{Z}–\mathrm{axis}\\ \mathrm{respectively}.\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{plane}\mathrm{having}\mathrm{intercept}\mathrm{a},\mathrm{b},\mathrm{c}\mathrm{on}\\ \mathrm{coordinate}\mathrm{axes}\mathrm{is}\\ \frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}+\frac{\mathrm{z}}{\mathrm{c}}=1\\ ⇒\frac{\mathrm{bcx}+\mathrm{acy}+\mathrm{abz}}{\mathrm{abc}}=1\\ ⇒\mathrm{bcx}+\mathrm{acy}+\mathrm{abz}=\mathrm{abc}\\ ⇒\mathrm{bcx}+\mathrm{acy}+\mathrm{abz}-\mathrm{abc}=0\\ \mathrm{Now},\mathrm{distance}\mathrm{of}\mathrm{the}\mathrm{plane}\mathrm{from}\mathrm{origin}\mathrm{is}\\ \mathrm{d}=\left|\frac{0+0+0-\mathrm{abc}}{\sqrt{{\mathrm{b}}^{2}{\mathrm{c}}^{2}+{\mathrm{a}}^{2}{\mathrm{c}}^{2}+{\mathrm{a}}^{2}{\mathrm{b}}^{2}}}\right|\\ \mathrm{}=\frac{\mathrm{abc}}{\sqrt{{\mathrm{b}}^{2}{\mathrm{c}}^{2}+{\mathrm{a}}^{2}{\mathrm{c}}^{2}+{\mathrm{a}}^{2}{\mathrm{b}}^{2}}}\\ \end{array}$

Q.35

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{in}\mathrm{vector}\mathrm{and}\mathrm{cartesian}\\ \mathrm{form}\mathrm{that}\mathrm{passes}\mathrm{through}\mathrm{the}\mathrm{point}\mathrm{with}\mathrm{position}\mathrm{vector}\\ \mathrm{2}\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}\mathrm{and}\mathrm{in}\mathrm{the}\mathrm{direction}\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}.\end{array}$

Ans

$\begin{array}{l}\mathrm{Here},\mathrm{}\stackrel{\to }{\mathrm{a}}=\mathrm{2}\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}\mathrm{and}\stackrel{\to }{\mathrm{b}}=\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}.\\ \therefore \mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{line}\mathrm{in}\mathrm{vector}\mathrm{form}\mathrm{is}\\ \stackrel{\to }{\mathrm{r}}=\stackrel{\to }{\mathrm{a}}+\mathrm{\lambda }\stackrel{\to }{\mathrm{b}}\\ \therefore \stackrel{\to }{\mathrm{r}}=\mathrm{2}\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}+\mathrm{\lambda }\left(\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}.\right)\\ \\ \mathrm{The}\mathrm{standard}\mathrm{equation}\mathrm{of}\mathrm{line}\mathrm{in}\mathrm{cartesian}\mathrm{form}\mathrm{is}\\ \frac{\mathrm{x}-{\mathrm{x}}_{1}}{\mathrm{a}}=\frac{\mathrm{y}-{\mathrm{y}}_{1}}{\mathrm{b}}=\frac{\mathrm{z}-{\mathrm{z}}_{1}}{\mathrm{c}}...\left(1\right)\\ \mathrm{we}\mathrm{have},\\ \mathrm{a}=1,\mathrm{b}=2,\mathrm{c}=-1\mathrm{and}\\ {\mathrm{x}}_{1}=2\\ {\mathrm{y}}_{1}=-1\\ {\mathrm{z}}_{1}=4\\ \mathrm{On}\mathrm{putting}\mathrm{the}\mathrm{values}\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get}\\ \mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{line}\mathrm{in}\mathrm{cartesian}\mathrm{form}\mathrm{is}\\ \frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}+1}{2}=\frac{\mathrm{z}-4}{-1}\end{array}$

Q.36

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{in}\mathrm{vector}\mathrm{form}\mathrm{that}\mathrm{passes}\\ \mathrm{through}\mathrm{the}\mathrm{point}\left(1,2,3\right)\mathrm{and}\mathrm{perpendicular}\mathrm{to}\mathrm{the}\\ \mathrm{plane}\mathrm{x}+2\mathrm{y}-5\mathrm{z}+9=0.\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{a}\mathrm{plane}\mathrm{in}\mathrm{cartesian}\mathrm{form}\mathrm{is}\\ \mathrm{x}+2\mathrm{y}-5\mathrm{z}+9=0.\\ \mathrm{The}\mathrm{direction}\mathrm{ratio}\mathrm{of}\mathrm{this}\mathrm{plane}\mathrm{is}\left(1,2,-5\right),\mathrm{which}\mathrm{is}\mathrm{directed}\\ \mathrm{along}\mathrm{a}\mathrm{line}\mathrm{perpendicular}\mathrm{to}\mathrm{the}\mathrm{plane}.\\ \therefore \mathrm{The}\mathrm{line}\mathrm{perpendicular}\mathrm{to}\mathrm{the}\mathrm{plane}\mathrm{is}\mathrm{along}\stackrel{\to }{\mathrm{b}}=\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-5\stackrel{^}{\mathrm{k}.}\\ \mathrm{Since}\mathrm{ }\mathrm{the}\mathrm{line}\mathrm{is}\mathrm{passing}\mathrm{through}\mathrm{the}\mathrm{point}\left(1,2,3\right),\mathrm{therefore}\mathrm{its}\\ \mathrm{position}\mathrm{vector}\mathrm{is}\stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}.\\ \mathrm{Now},\mathrm{we}\mathrm{have}–\\ \mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{line}\mathrm{in}\mathrm{vector}\mathrm{form}\mathrm{is}\\ \stackrel{\to }{\mathrm{r}}=\stackrel{\to }{\mathrm{a}}+\mathrm{\lambda }\stackrel{\to }{\mathrm{b},}\mathrm{where}\stackrel{\to }{\mathrm{a}}\mathrm{represents}\mathrm{point}\mathrm{and}\stackrel{\to }{\mathrm{b}}\mathrm{direction}.\\ ⇒\stackrel{\to }{\mathrm{r}}=\left(\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\right)+\mathrm{\lambda }\left(\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-5\stackrel{^}{\mathrm{k}}\right)\end{array}$

Q.37

$\begin{array}{l}\mathrm{If}\mathrm{\alpha },\mathrm{\beta }\mathrm{and}\mathrm{\gamma }\mathrm{are}\mathrm{the}\mathrm{angles}\mathrm{made}\mathrm{by}\mathrm{a}\mathrm{line}\mathrm{OP}\mathrm{with}\\ \mathrm{coordinate}\mathrm{axes},\mathrm{then}\mathrm{prove}\mathrm{that}{\mathrm{cos}}^{2}\mathrm{\alpha }+{\mathrm{cos}}^{2}\mathrm{\beta }+{\mathrm{cos}}^{2}\mathrm{\gamma }=1.\end{array}$

Ans

$\begin{array}{l}\mathrm{Since}\mathrm{\alpha },\mathrm{\beta }\mathrm{and}\mathrm{\gamma }\mathrm{are}\mathrm{the}\mathrm{angles}\mathrm{made}\mathrm{by}\mathrm{the}\mathrm{line}\mathrm{OP}\mathrm{with}\\ \mathrm{coordinate}\mathrm{axes},\mathrm{let}\mathrm{P}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)\mathrm{be}\mathrm{any}\mathrm{point}\mathrm{such}\mathrm{that}\\ \mathrm{OP}=\mathrm{r}\\ \mathrm{OR}=\mathrm{x}\\ \mathrm{QR}=\mathrm{y}\\ ⇒\mathrm{OQ}=\sqrt{{\mathrm{x}}^{2}+{\mathrm{y}}^{2}}\\ \mathrm{PQ}=\mathrm{z}\\ \therefore \mathrm{OP}=\mathrm{r}=\sqrt{{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}}\end{array}$ $\begin{array}{l}\therefore \mathrm{cos\alpha }=\frac{\mathrm{x}}{\mathrm{r}}\\ ⇒\mathrm{x}=\mathrm{rcos\alpha }\\ ⇒\mathrm{y}=\mathrm{rcos\beta }\\ ⇒\mathrm{z}=\mathrm{rcos\gamma }\\ \mathrm{We}\mathrm{have}\mathrm{OP}=\mathrm{r}\\ ⇒\sqrt{{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}}=\mathrm{r}\\ ⇒{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}={\mathrm{r}}^{2}\\ ⇒{\left(\mathrm{rcos\alpha }\right)}^{2}+{\left(\mathrm{rcos\beta }\right)}^{2}+{\left(\mathrm{rcos\gamma }\right)}^{2}={\mathrm{r}}^{2}\\ ⇒{\mathrm{r}}^{2}\left({\mathrm{cos}}^{2}\mathrm{\alpha }+{\mathrm{cos}}^{2}\mathrm{\beta }+{\mathrm{cos}}^{2}\mathrm{\gamma }\right)={\mathrm{r}}^{2}\\ ⇒{\mathrm{cos}}^{2}\mathrm{\alpha }+{\mathrm{cos}}^{2}\mathrm{\beta }+{\mathrm{cos}}^{2}\mathrm{\gamma }=1\end{array}$

Q.38

$\text{Find the vector equation of the line}\frac{\mathrm{x}+3}{5}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}+4}{3}.$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have},\\ \mathrm{}\frac{\mathrm{x}+3}{5}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}+4}{3}\\ \mathrm{The}\mathrm{standard}\mathrm{equation}\mathrm{of}\mathrm{line}\mathrm{in}\mathrm{cartesian}\mathrm{form}\mathrm{is}\\ \frac{\mathrm{x}-{\mathrm{x}}_{1}}{\mathrm{a}}=\frac{\mathrm{y}-{\mathrm{y}}_{1}}{\mathrm{b}}=\frac{\mathrm{z}-{\mathrm{z}}_{1}}{\mathrm{c}}\\ \mathrm{On}\mathrm{comparing}\mathrm{the}\mathrm{equations},\mathrm{we}\mathrm{get}\\ {\mathrm{x}}_{1}=-3\\ {\mathrm{y}}_{1}=1\\ {\mathrm{z}}_{1}=-4\\ \mathrm{a}=5\\ \mathrm{b}=2\\ \mathrm{c}=3\\ \therefore \mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{line}\mathrm{in}\mathrm{vector}\mathrm{form}\mathrm{is}\\ \stackrel{\to }{\mathrm{r}}=\stackrel{\to }{\mathrm{a}}+\mathrm{\lambda }\stackrel{\to }{\mathrm{b}},\mathrm{where}\mathrm{a}\mathrm{represents}\mathrm{point}\mathrm{and}\mathrm{b}\mathrm{direction}.\\ ⇒\stackrel{\to }{\mathrm{r}}=\left(-3\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}-4\stackrel{^}{\mathrm{k}}\right)+\mathrm{\lambda }\left(5\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\right)\end{array}$

Q.39

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{perpedicular}\mathrm{distance}\mathrm{of}\mathrm{a}\mathrm{plane}\mathrm{from}\mathrm{the}\\ \mathrm{origin}\mathrm{that}\mathrm{makes}\mathrm{intercepts}\mathrm{a},\mathrm{b},\mathrm{c}\mathrm{with}\mathrm{X},\mathrm{Y},\mathrm{Z}–\mathrm{axis}\\ \mathrm{respectively}.\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{plane}\mathrm{having}\mathrm{intercept}\mathrm{a},\mathrm{b},\mathrm{c}\mathrm{on}\\ \mathrm{coordinate}\mathrm{axes}\mathrm{is}\\ \frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}+\frac{\mathrm{z}}{\mathrm{c}}=1\\ ⇒\frac{\mathrm{bcx}+\mathrm{acy}+\mathrm{abz}}{\mathrm{abc}}=1\\ ⇒\mathrm{bcx}+\mathrm{acy}+\mathrm{abz}=\mathrm{abc}\\ ⇒\mathrm{bcx}+\mathrm{acy}+\mathrm{abz}-\mathrm{abc}=0\\ \mathrm{Now},\mathrm{distance}\mathrm{of}\mathrm{the}\mathrm{plane}\mathrm{from}\mathrm{origin}\mathrm{is}\\ \mathrm{d}=\left|\frac{0+0+0-\mathrm{abc}}{\sqrt{{\mathrm{b}}^{2}{\mathrm{c}}^{2}+{\mathrm{a}}^{2}{\mathrm{c}}^{2}+{\mathrm{a}}^{2}{\mathrm{b}}^{2}}}\right|\\ \mathrm{}=\frac{\mathrm{abc}}{\sqrt{{\mathrm{b}}^{2}{\mathrm{c}}^{2}+{\mathrm{a}}^{2}{\mathrm{c}}^{2}+{\mathrm{a}}^{2}{\mathrm{b}}^{2}}}\\ \end{array}$

Q.40

$\begin{array}{l}\mathrm{If}\mathrm{\alpha },\mathrm{\beta }\mathrm{and}\mathrm{\gamma }\mathrm{are}\mathrm{the}\mathrm{angles}\mathrm{made}\mathrm{by}\mathrm{a}\mathrm{line}\mathrm{OP}\mathrm{with}\\ \mathrm{coordinate}\mathrm{axes},\mathrm{then}\mathrm{prove}\mathrm{that}{\mathrm{sin}}^{2}\mathrm{\alpha }+{\mathrm{sin}}^{2}\mathrm{\beta }+{\mathrm{sin}}^{2}\mathrm{\gamma }=2.\end{array}$

Ans

$\begin{array}{l}\mathrm{Since}\mathrm{\alpha },\mathrm{\beta }\mathrm{and}\mathrm{\gamma }\mathrm{are}\mathrm{the}\mathrm{angles}\mathrm{made}\mathrm{by}\mathrm{a}\mathrm{line}\mathrm{OP}\mathrm{with}\\ \mathrm{coordinate}\mathrm{axes},\mathrm{let}\mathrm{P}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)\mathrm{be}\mathrm{any}\mathrm{point}\mathrm{such}\mathrm{that}\mathrm{OP}=\mathrm{r}.\\ \mathrm{OP}=\mathrm{r}\\ \mathrm{OR}=\mathrm{x}\\ \mathrm{QR}=\mathrm{y}\\ ⇒\mathrm{OQ}=\sqrt{{\mathrm{x}}^{2}+{\mathrm{y}}^{2}}\\ \mathrm{PQ}=\mathrm{z}\\ \therefore \mathrm{OP}=\mathrm{r}=\sqrt{{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}}\\ \therefore \mathrm{cos\alpha }=\frac{\mathrm{x}}{\mathrm{r}}\\ ⇒\mathrm{x}=\mathrm{rcos\alpha }\\ ⇒\mathrm{y}=\mathrm{rcos\beta }\\ ⇒\mathrm{z}=\mathrm{rcos\gamma }\end{array}$ $\begin{array}{l}\mathrm{We}\mathrm{have},\\ ⇒\sqrt{{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}}=\mathrm{r}\\ ⇒{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}={\mathrm{r}}^{2}\\ ⇒{\left(\mathrm{rcos\alpha }\right)}^{2}+{\left(\mathrm{rcos\beta }\right)}^{2}+{\left(\mathrm{rcos\gamma }\right)}^{2}={\mathrm{r}}^{2}\\ ⇒{\mathrm{r}}^{2}\left({\mathrm{cos}}^{2}\mathrm{\alpha }+{\mathrm{cos}}^{2}\mathrm{\beta }+{\mathrm{cos}}^{2}\mathrm{\gamma }\right)={\mathrm{r}}^{2}\\ ⇒{\mathrm{cos}}^{2}\mathrm{\alpha }+{\mathrm{cos}}^{2}\mathrm{\beta }+{\mathrm{cos}}^{2}\mathrm{\gamma }=1\\ ⇒1-{\mathrm{sin}}^{2}\mathrm{\alpha }+1-{\mathrm{sin}}^{2}\mathrm{\beta }+1-{\mathrm{sin}}^{2}\mathrm{\gamma }=1\\ ⇒3-\left({\mathrm{sin}}^{2}\mathrm{\alpha }+{\mathrm{sin}}^{2}\mathrm{\beta }+{\mathrm{sin}}^{2}\mathrm{\gamma }\right)=1\\ ⇒{\mathrm{sin}}^{2}\mathrm{\alpha }+{\mathrm{sin}}^{2}\mathrm{\beta }+{\mathrm{sin}}^{2}\mathrm{\gamma }=2\end{array}$

Q.41

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{in}\mathrm{vector}\mathrm{and}\mathrm{cartesian}\\ \mathrm{form}\mathrm{that}\mathrm{passes}\mathrm{through}\mathrm{the}\mathrm{point}\mathrm{with}\mathrm{position}\mathrm{vector}\\ \mathrm{2}\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}\mathrm{and}\mathrm{in}\mathrm{the}\mathrm{direction}\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}.\end{array}$

Ans

$\begin{array}{l}\mathrm{Here},\mathrm{}\stackrel{\to }{\mathrm{a}}=\mathrm{2}\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}\mathrm{and}\stackrel{\to }{\mathrm{b}}=\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}.\\ \therefore \mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{line}\mathrm{in}\mathrm{vector}\mathrm{form}\mathrm{is}\\ \stackrel{\to }{\mathrm{r}}=\stackrel{\to }{\mathrm{a}}+\mathrm{\lambda }\stackrel{\to }{\mathrm{b}}\\ \therefore \stackrel{\to }{\mathrm{r}}=\mathrm{2}\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}+\mathrm{\lambda }\left(\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}.\right)\\ \\ \mathrm{The}\mathrm{standard}\mathrm{equation}\mathrm{of}\mathrm{line}\mathrm{in}\mathrm{cartesian}\mathrm{form}\mathrm{is}\\ \frac{\mathrm{x}-{\mathrm{x}}_{1}}{\mathrm{a}}=\frac{\mathrm{y}-{\mathrm{y}}_{1}}{\mathrm{b}}=\frac{\mathrm{z}-{\mathrm{z}}_{1}}{\mathrm{c}}...\left(1\right)\\ \mathrm{we}\mathrm{have},\\ \mathrm{a}=1,\mathrm{b}=2,\mathrm{c}=-1\mathrm{and}\\ {\mathrm{x}}_{1}=2\\ {\mathrm{y}}_{1}=-1\\ {\mathrm{z}}_{1}=4\\ \mathrm{On}\mathrm{putting}\mathrm{the}\mathrm{values}\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get}\\ \mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{line}\mathrm{in}\mathrm{cartesian}\mathrm{form}\mathrm{is}\\ \frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}+1}{2}=\frac{\mathrm{z}-4}{-1}\end{array}$

Q.42 Find the equation of the plane through the intersection of the planes 3x+y+2z-8=0 and x+2y+5z-1=0 and the point (0,0,1).

Ans

$\begin{array}{l}\text{Let equation of the plane passing through the intersection of the}\\ \text{planes 3x+y+2z-8 = 0 and}\\ \text{x+2y+5z-1 = 0 is}\\ \text{x+2y+5z-1+}\lambda \left(\text{3x+y+2z-8}\right)\text{= 0}\\ \text{(1+3}\lambda \text{)x+(2+}\lambda \text{)y + (5+2}\lambda \text{)z = 8}\lambda \text{+1 …}\left(\text{1}\right)\\ \text{Since this plane passes through point}\left(\text{0, 0, 1}\right)\text{,}\\ \therefore \text{0+0+5+2}\lambda \text{= 8}\lambda \text{+1}\\ ⇒\text{6}\lambda \text{=4}\\ ⇒\lambda \text{=4/6}\\ ⇒\lambda \text{=2/3}\\ \text{On putting the value of}\lambda \text{in equation}\left(\text{1}\right)\text{, we get}\\ \left(\text{1+2}\right)\text{x+}\left(\text{2+2/3}\right)\text{y +}\left(\text{5+4/3}\right)\text{z=16/3+1 …}\left(\text{1}\right)\\ ⇒\text{3x+}\left(\text{8/3}\right)\text{y+}\left(\text{19/3}\right)\text{z =19/3}\\ \text{⇒ 9x+8y+19z =19}\\ \text{⇒ 9x+8y+19z-19 = 0}\end{array}$

Q.43

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{Cartesian}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{plane}\mathrm{which}\\ \mathrm{passing}\mathrm{through}\mathrm{the}\mathrm{intersection}\mathrm{of}\mathrm{the}\mathrm{planes}\\ \stackrel{\to }{\mathrm{r}}.\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)=1\mathrm{and}\stackrel{\to }{\mathrm{r}}.\left(3\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right)=8\mathrm{and}\mathrm{the}\mathrm{point}\\ \left(1,1,1\right).\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have}\stackrel{\to }{\mathrm{r}}=\mathrm{x}\stackrel{^}{\mathrm{i}}+\mathrm{y}\stackrel{^}{\mathrm{j}}+\mathrm{z}\stackrel{^}{\mathrm{k}}\\ \mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{planes}\mathrm{are}\\ \left(\mathrm{x}\stackrel{^}{\mathrm{i}}+\mathrm{y}\stackrel{^}{\mathrm{j}}+\mathrm{z}\stackrel{^}{\mathrm{k}}\right).\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)=1\mathrm{and}\left(\mathrm{x}\stackrel{^}{\mathrm{i}}+\mathrm{y}\stackrel{^}{\mathrm{j}}+\mathrm{z}\stackrel{^}{\mathrm{k}}\right).\left(3\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right)=8\\ ⇒\mathrm{x}+\mathrm{y}+\mathrm{z}-\mathrm{1}=0\mathrm{and}3\mathrm{x}+2\mathrm{y}+2\mathrm{z}-\mathrm{8}=0\\ \mathrm{Let}\mathrm{Equation}\mathrm{of}\mathrm{the}\mathrm{plane}\mathrm{passing}\mathrm{through}\mathrm{the}\mathrm{intersection}\\ \mathrm{of}\mathrm{the}\mathrm{planes}3\mathrm{x}+2\mathrm{y}+2\mathrm{z}-\mathrm{8}=0\mathrm{and}\mathrm{x}+\mathrm{y}+\mathrm{z}-\mathrm{1}=0\mathrm{is}\\ 3\mathrm{x}+2\mathrm{y}+2\mathrm{z}-\mathrm{8}+\mathrm{\lambda }\left(\mathrm{x}+\mathrm{y}+\mathrm{z}-\mathrm{1}\right)=0\\ \left(\mathrm{3}+\mathrm{\lambda }\right)\mathrm{x}+\left(2+\mathrm{\lambda }\right)\mathrm{y}+\mathrm{}\left(\mathrm{2}+\mathrm{\lambda }\right)\mathrm{z}=\mathrm{8}+\mathrm{\lambda }\dots \left(\mathrm{1}\right)\\ \mathrm{Since}\mathrm{this}\mathrm{plane}\mathrm{passes}\mathrm{through}\mathrm{point}\left(\mathrm{1},1,1\right)\\ \therefore \left(\mathrm{3}+\mathrm{\lambda }\right)+\left(2+\mathrm{\lambda }\right)+\mathrm{}\left(\mathrm{2}+\mathrm{\lambda }\right)=\mathrm{8}+\mathrm{\lambda }\\ ⇒7+3\mathrm{\lambda }=8+\mathrm{\lambda }\\ ⇒2\mathrm{\lambda }=1\\ ⇒\mathrm{\lambda }=\frac{1}{2}\\ \mathrm{On}\mathrm{putting}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{\lambda }\mathrm{in}\mathrm{equation}\left(\mathrm{1}\right)\mathrm{we}\mathrm{get},\\ \left(\mathrm{3}+\frac{1}{2}\right)\mathrm{x}+\left(2+\frac{1}{2}\right)\mathrm{y}+\left(\mathrm{2}+\frac{1}{2}\right)\mathrm{z}=\mathrm{8}+\frac{1}{2}\\ ⇒\frac{7}{2}\mathrm{x}+\frac{5}{2}\mathrm{y}+\frac{5}{2}\mathrm{z}=\frac{17}{2}\\ ⇒7\mathrm{x}+5\mathrm{y}+5\mathrm{z}-17=0\end{array}$

Q.44

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{distance}\mathrm{of}\mathrm{a}\mathrm{point}\left(1,0,1\right)\mathrm{from}\mathrm{the}\mathrm{plane}\\ \stackrel{\to }{\mathrm{r}}.\left(6\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right)=4.\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{plane}\mathrm{in}\mathrm{vector}\mathrm{form}.\\ \stackrel{\to }{\mathrm{r}}.\left(6\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right)=4,\mathrm{where}\mathrm{}\stackrel{\to }{\mathrm{r}}=\mathrm{x}\stackrel{^}{\mathrm{i}}+\mathrm{y}\stackrel{^}{\mathrm{j}}+\mathrm{z}\stackrel{^}{\mathrm{k}}\\ \mathrm{Equation}\mathrm{of}\mathrm{the}\mathrm{plane}\mathrm{in}\mathrm{cartesian}\mathrm{form}\mathrm{is}\\ \left(\mathrm{x}\stackrel{^}{\mathrm{i}}+\mathrm{y}\stackrel{^}{\mathrm{j}}+\mathrm{z}\stackrel{^}{\mathrm{k}}\right).\left(6\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right)=4\\ ⇒6\mathrm{x}-3\mathrm{y}+2\mathrm{z}=4\\ ⇒6\mathrm{x}-3\mathrm{y}+2\mathrm{z}-4=0\\ \mathrm{Now},\mathrm{the}\mathrm{distance}\mathrm{of}\mathrm{plane}\mathrm{from}\mathrm{point}\left(1,0,1\right)–\\ \mathrm{}=\left|\frac{6\left(1\right)-3\left(0\right)+2\left(1\right)-4}{\sqrt{{6}^{2}+{\left(-3\right)}^{2}+{2}^{2}}}\right|\\ \mathrm{}=\left|\frac{6-0+2-4}{\sqrt{36+9+4}}\right|\\ \mathrm{}=\left|\frac{4}{\sqrt{49}}\right|\\ \mathrm{}=\frac{4}{7}\mathrm{units}\\ \end{array}$

Q.45

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{in}\mathrm{cartesian}\mathrm{and}\mathrm{vector}\mathrm{form}\\ \mathrm{that}\mathrm{passes}\mathrm{through}\mathrm{the}\mathrm{point}\left(1,2,3\right)\mathrm{and}\mathrm{perpendicular}\\ \mathrm{to}\mathrm{the}\mathrm{plane}\stackrel{\to }{\mathrm{r}}.\left(\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-5\stackrel{^}{\mathrm{k}}\right)+9=0.\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{plane}\mathrm{in}\mathrm{vector}\mathrm{form}\mathrm{is}\\ \stackrel{\to }{\mathrm{r}}.\left(\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-5\stackrel{^}{\mathrm{k}}\right)+9=0,\mathrm{where}\stackrel{\to }{\mathrm{r}}=\mathrm{x}\stackrel{^}{\mathrm{i}}+\mathrm{y}\stackrel{^}{\mathrm{j}}+\mathrm{z}\stackrel{^}{\mathrm{k}}\\ ⇒\left(\mathrm{x}\stackrel{^}{\mathrm{i}}+\mathrm{y}\stackrel{^}{\mathrm{j}}+\mathrm{z}\stackrel{^}{\mathrm{k}}\right).\left(\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-5\stackrel{^}{\mathrm{k}}\right)+9=0\\ \mathrm{In}\mathrm{cartesian}\mathrm{form},\mathrm{the}\mathrm{equation}\mathrm{becomes}\\ \mathrm{x}+2\mathrm{y}-5\mathrm{z}+9=0\\ \mathrm{The}\mathrm{direction}\mathrm{ratio}\mathrm{of}\mathrm{this}\mathrm{plane}\mathrm{is}\left(1,2,-5\right),\mathrm{which}\mathrm{is}\mathrm{directed}\\ \mathrm{along}\mathrm{a}\mathrm{line}\mathrm{perpendicular}\mathrm{to}\mathrm{the}\mathrm{plane}.\\ \therefore \mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{a}\mathrm{line}\mathrm{that}\mathrm{passes}\mathrm{through}\mathrm{the}\mathrm{point}\\ \left(1,2,3\right)\mathrm{and}\mathrm{having}\mathrm{direction}\mathrm{ratio}\left(1,2,-5\right)\mathrm{is}\\ \frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}-2}{2}=\frac{\mathrm{z}-3}{-5}\\ \mathrm{In}\mathrm{vector}\mathrm{form},\\ \mathrm{the}\mathrm{position}\mathrm{vector}\mathrm{of}\mathrm{a}\mathrm{point}\mathrm{P}\left(1,2,3\right)\mathrm{is}\stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}.\\ \mathrm{The}\mathrm{vector}\mathrm{parallel}\mathrm{to}\mathrm{line}\mathrm{is}\stackrel{\to }{\mathrm{b}}=\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-5\stackrel{^}{\mathrm{k}}.\\ \mathrm{We}\mathrm{have}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{a}\mathrm{line}\mathrm{which}\mathrm{passes}\mathrm{through}\mathrm{the}\mathrm{point}\\ \mathrm{whose}\mathrm{position}\mathrm{vector}\mathrm{is}\stackrel{\to }{\mathrm{a}}\mathrm{and}\mathrm{parallel}\mathrm{to}\stackrel{\to }{\mathrm{b}}\mathrm{is}\\ \stackrel{\to }{\mathrm{r}}=\stackrel{\to }{\mathrm{a}}+\mathrm{\lambda }\stackrel{\to }{\mathrm{b}}\\ \mathrm{Thus},\mathrm{equation}\mathrm{of}\mathrm{line}\mathrm{in}\mathrm{vector}\mathrm{form}\mathrm{is}\\ ⇒\stackrel{\to }{\mathrm{r}}=\left(\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\right)+\mathrm{\lambda }\left(\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-5\stackrel{^}{\mathrm{k}}\right)\end{array}$

Q.46 Find the angle between the two planes 3x-6y+2z-7 = 0 and 2x+2y-2z = 5.

Ans

$\begin{array}{l}\begin{array}{l}\text{The equation of}are\text{3x-6y+2z = 0 and 6x+3y-2z = 5}\\ \end{array}\\ \\ \mathrm{Here},{\mathrm{a}}_{1}=3,\mathrm{}{\mathrm{b}}_{1}=-6,\mathrm{}{\mathrm{c}}_{1}=2\\ {\mathrm{a}}_{2}=6,\mathrm{}{\mathrm{b}}_{2}=3,\mathrm{}{\mathrm{c}}_{2}=-2\\ \mathrm{Let}\mathrm{}\mathrm{\theta }\mathrm{be}\mathrm{the}\mathrm{angle}\mathrm{between}\mathrm{the}\mathrm{planes}.\\ \mathrm{cos\theta }\mathrm{}=\mathrm{}\left|\frac{{\mathrm{a}}_{1}{\mathrm{a}}_{2}+{\mathrm{b}}_{1}{\mathrm{b}}_{2}+{\mathrm{c}}_{1}{\mathrm{c}}_{2}}{\sqrt{{\left({\mathrm{a}}_{1}\right)}^{2}+{\left({\mathrm{b}}_{1}\right)}^{2}+{\left({\mathrm{c}}_{1}\right)}^{2}}\sqrt{{\left({\mathrm{a}}_{2}\right)}^{2}+{\left({\mathrm{b}}_{2}\right)}^{2}+{\left({\mathrm{c}}_{2}\right)}^{2}}}\right|\\ =\left|\frac{\mathrm{3}×6-6×3+2×-2}{\sqrt{{\left(3\right)}^{2}+{\left(-6\right)}^{2}+{\left(2\right)}^{2}}\sqrt{{\left(6\right)}^{2}+{\left(3\right)}^{2}+{\left(-2\right)}^{2}}}\mathrm{}\right|\\ =\left|\frac{18-18-4}{\sqrt{{\left(3\right)}^{2}+{\left(-6\right)}^{2}+{\left(2\right)}^{2}}\sqrt{{\left(6\right)}^{2}+{\left(3\right)}^{2}+{\left(-2\right)}^{2}}}\mathrm{}\right|\\ =\frac{\left|-4\right|}{\sqrt{9+36+4}\sqrt{9+36+4}}\\ =\frac{4}{\sqrt{49\sqrt{49}}}\\ =\frac{4}{7×7}\\ =\frac{4}{49}\\ \mathrm{cos\theta }\mathrm{}=\frac{4}{49}\\ ⇒\mathrm{\theta }={\mathrm{cos}}^{-1}\left(\frac{4}{49}\right)\end{array}$

Q.47

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{foot}\mathrm{of}\mathrm{the}\mathrm{perpendicular}\mathrm{from}\left(0, 2, 3\right)\mathrm{to}\mathrm{the}\\ \mathrm{line}\frac{\mathrm{x}+3}{5}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}+4}{3}.\end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{}\frac{\mathrm{x}+3}{5}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}+4}{3}=\mathrm{k}\left(\mathrm{say}\right)\\ \therefore \mathrm{Any}\mathrm{point}\mathrm{P}\mathrm{on}\mathrm{the}\mathrm{line}\mathrm{is}\mathrm{given}\mathrm{by}\\ \frac{\mathrm{x}+3}{5}=\mathrm{k}\\ ⇒\mathrm{x}=5\mathrm{k}-3\\ \frac{\mathrm{y}-1}{2}=\mathrm{k}\\ ⇒\mathrm{y}=2\mathrm{k}+1\\ \frac{\mathrm{z}+4}{3}=\mathrm{k}\\ ⇒\mathrm{z}=3\mathrm{k}-4\\ \therefore \mathrm{The}\mathrm{point}\mathrm{P}\mathrm{is}\left(5\mathrm{k}-3,2\mathrm{k}+1,3\mathrm{k}-4\right)\mathrm{}...\left(1\right)\\ \mathrm{Let}\mathrm{the}\mathrm{given}\mathrm{point}\mathrm{be}\mathrm{Q}\left(0,2,3\right).\\ \mathrm{Direction}\mathrm{ratio}\mathrm{of}\mathrm{line}\mathrm{PQ}\mathrm{is}\\ \left(5\mathrm{k}-3-0,2\mathrm{k}+1-2,3\mathrm{k}-4-3\right)=\left(5\mathrm{k}-3,2\mathrm{k}-1,3\mathrm{k}-7\right)\\ \mathrm{PQ}\mathrm{is}\mathrm{perpendicular}\mathrm{to}\mathrm{the}\mathrm{line}\\ \therefore 5\left(5\mathrm{k}-3\right)+2\left(2\mathrm{k}-1\right)+3\left(3\mathrm{k}-7\right)=0\\ ⇒25\mathrm{k}-15+4\mathrm{k}-2+9\mathrm{k}-21=0\\ ⇒38\mathrm{k}=38\\ ⇒\mathrm{k}=1,\mathrm{putting}\mathrm{in}\left(1\right)\mathrm{we}\mathrm{get},\\ \mathrm{Point}\mathrm{P}\mathrm{is}\left(2,3,-1\right).\end{array}$

Q.48

$\begin{array}{l}\mathrm{Prove}\mathrm{that}\mathrm{if}\mathrm{a}\mathrm{plane}\mathrm{has}\mathrm{the}\mathrm{intercepts}\mathrm{a},\mathrm{b},\mathrm{c}\mathrm{and}\mathrm{is}\mathrm{at}\\ \mathrm{a}\mathrm{distance}\mathrm{of}\mathrm{p}\mathrm{units}\mathrm{from}\mathrm{the}\mathrm{origin},\mathrm{then}\\ \frac{1}{{\mathrm{a}}^{2}}\mathrm{+}\frac{1}{{\mathrm{b}}^{2}}\mathrm{+}\frac{1}{{\mathrm{c}}^{2}}\mathrm{=}\frac{1}{{\mathrm{p}}^{2}}.\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{plane}\mathrm{having}\mathrm{intercept}\mathrm{a},\mathrm{b},\mathrm{c}\mathrm{on}\\ \mathrm{coordinate}\mathrm{axes}\mathrm{is}\\ \frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}+\frac{\mathrm{z}}{\mathrm{c}}=1\\ ⇒\frac{\mathrm{bcx}+\mathrm{acy}+\mathrm{abz}}{\mathrm{abc}}=1\\ ⇒\mathrm{bcx}+\mathrm{acy}+\mathrm{abz}=\mathrm{abc}\\ ⇒\mathrm{bcx}+\mathrm{acy}+\mathrm{abz}-\mathrm{abc}=0\\ \mathrm{Given}\mathrm{that}\mathrm{distance}\mathrm{of}\mathrm{this}\mathrm{plane}\mathrm{from}\mathrm{origin}\mathrm{is}\mathrm{p}.\\ ⇒\mathrm{p}=\left|\frac{0+0+0-\mathrm{abc}}{\sqrt{{\mathrm{b}}^{2}{\mathrm{c}}^{2}+{\mathrm{a}}^{2}{\mathrm{c}}^{2}+{\mathrm{a}}^{2}{\mathrm{b}}^{2}}}\right|\\ ⇒\mathrm{p}=\frac{\mathrm{abc}}{\sqrt{{\mathrm{b}}^{2}{\mathrm{c}}^{2}+{\mathrm{a}}^{2}{\mathrm{c}}^{2}+{\mathrm{a}}^{2}{\mathrm{b}}^{2}}}\\ \mathrm{On}\mathrm{squaring}\mathrm{both}\mathrm{sides},\\ {\mathrm{p}}^{2}=\frac{{\mathrm{a}}^{2}{\mathrm{b}}^{2}{\mathrm{c}}^{2}}{{\mathrm{b}}^{2}{\mathrm{c}}^{2}+{\mathrm{a}}^{2}{\mathrm{c}}^{2}+{\mathrm{a}}^{2}{\mathrm{b}}^{2}}\\ \mathrm{On}\mathrm{taking}\mathrm{reciprocal},\\ ⇒\frac{1}{{\mathrm{p}}^{2}}=\frac{{\mathrm{b}}^{2}{\mathrm{c}}^{2}+{\mathrm{a}}^{2}{\mathrm{c}}^{2}+{\mathrm{a}}^{2}{\mathrm{b}}^{2}}{{\mathrm{a}}^{2}{\mathrm{b}}^{2}{\mathrm{c}}^{2}}\\ ⇒\frac{1}{{\mathrm{p}}^{2}}=\frac{{\mathrm{b}}^{2}{\mathrm{c}}^{2}}{{\mathrm{a}}^{2}{\mathrm{b}}^{2}{\mathrm{c}}^{2}}+\frac{{\mathrm{a}}^{2}{\mathrm{c}}^{2}}{{\mathrm{a}}^{2}{\mathrm{b}}^{2}{\mathrm{c}}^{2}}+\frac{{\mathrm{a}}^{2}{\mathrm{b}}^{2}}{{\mathrm{a}}^{2}{\mathrm{b}}^{2}{\mathrm{c}}^{2}}\\ ⇒\frac{1}{{\mathrm{a}}^{2}}+\frac{1}{{\mathrm{b}}^{2}}+\frac{1}{{\mathrm{c}}^{2}}=\frac{1}{{\mathrm{p}}^{2}}\mathrm{}\\ \mathrm{Hence},\mathrm{proved}.\end{array}$

Q.49

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{distance}\mathrm{between}\mathrm{the}\mathrm{two}\mathrm{skew}\mathrm{lines}.\\ \stackrel{\to }{\mathrm{r}}=\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)+\mathrm{\lambda }\left(3\stackrel{^}{\mathrm{i}}+5\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right)\mathrm{and}\\ \stackrel{\to }{\mathrm{r}}=\left(3\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}\right)+\mathrm{\mu }\left(3\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right)\end{array}$

Ans

$\begin{array}{l}We\text{}have\text{the vector equation of lines}\\ \stackrel{\to }{r}=\left(\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}\right)+\lambda \left(3\stackrel{^}{i}+5\stackrel{^}{j}+2\stackrel{^}{k}\right)\text{and}\\ \stackrel{\to }{r}=\left(3\stackrel{^}{i}+2\stackrel{^}{j}+4\stackrel{^}{k}\right)+\mu \left(3\stackrel{^}{i}+4\stackrel{^}{j}+2\stackrel{^}{k}\right)\\ \text{Here,}\stackrel{\to }{{\text{a}}_{1}}=\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}\text{,}\\ \text{}{\stackrel{\to }{\text{a}}}_{2}=3\stackrel{^}{i}+2\stackrel{^}{j}+4\stackrel{^}{k}\\ ⇒{\stackrel{\to }{\text{a}}}_{2}-\stackrel{\to }{{\text{a}}_{1}}=2\stackrel{^}{i}+\stackrel{^}{j}+3\stackrel{^}{k}\\ and\text{}{\stackrel{\to }{\text{b}}}_{1}=3\stackrel{^}{i}+5\stackrel{^}{j}+2\stackrel{^}{k},\\ \text{}{\stackrel{\to }{\text{b}}}_{2}=3\stackrel{^}{i}+4\stackrel{^}{j}+2\stackrel{^}{k}\\ \therefore {\stackrel{\to }{\text{b}}}_{1}×{\stackrel{\to }{\text{b}}}_{2}=\left|\begin{array}{ccc}\stackrel{^}{i}& \stackrel{^}{j}& \stackrel{^}{k}\\ 3& 5& 2\\ 3& 4& 2\end{array}\right|\\ \text{}=\stackrel{^}{i}\left(10-8\right)-\stackrel{^}{j}\left(6-6\right)+\stackrel{^}{k}\left(12-15\right)\\ \text{}=\stackrel{^}{i}\left(2\right)-\stackrel{^}{j}\left(0\right)+\stackrel{^}{k}\left(-3\right)\\ \text{}=2\stackrel{^}{i}-3\stackrel{^}{k}\\ \left|{\stackrel{\to }{\text{b}}}_{1}×{\stackrel{\to }{\text{b}}}_{2}\right|=\sqrt{4+0+9}\\ \text{}=\sqrt{13}\\ \left({\stackrel{\to }{\text{a}}}_{2}-\stackrel{\to }{{\text{a}}_{1}}\right).\left({\stackrel{\to }{\text{b}}}_{1}×{\stackrel{\to }{\text{b}}}_{2}\right)=\left(2\stackrel{^}{i}+\stackrel{^}{j}+3\stackrel{^}{k}\right).\left(2\stackrel{^}{i}-3\stackrel{^}{k}\right)\\ \text{=4+0-9}\\ \text{=-5}\end{array}$ $\begin{array}{l}\mathrm{The}\mathrm{}\mathrm{distance}\mathrm{between}\mathrm{two}\mathrm{skew}\mathrm{lines}\mathrm{is}\mathrm{given}\mathrm{by},\\ \mathrm{d}=\frac{\left|\left({\stackrel{\to }{\mathrm{a}}}_{2}-\stackrel{\to }{{\mathrm{a}}_{1}}\right).\left({\stackrel{\to }{\mathrm{b}}}_{1}×{\stackrel{\to }{\mathrm{b}}}_{2}\right)\right|}{\left|\left({\stackrel{\to }{\mathrm{b}}}_{1}×{\stackrel{\to }{\mathrm{b}}}_{2}\right)\right|}\\ =\frac{\left|-5\right|}{13}\\ =\frac{5}{13}\end{array}$

Q.50

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{Cartesian}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{plane}\mathrm{which}\\ \mathrm{passing}\mathrm{through}\mathrm{the}\mathrm{intersection}\mathrm{of}\mathrm{the}\mathrm{planes}\\ \stackrel{\to }{\mathrm{r}}.\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)=1\mathrm{and}\stackrel{\to }{\mathrm{r}}.\left(3\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right)=8\mathrm{and}\mathrm{the}\mathrm{point}\\ \left(1,1,1\right).\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have}\stackrel{\to }{\mathrm{r}}=\mathrm{x}\stackrel{^}{\mathrm{i}}+\mathrm{y}\stackrel{^}{\mathrm{j}}+\mathrm{z}\stackrel{^}{\mathrm{k}}\\ \mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{planes}\mathrm{are}\\ \left(\mathrm{x}\stackrel{^}{\mathrm{i}}+\mathrm{y}\stackrel{^}{\mathrm{j}}+\mathrm{z}\stackrel{^}{\mathrm{k}}\right).\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)=1\mathrm{and}\left(\mathrm{x}\stackrel{^}{\mathrm{i}}+\mathrm{y}\stackrel{^}{\mathrm{j}}+\mathrm{z}\stackrel{^}{\mathrm{k}}\right).\left(3\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right)=8\\ ⇒\mathrm{x}+\mathrm{y}+\mathrm{z}-\mathrm{1}=0\mathrm{and}3\mathrm{x}+2\mathrm{y}+2\mathrm{z}-\mathrm{8}=0\\ \mathrm{Let}\mathrm{Equation}\mathrm{of}\mathrm{the}\mathrm{plane}\mathrm{passing}\mathrm{through}\mathrm{the}\mathrm{intersection}\\ \mathrm{of}\mathrm{the}\mathrm{planes}3\mathrm{x}+2\mathrm{y}+2\mathrm{z}-\mathrm{8}=0\mathrm{and}\mathrm{x}+\mathrm{y}+\mathrm{z}-\mathrm{1}=0\mathrm{is}\\ 3\mathrm{x}+2\mathrm{y}+2\mathrm{z}-\mathrm{8}+\mathrm{\lambda }\left(\mathrm{x}+\mathrm{y}+\mathrm{z}-\mathrm{1}\right)=0\\ \left(\mathrm{3}+\mathrm{\lambda }\right)\mathrm{x}+\left(2+\mathrm{\lambda }\right)\mathrm{y}+\mathrm{}\left(\mathrm{2}+\mathrm{\lambda }\right)\mathrm{z}=\mathrm{8}+\mathrm{\lambda }\dots \left(\mathrm{1}\right)\\ \mathrm{Since}\mathrm{this}\mathrm{plane}\mathrm{passes}\mathrm{through}\mathrm{point}\left(\mathrm{1},1,1\right)\\ \therefore \left(\mathrm{3}+\mathrm{\lambda }\right)+\left(2+\mathrm{\lambda }\right)+\mathrm{}\left(\mathrm{2}+\mathrm{\lambda }\right)=\mathrm{8}+\mathrm{\lambda }\\ ⇒7+3\mathrm{\lambda }=8+\mathrm{\lambda }\\ ⇒2\mathrm{\lambda }=1\\ ⇒\mathrm{\lambda }=\frac{1}{2}\\ \mathrm{On}\mathrm{putting}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{\lambda }\mathrm{in}\mathrm{equation}\left(\mathrm{1}\right)\mathrm{we}\mathrm{get},\\ \left(\mathrm{3}+\frac{1}{2}\right)\mathrm{x}+\left(2+\frac{1}{2}\right)\mathrm{y}+\left(\mathrm{2}+\frac{1}{2}\right)\mathrm{z}=\mathrm{8}+\frac{1}{2}\\ ⇒\frac{7}{2}\mathrm{x}+\frac{5}{2}\mathrm{y}+\frac{5}{2}\mathrm{z}=\frac{17}{2}\\ ⇒7\mathrm{x}+5\mathrm{y}+5\mathrm{z}-17=0\end{array}$

Q.51

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{in}\mathrm{cartesian}\mathrm{and}\mathrm{vector}\mathrm{form}\\ \mathrm{that}\mathrm{passes}\mathrm{through}\mathrm{the}\mathrm{point}\left(1,\mathrm{}2,\mathrm{}3\right)\mathrm{and}\mathrm{perpendicular}\\ \mathrm{to}\mathrm{the}\mathrm{plane}\stackrel{\to }{\mathrm{r}.}\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{2\mathrm{j}}-\stackrel{^}{5\mathrm{k}}\right)\mathrm{}+9\mathrm{}=\mathrm{}0.\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{plane}\mathrm{in}\mathrm{vector}\mathrm{form}\mathrm{is}\\ \mathrm{}\stackrel{\to }{\mathrm{r}.}\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{2\mathrm{j}}-\stackrel{^}{5\mathrm{k}}\right)\mathrm{}+9\mathrm{}=\mathrm{}0,\mathrm{where}\stackrel{\to }{\mathrm{r}.}=\stackrel{^}{\mathrm{xi}}+\stackrel{^}{\mathrm{yj}}+\stackrel{^}{\mathrm{zk}}\\ ⇒\mathrm{}\left(\stackrel{^}{\mathrm{xi}}+\stackrel{^}{\mathrm{yj}}+\stackrel{^}{\mathrm{zk}}\right),\mathrm{}\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{2\mathrm{j}}-\stackrel{^}{5\mathrm{k}}\right)\mathrm{}+\mathrm{}9\mathrm{}=\mathrm{}0\\ \mathrm{In}\mathrm{cartesian}\mathrm{form},\mathrm{the}\mathrm{equation}\mathrm{becomes}\\ \mathrm{x}+2\mathrm{y}-5\mathrm{z}+ 9 = 0\\ \mathrm{The}\mathrm{direction}\mathrm{ratio}\mathrm{of}\mathrm{this}\mathrm{plane}\mathrm{is}\left(1,\mathrm{}2,\mathrm{}-5\right),\mathrm{which}\mathrm{is}\mathrm{directed}\\ \mathrm{along}\mathrm{a}\mathrm{line}\mathrm{perpendicular}\mathrm{to}\mathrm{the}\mathrm{plane}.\\ \therefore \mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{a}\mathrm{line}\mathrm{that}\mathrm{passes}\mathrm{through}\mathrm{the}\mathrm{point}\\ \left(1,\mathrm{}2,\mathrm{}3\right)\mathrm{and}\mathrm{having}\mathrm{direction}\mathrm{ratio}\left(1,\mathrm{}2,\mathrm{}-5\right)\mathrm{is}\\ \frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}-2}{2}\mathrm{}=\mathrm{}\frac{\mathrm{z}-3}{-5}\\ \mathrm{In}\mathrm{vector}\mathrm{from},\\ \mathrm{the}\mathrm{position}\mathrm{vector}\mathrm{of}\mathrm{a}\mathrm{point}\mathrm{P}\left(1,\mathrm{}2,3\right)\mathrm{is}\stackrel{\to }{\mathrm{a}}=\mathrm{}\stackrel{^}{\mathrm{i}}+\stackrel{^}{2\mathrm{j}}+\stackrel{^}{3\mathrm{k}}\\ \mathrm{The}\mathrm{vector}\mathrm{parallel}\mathrm{to}\mathrm{line}\mathrm{is}\stackrel{\to }{\mathrm{b}}=\stackrel{^}{\mathrm{i}}+\stackrel{^}{2\mathrm{j}}-\stackrel{^}{5\mathrm{k}}.\\ \mathrm{We}\mathrm{have}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{a}\mathrm{line}\mathrm{which}\mathrm{passes}\mathrm{through}\mathrm{the}\mathrm{point}\\ \mathrm{whose}\mathrm{position}\mathrm{vector}\mathrm{is}\stackrel{\to }{\mathrm{a}}\mathrm{and}\mathrm{parallel}\mathrm{to}\stackrel{\to }{\mathrm{b}}\mathrm{is}\\ \stackrel{\to }{\mathrm{r}}\mathrm{}=\mathrm{}\stackrel{\to }{\mathrm{a}}\mathrm{}+\mathrm{\lambda }\mathrm{}\stackrel{\to }{\mathrm{b}}\mathrm{}\\ \mathrm{Thus},\mathrm{equation}\mathrm{of}\mathrm{line}\mathrm{in}\mathrm{vector}\mathrm{form}\mathrm{is}\\ ⇒\mathrm{}\stackrel{\to }{\mathrm{r}}\mathrm{}=\mathrm{}\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{2\mathrm{j}}+\stackrel{^}{3\mathrm{k}}\right)\mathrm{}+\mathrm{\lambda }\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{2\mathrm{j}}-\stackrel{^}{5\mathrm{k}}\right)\end{array}$

Q.52 Find the angle between the two planes 3x-6y+2z-7 = 0 and 2x+2y-2z = 5.

Ans

$\begin{array}{l}\mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{planes}\mathrm{are}3\mathrm{x}-6\mathrm{y}+2\mathrm{z}-7 = 0\mathrm{and}\\ 6\mathrm{x}+3\mathrm{y}-2\mathrm{z}= 5.\\ \mathrm{Here}{\mathrm{a}}_{1}=3,{\mathrm{b}}_{1}=-6,{\mathrm{c}}_{1}=\mathrm{}2\\ {\mathrm{a}}_{1}=6,{\mathrm{b}}_{1}=3,{\mathrm{c}}_{1}=\mathrm{}-2\\ \mathrm{Let}\mathrm{\theta }\mathrm{be}\mathrm{the}\mathrm{angle}\mathrm{between}\mathrm{the}\mathrm{planes}.\\ \mathrm{cos\theta }=\left|\frac{{\mathrm{a}}_{1}{\mathrm{a}}_{2}+{\mathrm{b}}_{1}{\mathrm{b}}_{2}+{\mathrm{c}}_{1}{\mathrm{c}}_{2}}{\sqrt{{\left({\mathrm{a}}_{1}\right)}^{2}+{\left({\mathrm{b}}_{1}\right)}^{2}+{\left({\mathrm{c}}_{1}\right)}^{2}}\mathrm{}\sqrt{}{\left({\mathrm{a}}_{2}\right)}^{2}+{\left({\mathrm{b}}_{2}\right)}^{2}+{\left({\mathrm{c}}_{3}\right)}^{2}}\right|\\ =\left|\frac{3×6-6×3+2×-2}{\sqrt{{\left(3\right)}^{2}+{\left(-6\right)}^{2}+{\left(2\right)}^{2}}\mathrm{}\sqrt{{\left(6\right)}^{2}+{\left(3\right)}^{2}+{\left(-2\right)}^{2}}}\right|\\ =\left|\frac{18-18-4}{\sqrt{{\left(3\right)}^{2}+{\left(-6\right)}^{2}+{\left(2\right)}^{2}}\mathrm{}\sqrt{{\left(6\right)}^{2}+{\left(3\right)}^{2}+{\left(-2\right)}^{2}}}\right|\\ =\left|\frac{-4}{\sqrt{9+36+4}\mathrm{}\sqrt{9+36+4}}\right|\\ =\left|\frac{4}{\sqrt{49}\mathrm{}\sqrt{49}}\right|\\ =\left|\frac{4}{7×7}\right|\\ =\left|\frac{4}{49}\right|\\ \mathrm{}\mathrm{cos\theta }\mathrm{}=\mathrm{}\frac{4}{49}\\ ⇒\mathrm{}\mathrm{\theta }\mathrm{}=\mathrm{}{\mathrm{cos}}^{-1}\left(\frac{4}{49}\right)\end{array}$

Q.53

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{foot}\mathrm{of}\mathrm{the}\mathrm{perpendicular}\mathrm{from}\left(0, 2, 3\right)\mathrm{to}\mathrm{the}\\ \mathrm{line}\frac{\mathrm{x}+3}{5}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}+4}{3}.\end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{}\frac{\mathrm{x}+3}{5}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}+4}{3}=\mathrm{k}\left(\mathrm{say}\right)\\ \therefore \mathrm{Any}\mathrm{point}\mathrm{P}\mathrm{on}\mathrm{the}\mathrm{line}\mathrm{is}\mathrm{given}\mathrm{by}\\ \frac{\mathrm{x}+3}{5}=\mathrm{k}\\ ⇒\mathrm{x}=5\mathrm{k}-3\\ \frac{\mathrm{y}-1}{2}=\mathrm{k}\\ ⇒\mathrm{y}=2\mathrm{k}+1\\ \frac{\mathrm{z}+4}{3}=\mathrm{k}\\ ⇒\mathrm{z}=3\mathrm{k}-4\\ \therefore \mathrm{The}\mathrm{point}\mathrm{P}\mathrm{is}\left(5\mathrm{k}-3,2\mathrm{k}+1,3\mathrm{k}-4\right)\mathrm{}...\left(1\right)\\ \mathrm{Let}\mathrm{the}\mathrm{given}\mathrm{point}\mathrm{be}\mathrm{Q}\left(0,2,3\right).\\ \mathrm{Direction}\mathrm{ratio}\mathrm{of}\mathrm{line}\mathrm{PQ}\mathrm{is}\\ \left(5\mathrm{k}-3-0,2\mathrm{k}+1-2,3\mathrm{k}-4-3\right)=\left(5\mathrm{k}-3,2\mathrm{k}-1,3\mathrm{k}-7\right)\\ \mathrm{PQ}\mathrm{is}\mathrm{perpendicular}\mathrm{to}\mathrm{the}\mathrm{line}\\ \therefore 5\left(5\mathrm{k}-3\right)+2\left(2\mathrm{k}-1\right)+3\left(3\mathrm{k}-7\right)=0\\ ⇒25\mathrm{k}-15+4\mathrm{k}-2+9\mathrm{k}-21=0\\ ⇒38\mathrm{k}=38\\ ⇒\mathrm{k}=1,\mathrm{putting}\mathrm{in}\left(1\right)\mathrm{we}\mathrm{get},\\ \mathrm{Point}\mathrm{P}\mathrm{is}\left(2,3,-1\right).\end{array}$

Q.54

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{distance}\mathrm{between}\mathrm{the}\mathrm{lines}\\ \stackrel{\to }{\mathrm{r}}=\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)+\mathrm{\lambda }\left(3\stackrel{^}{\mathrm{i}}+5\stackrel{^}{\mathrm{j}}-2\stackrel{^}{\mathrm{k}}\right)\mathrm{and}\\ \stackrel{\to }{\mathrm{r}}=\left(3\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}\right)+\mathrm{\mu }\left(3\stackrel{^}{\mathrm{i}}+5\stackrel{^}{\mathrm{j}}-2\stackrel{^}{\mathrm{k}}\right)\end{array}$

Ans

$\begin{array}{l}\stackrel{\to }{\mathrm{r}}=\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)+\mathrm{\lambda }\left(3\stackrel{^}{\mathrm{i}}+5\stackrel{^}{\mathrm{j}}-2\stackrel{^}{\mathrm{k}}\right)\mathrm{and}\\ \stackrel{\to }{\mathrm{r}}=\left(3\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}\right)+\mathrm{\mu }\left(3\stackrel{^}{\mathrm{i}}+5\stackrel{^}{\mathrm{j}}-2\stackrel{^}{\mathrm{k}}\right)\\ \mathrm{Since}\mathrm{coefficient}\mathrm{of}\mathrm{\lambda }\mathrm{and}\mathrm{\mu }\mathrm{are}\mathrm{same},\mathrm{therefore},\mathrm{lines}\\ \mathrm{are}\mathrm{parallel}.\\ \mathrm{Here},\stackrel{\to }{{\mathrm{a}}_{1}}=\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}},{\stackrel{\to }{\mathrm{a}}}_{2}=3\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}\\ \mathrm{and}\mathrm{}\stackrel{\to }{\mathrm{b}}=3\stackrel{^}{\mathrm{i}}+5\stackrel{^}{\mathrm{j}}-2\stackrel{^}{\mathrm{k}}\\ {\stackrel{\to }{\mathrm{a}}}_{2}-\stackrel{\to }{{\mathrm{a}}_{1}}=2\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\\ \therefore \stackrel{\to }{\mathrm{b}}×\left({\stackrel{\to }{\mathrm{a}}}_{2}-\stackrel{\to }{{\mathrm{a}}_{1}}\right)=\left|\begin{array}{ccc}\stackrel{^}{\mathrm{i}}& \stackrel{^}{\mathrm{j}}& \stackrel{^}{\mathrm{k}}\\ 3& 5& -2\\ 2& 1& 3\end{array}\right|\\ =\stackrel{^}{\mathrm{i}}\left(15+2\right)-\stackrel{^}{\mathrm{j}}\left(9+4\right)+\stackrel{^}{\mathrm{k}}\left(3-10\right)\\ =\stackrel{^}{\mathrm{i}}\left(17\right)-\stackrel{^}{\mathrm{j}}\left(13\right)+\stackrel{^}{\mathrm{k}}\left(-7\right)\\ \left|\stackrel{\to }{\mathrm{b}}×\left({\stackrel{\to }{\mathrm{a}}}_{2}-\stackrel{\to }{{\mathrm{a}}_{1}}\right)\right|=\sqrt{289+169+49}\\ =\sqrt{507}\\ \stackrel{\to }{\mathrm{b}}=3\stackrel{^}{\mathrm{i}}+5\stackrel{^}{\mathrm{j}}-2\stackrel{^}{\mathrm{k}}\\ \therefore \left|\stackrel{\to }{\mathrm{b}}\right|=\sqrt{9+25+4}\\ =\sqrt{38}\end{array}$ $\begin{array}{l}\mathrm{Distance}\mathrm{between}\mathrm{lines},\mathrm{d}=\frac{\left|\stackrel{\to }{\mathrm{b}}×\left({\stackrel{\to }{\mathrm{a}}}_{2}-{\stackrel{\to }{\mathrm{a}}}_{1}\right)\right|}{\left|\stackrel{\to }{\mathrm{b}}\right|}\\ =\frac{\sqrt{507}}{\sqrt{38}}\\ =\sqrt{\frac{507}{38}}\end{array}$

Q.55

$\mathrm{Find}\mathrm{the}\mathrm{image}\mathrm{of}\mathrm{the}\mathrm{point}\left(3,\mathrm{}-2,\mathrm{}1\right)\mathrm{in}\mathrm{the}\mathrm{plane}3\mathrm{x}–\mathrm{y}+4\mathrm{z}= 2.$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{Q}\mathrm{be}\mathrm{the}\mathrm{image}\mathrm{of}\mathrm{the}\mathrm{point}\mathrm{P}\left(3,\mathrm{}-2,\mathrm{}1\right)\mathrm{in}\mathrm{the}\mathrm{plane}\\ 3\mathrm{x}–\mathrm{y}+4\mathrm{z}= 2.\\ \mathrm{Then}\mathrm{PQ}\mathrm{is}\mathrm{normal}\mathrm{to}\mathrm{the}\mathrm{plane}.\mathrm{Therefore}\mathrm{direction}\mathrm{ratios}\mathrm{of}\\ \mathrm{PQ}\mathrm{are}3, -1, 4.\\ \mathrm{Since}\mathrm{PQ}\mathrm{passes}\mathrm{through}\mathrm{P}\left(3,\mathrm{}-2,\mathrm{}1\right)\mathrm{and}\mathrm{has}\mathrm{direction}\mathrm{ratios}3, -1, 4.\\ \mathrm{Therefore}\mathrm{equation}\mathrm{of}\mathrm{PQ}\mathrm{is}\\ \frac{\mathrm{x}-3}{3}\mathrm{}=\mathrm{}\frac{\mathrm{y}+2}{-1}\mathrm{}=\mathrm{}\frac{\mathrm{z}-1}{4}\mathrm{}=\mathrm{r}\mathrm{}\left(\mathrm{say}\right)\\ \mathrm{Let}\mathrm{the}\mathrm{coordinates}\mathrm{of}\mathrm{Q}\mathrm{are}\left(3\mathrm{r}+3,-\mathrm{r}-2,\mathrm{}4\mathrm{r}+1\right).\\ \mathrm{Let}\mathrm{R}\mathrm{be}\mathrm{the}\mathrm{mid}–\mathrm{point}\mathrm{of}\mathrm{PQ}.\\ \mathrm{Then}\mathrm{R}\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{plane}3\mathrm{x}–\mathrm{y}+4\mathrm{z}+2.\\ \mathrm{The}\mathrm{coordinates}\mathrm{of}\mathrm{R}\mathrm{are}\\ \left(\frac{3\mathrm{r}+3}{3},\mathrm{}\frac{-\mathrm{r}-2-2}{2},\mathrm{}\frac{4\mathrm{r}+1+1}{2}\right)⇒\left(\frac{3\mathrm{r}+6}{2},\mathrm{}\frac{-\mathrm{r}-4}{2},\mathrm{}2\mathrm{r}+1\right).\\ \mathrm{Since}\mathrm{R}\mathrm{lies}\mathrm{on}3\mathrm{x}–\mathrm{y}+4\mathrm{z}= 2,\mathrm{therefore}\\ \mathrm{3}\left(\frac{3\mathrm{r}+6}{2}\right)-\left(\frac{-\mathrm{r}-4}{2}\right)+4\left(2\mathrm{r}+1\right)\mathrm{}=\mathrm{}2\\ ⇒\mathrm{}13\mathrm{r}=-13\mathrm{}⇒\mathrm{}\mathrm{r}=-1\\ \mathrm{So},\mathrm{the}\mathrm{coordinates}\mathrm{of}\mathrm{Q}\mathrm{are}\\ \left\{3\left(-1\right)+3,\mathrm{}-\left(-1\right)-2,\mathrm{}4\mathrm{}\left(-1\right)+1\right\}⇒\left(0,\mathrm{}-1,\mathrm{}-3\right).\end{array}$

Q.56

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{distance}\mathrm{between}\mathrm{the}\mathrm{two}\mathrm{skew}\mathrm{lines}.\\ \stackrel{\to }{\mathrm{r}}=\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)+\mathrm{\lambda }\left(3\stackrel{^}{\mathrm{i}}+5\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right)\mathrm{and}\\ \stackrel{\to }{\mathrm{r}}=\left(3\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}\right)+\mathrm{\mu }\left(3\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right)\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{}\mathrm{have}\mathrm{the}\mathrm{vector}\mathrm{equation}\mathrm{of}\mathrm{lines}\\ \stackrel{\to }{\mathrm{r}}=\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)+\mathrm{\lambda }\left(3\stackrel{^}{\mathrm{i}}+5\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right)\mathrm{and}\\ \stackrel{\to }{\mathrm{r}}=\left(3\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}\right)+\mathrm{\mu }\left(3\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right)\\ \mathrm{Here},\stackrel{\to }{{\mathrm{a}}_{1}}=\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}},\\ {\stackrel{\to }{\mathrm{a}}}_{2}=3\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+4\stackrel{^}{\mathrm{k}}\\ ⇒{\stackrel{\to }{\mathrm{a}}}_{2}-\stackrel{\to }{{\mathrm{a}}_{1}}=2\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\\ \mathrm{and}\mathrm{}{\stackrel{\to }{\mathrm{b}}}_{1}=3\stackrel{^}{\mathrm{i}}+5\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}},\\ {\stackrel{\to }{\mathrm{b}}}_{2}=3\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\\ \therefore {\stackrel{\to }{\mathrm{b}}}_{1}×{\stackrel{\to }{\mathrm{b}}}_{2}=\left|\begin{array}{ccc}\stackrel{^}{\mathrm{i}}& \stackrel{^}{\mathrm{j}}& \stackrel{^}{\mathrm{k}}\\ 3& 5& 2\\ 3& 4& 2\end{array}\right|\\ =\stackrel{^}{\mathrm{i}}\left(10-8\right)-\stackrel{^}{\mathrm{j}}\left(6-6\right)+\stackrel{^}{\mathrm{k}}\left(12-15\right)\\ =\stackrel{^}{\mathrm{i}}\left(2\right)-\stackrel{^}{\mathrm{j}}\left(0\right)+\stackrel{^}{\mathrm{k}}\left(-3\right)\\ =2\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{k}}\\ \left|{\stackrel{\to }{\mathrm{b}}}_{1}×{\stackrel{\to }{\mathrm{b}}}_{2}\right|=\sqrt{4+0+9}\\ =\sqrt{13}\\ \left({\stackrel{\to }{\mathrm{a}}}_{2}-\stackrel{\to }{{\mathrm{a}}_{1}}\right).\left({\stackrel{\to }{\mathrm{b}}}_{1}×{\stackrel{\to }{\mathrm{b}}}_{2}\right)=\left(2\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\right).\left(2\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{k}}\right)\\ =4+0-9\\ =-5\end{array}$ $\begin{array}{l}The\text{}dis\mathrm{tan}ce\text{between two skew lines is given by,}\\ \text{d=}\frac{\left|\left({\stackrel{\to }{\text{a}}}_{2}-\stackrel{\to }{{\text{a}}_{1}}\right).\left({\stackrel{\to }{\text{b}}}_{1}×{\stackrel{\to }{\text{b}}}_{2}\right)\right|}{\left|\left({\stackrel{\to }{\text{b}}}_{1}×{\stackrel{\to }{\text{b}}}_{2}\right)\right|}\\ \text{}=\frac{\left|-5\right|}{13}\\ \text{}=\frac{5}{13}\\ \end{array}$