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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.

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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= (b–a) = (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= (b–a) = (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= x2–x1PQ , cosβ = y2–y1PQ and cosγ = z2–z1PQ, where PQ=(x2–x1)2+(y2–y1)2+(z2–z1)2

The direction ratios of a line segment can also be considered as (x2–x1, y2–y1, z2–z1 or x1–x2, y1–y2, z1–z2).

Equation of a line in Space

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 +z1k̂

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

a and b = position vectors

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

= real number.

In Cartesian form:

x-x1x2–x1=y-y1y2–y1=z-z1z2–z1, where x = x1 + (x2–x1), y = y1 + (y2–y1) and z = z1 + (z2–z1)

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 = (a1b2–a2b1)2+(b1c2–b2c1)2+(c1a2–c2a1)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:

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 . (a1 – a2)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 (r – a). 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 (r – a) [(b – a) x (c – a)] = 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 (a2–a1).(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

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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

$\begin{array}{l} We know \vec{r} = \vec{a} + λ (\vec{b} - \vec{a}) \\ ∴ \vec{r} = - \hat{i} + 2 \hat{k} + λ [3 \hat{i} + 4 \hat{j} + 6 \hat{k} - (- \hat{i} + 2 \hat{k})] \\ \vec{r} = - \hat{i} + 2 \hat{k} + λ [4 \hat{i} + 4 \hat{j} + 4 \hat{k}] \end{array}$

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

$\begin{array}{l} \frac{x - 0}{2 - 0} = \frac{y + 1}{- 3 + 1} = \frac{z - 3}{- 1 - 3} \\ \frac{x}{2} = \frac{y + 1}{- 2} = \frac{z - 3}{- 4} \\ \frac{x}{1} = \frac{y + 1}{- 1} = \frac{z - 3}{- 2} = λ (say) \\ x = λ, y = - λ - 1, z = - 2 λ + 3 \\ Let the coordinates of point M be \\ M (λ, - λ - 1, - 2 λ + 3) \\ Now d.r’s of AM are \\ λ - 1, - λ - 1 - 8, - 2 λ + 3 - 4 \\ λ - 1, - λ - 9, - 2 λ - 1 \\ AM ⊥ BC \\ (λ - 1) 1 + (- λ - 9) . (- 1) + (- 2 λ - 1) . (- 2) = 0 \\ 6 λ + 10 = 0 \\ λ = - \frac{5}{3} \\ Thus, M (- \frac{5}{3}, \frac{2}{3}, \frac{19}{3}) \end{array}$

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

Ans

Direction cosines are

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

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

Ans

$\begin{array}{l} Direction cosines of the line passing through two points \\ P (x_{1}, y_{1}, z_{1}) and Q (x_{2}, y_{2}, z_{2}) \\ \frac{x_{2} - x_{1}}{PQ}, \frac{y_{2} - y_{1}}{PQ}, \frac{z_{2} - z_{1}}{PQ}, Where, PQ \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2} + {(z_{2} - z_{1})}^{2}} \\ ∴ PQ = \sqrt{{(1 - 2)}^{2} + {(2 - 4)}^{2} + {(3 + 5)}^{2}} = \sqrt{77} \\ Thus, the direction cosines \\ ∴ \frac{3}{\sqrt{77}}, \frac{- 2}{\sqrt{77}}, \frac{8}{\sqrt{77}} \end{array}$

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

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

Ans

$\vec{r} = \hat{a} + λ \hat{b}, where \hat{a} is the position vector of the given point A .$

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

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

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

Ans

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

Q.9 The Cartesian equation of a line is

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

. Find the vector equation for the line.

Ans

Comparing the given equation with the standard form

$\frac{x - x_{1}}{a}, \frac{y - y_{1}}{b}, \frac{z - z_{1}}{c}$

, 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

$2 \hat{i} + 4 \hat{j} + 2 \hat{k}$

Then, vector equation of the line

$\vec{r} = - 3 \hat{i} + 5 \hat{j} - 6 \hat{k} + λ (2 \hat{i} + 4 \hat{j} + 2 \hat{k})$

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

$\vec{r} = 3 \hat{i} + 2 \hat{j} - 4 \hat{k} + λ (\hat{i} + 2 \hat{j} + 2 \hat{k}) and \vec{r} = 5 \hat{i} - 2 \hat{j} + μ (3 \hat{i} + 2 \hat{j} + 6 \hat{k})$

Ans

$\begin{array}{l} Here, \vec{b_{1}} = \hat{i} + 2 \hat{j} + 2 \hat{k} and \vec{b_{2}} = 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \\ Therefore, \\ \cos θ = \frac{{\vec{b}}_{1} . {\vec{b}}_{2}}{|{\vec{b}}_{1}| |{\vec{b}}_{2}|} \\ = \frac{(\hat{i} + 2 \hat{j} + 2 \hat{k}) . (3 \hat{i} + 2 \hat{j} + 6 \hat{k})}{\sqrt{1 + 4 + 4} \sqrt{9 + 4 + 36}} \\ = \frac{3 + 4 + 12}{\sqrt{9} \sqrt{49}} = \frac{19}{21} \\ \Rightarrow θ = \cos^{- 1} (\frac{19}{21}) \end{array}$

Q.11

$\begin{array}{l} If the line \vec{r} = (\hat{i} - 2 \hat{j} + \hat{k}) + λ (2 \hat{i} + \hat{j} + 2 \hat{k}) and the plane \\ \vec{r} . (3 \hat{i} - 2 \hat{j} + m \hat{k}) = 14 . Find the value of m . \end{array}$

Ans

$\begin{array}{l} The given line is parallel to the vector \vec{b} = 2 \hat{i} + \hat{j} + 2 \hat{k} \\ and the given vector is normal to the vector \vec{n} = 3 \hat{i} - 2 \hat{j} + m \hat{k} \\ If the line is parallel to th the plane then normal is perpendicular \\ to the line . \\ ∴ \vec{b} ⊥ \vec{n} \\ \Rightarrow \vec{b} . \vec{n} =0 \\ \Rightarrow (2 \hat{i} + \hat{j} + 2 \hat{k}) . (3 \hat{i} - 2 \hat{j} + m \hat{k}) = 0 \\ \Rightarrow 6 - 2 + 2 m = 0 \\ \Rightarrow m = - 2 \end{array}$

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

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

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

$\sqrt{29}$

, we get

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

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

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

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

$\vec{r} . (3 \hat{i} + 4 \hat{j} - 12 \hat{k}) + 13 = 0$

, then find the value of p.

Ans

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

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

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

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

$If \vec{r} . {\vec{n}}_{_{1}} = d_{1} and \vec{r} . {\vec{n}}_{_{2}} = d_{2}$

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

$\vec{r} . ({\vec{n}}_{_{1}} + λ {\vec{n}}_{2}) = d_{1} + {λd}_{2}$

Q.19

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

Ans

$\begin{array}{l} If the plane 4 x + 4 y - λz = 0 contains the line \\ (x - 1) / 2 = (y + 1) / 3 = z / 4, then \\ 2 \times 4 + 3 \times 4 - 4 \times λ = 0 \\ \Rightarrow 20 - 4 \times λ = 0 \\ \Rightarrow λ = 5 \end{array}$

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

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

Ans

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

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

Ans

$\begin{array}{l} The angle between two planes A_{1} x + B_{1} y + C_{1} z + D_{1} = 0 and \\ A_{2} x + B_{2} y + C_{2} z + D_{2} = 0 is given by \\ θ = \cos^{- 1} |\frac{A_{1} A_{2} + B_{1} B_{2} + C_{1} C_{2}}{\sqrt{{A_{1}}^{2} + {B_{1}}^{2} + {C_{1}}^{2}} \sqrt{{A_{2}}^{2} + {B_{2}}^{2} + {C_{2}}^{2}}}| \\ Hence, the angle between the planes x+y+2z = 9 \\ and 2x-y+ z=15 is given by \\ θ = \cos^{- 1} |\frac{1.2 + 1 . (- 1) + 2.1}{\sqrt{1^{2} + 1^{2} + 2^{2}} \sqrt{2^{2} + {(- 1)}^{2} + 1^{2}}}| \\ θ = \cos^{- 1} |\frac{2 - 1 + 2}{6}| \\ θ = \cos^{- 1} |\frac{1}{2}| \\ θ = \frac{π}{3} \end{array}$

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

Ans

$\begin{array}{l} Converting the equations in vector form we have \\ \vec{r} = (- \hat{i} + 3 \hat{k}) + λ (2 \hat{i} + 3 \hat{j} + 6 \hat{k}) and \vec{r} . (10 \hat{i} + 2 \hat{j} - 11 \hat{k}) = 3 \\ Here \vec{b} = 2 \hat{i} + 3 \hat{j} + 6 \hat{k} and \vec{n} = 10 \hat{i} + 2 \hat{j} - 11 \hat{k} \\ \sin Ï• = |\frac{(2 \hat{i} + 3 \hat{j} + 6 \hat{k}) . (10 \hat{i} + 2 \hat{j} - 11 \hat{k})}{\sqrt{2^{2} + 3^{2} + 6^{2}} \sqrt{10^{2} + 2^{2} + 11^{2}}}| \\ = |\frac{20 + 6 - 66}{\sqrt{49} \sqrt{225}}| \\ = |\frac{- 40}{7 \times 15}| = |\frac{- 8}{21}| = \frac{8}{21} \\ \Rightarrow Ï• = \sin^{- 1} (\frac{8}{21}) \end{array}$

Q.23 The vector equations of two lines are

$\vec{r} = \hat{i} + 2 \hat{j} + 3 \hat{k} + λ (\hat{i} - 3 \hat{j} + 2 \hat{k}) and \vec{r} = 4 \hat{i} + 5 \hat{j} + 6 \hat{k} + μ (\hat{2 i} + 3 \hat{j} + \hat{k})$

.
Find the shortest distance between the above lines.

Ans

$\begin{array}{l} Comparing given lines with standard vector form of lines \\ \vec{r} = \vec{a} + λ (\vec{b_{1}}) and \vec{r} = \vec{a_{2}} + μ (\vec{b_{2}}) we get \\ {\vec{a}}_{1} = \hat{i} + 2 \hat{j} + 3 \hat{k}, \vec{b_{1}} = \hat{i} - 3 \hat{j} + 2 \hat{k}, {\vec{a}}_{2} = 4 \hat{i} + 5 \hat{j} + 6 \hat{k} \\ and {\vec{b}}_{2} = 2 \hat{i} + 3 \hat{j} + 3 \hat{k} \\ ∴ \vec{a_{2}} - {\vec{a}}_{1} = 4 \hat{i} + 5 \hat{j} + 6 \hat{k} - \hat{i} - 2 \hat{j} - 3 \hat{k} \\ = 3 \hat{i} + 3 \hat{j} + 3 \hat{k} \\ Now, \vec{b_{1}} \times {\vec{b}}_{2} = |\begin{matrix} \begin{array}{l} \hat{i} \\ 1 \\ 2 \end{array} & \begin{array}{l} \hat{j} \\ - 3 \\ 3 \end{array} & \hat{\begin{array}{l} k \\ 2 \\ 1 \end{array}} \end{matrix}| \\ = \hat{i} (- 3 - 6) - \hat{j} (1 - 4) + \hat{k} (3 + 6) \\ = - 9 \hat{i} + 3 \hat{j} + 9 \hat{k} \\ Therefore, \\ |\vec{b_{1}} \times {\vec{b}}_{2}| = \sqrt{81 + 9 + 81} \\ = \sqrt{171} \\ = 3 \sqrt{19} \\ Also, (\vec{a_{2}} - {\vec{a}}_{1}) . (\vec{b_{1}} \times {\vec{b}}_{2}) = (3 \hat{i} + 3 \hat{j} + 3 \hat{k}) . (- 9 \hat{i} + 3 \hat{j} + 9 \hat{k}) \\ = - 27 + 9 + 27 \\ = 9 \\ ∴ Required shortest distance \\ = |\frac{(\vec{a_{2}} - {\vec{a}}_{1}) . (\vec{b_{1}} \times {\vec{b}}_{2})}{|\vec{b_{1}} \times {\vec{b}}_{2}|}| \\ = |\frac{9}{3 \sqrt{19}}| \\ = \frac{3}{\sqrt{19}} \end{array}$

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

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

Q.25

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

Ans

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

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

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

Q.27

$\begin{array}{l} Find the distance between the lines \\ \vec{r} = (\hat{i} + \hat{j} + \hat{k}) + λ (3 \hat{i} + 5 \hat{j} - 2 \hat{k}) and \\ \vec{r} = (3 \hat{i} + 2 \hat{j} + 4 \hat{k}) + μ (3 \hat{i} + 5 \hat{j} - 2 \hat{k}) \end{array}$

Ans

$\begin{array}{l} \vec{r} = (\hat{i} + \hat{j} + \hat{k}) + λ (3 \hat{i} + 5 \hat{j} - 2 \hat{k}) and \\ \vec{r} = (3 \hat{i} + 2 \hat{j} + 4 \hat{k}) + μ (3 \hat{i} + 5 \hat{j} - 2 \hat{k}) \\ Since coefficient of λ and μ are same, therefore, lines \\ are parallel . \\ Here, \vec{a_{1}} = \hat{i} + \hat{j} + \hat{k}, {\vec{a}}_{2} = 3 \hat{i} + 2 \hat{j} + 4 \hat{k} \\ and \vec{b} = 3 \hat{i} + 5 \hat{j} - 2 \hat{k} \\ {\vec{a}}_{2} - \vec{a_{1}} = 2 \hat{i} + \hat{j} + 3 \hat{k} \\ ∴ \vec{b} \times ({\vec{a}}_{2} - \vec{a_{1}}) = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 5 & - 2 \\ 2 & 1 & 3 \end{matrix}| \\ = \hat{i} (15 + 2) - \hat{j} (9 + 4) + \hat{k} (3 - 10) \\ = \hat{i} (17) - \hat{j} (13) + \hat{k} (- 7) \\ |\vec{b} \times ({\vec{a}}_{2} - \vec{a_{1}})| = \sqrt{289 + 169 + 49} \\ = \sqrt{507} \\ \vec{b} = 3 \hat{i} + 5 \hat{j} - 2 \hat{k} \\ ∴ |\vec{b}| = \sqrt{9 + 25 + 4} \\ = \sqrt{38} \end{array}$ $\begin{array}{l} Distance between lines, d = \frac{|\vec{b} \times ({\vec{a}}_{2} - {\vec{a}}_{1})|}{|\vec{b}|} \\ = \frac{\sqrt{507}}{\sqrt{38}} \\ = \sqrt{\frac{507}{38}} \end{array}$

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

Ans

$\begin{array}{l} Let P (x, y, z) be any point such that OP = r, \\ Where O is origin . Let α, β and γ be the angle made \\ by a line OP with coordinate axes . \\ OP = r \\ OR = x \\ QR = y \\ \Rightarrow OQ = \sqrt{x^{2} + y^{2}} \\ PR = z \\ ∴ OP = r = \sqrt{x^{2} + y^{2} + z^{2}} \end{array}$

$\begin{array}{l} ∴ x = rcosα \\ y = rcosβ \\ z = rcosγ \\ We have, OP = r \\ \Rightarrow \sqrt{x^{2} + y^{2} + z^{2}} = r \\ \Rightarrow x^{2} + y^{2} + z^{2} = r^{2} \\ \Rightarrow {(rcosα)}^{2} + {(rcosβ)}^{2} + {(rcosγ)}^{2} = r^{2} \\ \Rightarrow r^{2} (\cos^{2} α + \cos^{2} β + \cos^{2} γ) r^{2} \\ \Rightarrow \cos^{2} α + \cos^{2} β + \cos^{2} γ = 1 \\ \Rightarrow l^{2} + m^{2} + n^{2} = 1 \end{array}$

Q.29

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

Ans

$\begin{array}{l} Since α, β and γ are the angles made by a line OP with \\ coordinate axes, let P (x, y, z) be any point such that OP = r . \\ OP = r \\ OR = x \\ QR = y \\ \Rightarrow OQ = \sqrt{x^{2} + y^{2}} \\ PQ = z \\ ∴ OP = r = \sqrt{x^{2} + y^{2} + z^{2}} \\ ∴ cosα = \frac{x}{r} \\ \Rightarrow x = rcosα \\ \Rightarrow y = rcosβ \\ \Rightarrow z = rcosγ \end{array}$

$\begin{array}{l} We have, \\ \Rightarrow \sqrt{x^{2} + y^{2} + z^{2}} = r \\ \Rightarrow x^{2} + y^{2} + z^{2} = r^{2} \\ \Rightarrow {(rcosα)}^{2} + {(rcosβ)}^{2} + {(rcosγ)}^{2} = r^{2} \\ \Rightarrow r^{2} (\cos^{2} α + \cos^{2} β + \cos^{2} γ) = r^{2} \\ \Rightarrow \cos^{2} α + \cos^{2} β + \cos^{2} γ = 1 \\ \Rightarrow 1 - \sin^{2} α + 1 - \sin^{2} β + 1 - \sin^{2} γ = 1 \\ \Rightarrow 3 - (\sin^{2} α + \sin^{2} β + \sin^{2} γ) = 1 \\ \Rightarrow \sin^{2} α + \sin^{2} β + \sin^{2} γ = 2 \end{array}$

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

Ans

$\begin{array}{l} Let P (x, y, z) be any point such that OP = r, \\ where O is origin . Let α, β and γ be the angle made \\ by a line OP with coordinate axes . \\ OP = r \\ OR = x \\ QR = y \\ \Rightarrow OQ = \sqrt{x^{2} + y^{2}} \\ PQ = z \\ ∴ OP = r = \sqrt{x^{2} + y^{2} + z^{2}} \end{array}$

$\begin{array}{l} ∴ x = rcosα \\ y = rcosβ \\ z = rcosγ \\ We have, OP = r \\ \Rightarrow \sqrt{x^{2} + y^{2} + z^{2}} = r \\ \Rightarrow x^{2} + y^{2} + z^{2} = r^{2} \\ \Rightarrow {(rcosα)}^{2} + {(rcosβ)}^{2} + {(rcosγ)}^{2} = r^{2} \\ \Rightarrow r^{2} (\cos^{2} α + \cos^{2} β + \cos^{2} γ) = r^{2} \\ \Rightarrow \cos^{2} α + \cos^{2} β + \cos^{2} γ = 1 \\ \Rightarrow l^{2} + m^{2} + n^{2} = 1 \end{array}$

Q.31

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

Ans

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

Q.32

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

Ans

$\begin{array}{l} Since α, β and γ are the angles made by the line OP with \\ coordinate axes, let P (x, y, z) be any point such that \\ OP = r \\ OR = x \\ QR = y \\ \Rightarrow OQ = \sqrt{x^{2} + y^{2}} \\ PQ = z \\ ∴ OP = r = \sqrt{x^{2} + y^{2} + z^{2}} \end{array}$

$\begin{array}{l} ∴ cosα = \frac{x}{r} \\ \Rightarrow x = rcosα \\ \Rightarrow y = rcosβ \\ \Rightarrow z = rcosγ \\ We have OP = r \\ \Rightarrow \sqrt{x^{2} + y^{2} + z^{2}} = r \\ \Rightarrow x^{2} + y^{2} + z^{2} = r^{2} \\ \Rightarrow {(rcosα)}^{2} + {(rcosβ)}^{2} + {(rcosγ)}^{2} = r^{2} \\ \Rightarrow r^{2} (\cos^{2} α + \cos^{2} β + \cos^{2} γ) = r^{2} \\ \Rightarrow \cos^{2} α + \cos^{2} β + \cos^{2} γ = 1 \end{array}$

Q.33

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

Ans

$\begin{array}{l} We have, \\ \frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3} \\ The standard equation of line in cartesian form is \\ \frac{x - x_{1}}{a} = \frac{y - y_{1}}{b} = \frac{z - z_{1}}{c} \\ On comparing the equations, we get \\ x_{1} = - 3 \\ y_{1} = 1 \\ z_{1} = - 4 \\ a = 5 \\ b = 2 \\ c = 3 \\ ∴ The equation of line in vector form is \\ \vec{r} = \vec{a} + λ \vec{b}, where a represents point and b direction . \\ \Rightarrow \vec{r} = (- 3 \hat{i} + \hat{j} - 4 \hat{k}) + λ (5 \hat{i} + 2 \hat{j} + 3 \hat{k}) \end{array}$

Q.34

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

Ans

Q.35

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

Ans

$\begin{array}{l} Here, \vec{a} = 2 \hat{i} - \hat{j} + 4 \hat{k} and \vec{b} = \hat{i} + 2 \hat{j} - \hat{k} . \\ ∴ The equation of line in vector form is \\ \vec{r} = \vec{a} + λ \vec{b} \\ ∴ \vec{r} = 2 \hat{i} - \hat{j} + 4 \hat{k} + λ (\hat{i} + 2 \hat{j} - \hat{k} .) \\ The standard equation of line in cartesian form is \\ \frac{x - x_{1}}{a} = \frac{y - y_{1}}{b} = \frac{z - z_{1}}{c} . .. (1) \\ we have, \\ a =1, b =2, c =-1 and \\ x_{1} = 2 \\ y_{1} = - 1 \\ z_{1} = 4 \\ On putting the values in (1), we get \\ the equation of line in cartesian form is \\ \frac{x - 2}{1} = \frac{y + 1}{2} = \frac{z - 4}{- 1} \end{array}$

Q.36

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

Ans

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

Q.37

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

Ans

Q.38

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

Ans

Q.39

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

Ans

Q.40

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

Ans

Q.41

Ans

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

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

Q.43

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

Ans

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

Q.44

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

Ans

Q.45

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

Ans

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

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

Ans

$\begin{array}{l} \begin{array}{l} The equation of a r e 3x-6y+2z = 0 and 6x+3y-2z = 5 \end{array} \\ Here, a_{1} = 3, b_{1} = - 6, c_{1} = 2 \\ a_{2} = 6, b_{2} = 3, c_{2} = - 2 \\ Let θ be the angle between the planes . \\ cosθ = |\frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt{{(a_{1})}^{2} + {(b_{1})}^{2} + {(c_{1})}^{2}} \sqrt{{(a_{2})}^{2} + {(b_{2})}^{2} + {(c_{2})}^{2}}}| \\ = |\frac{3 \times 6 - 6 \times 3 + 2 \times - 2}{\sqrt{{(3)}^{2} + {(- 6)}^{2} + {(2)}^{2}} \sqrt{{(6)}^{2} + {(3)}^{2} + {(- 2)}^{2}}}| \\ = |\frac{18 - 18 - 4}{\sqrt{{(3)}^{2} + {(- 6)}^{2} + {(2)}^{2}} \sqrt{{(6)}^{2} + {(3)}^{2} + {(- 2)}^{2}}}| \\ = \frac{|- 4|}{\sqrt{9 + 36 + 4} \sqrt{9 + 36 + 4}} \\ = \frac{4}{\sqrt{49 \sqrt{49}}} \\ = \frac{4}{7 \times 7} \\ = \frac{4}{49} \\ cosθ = \frac{4}{49} \\ \Rightarrow θ = \cos^{- 1} (\frac{4}{49}) \end{array}$

Q.47

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

Ans

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

Q.48

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

Ans

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

Q.49

$\begin{array}{l} Find the distance between the two skew lines . \\ \vec{r} = (\hat{i} + \hat{j} + \hat{k}) + λ (3 \hat{i} + 5 \hat{j} + 2 \hat{k}) and \\ \vec{r} = (3 \hat{i} + 2 \hat{j} + 4 \hat{k}) + μ (3 \hat{i} + 4 \hat{j} + 2 \hat{k}) \end{array}$

Ans

$\begin{array}{l} W e h a v e the vector equation of lines \\ \vec{r} = (\hat{i} + \hat{j} + \hat{k}) + λ (3 \hat{i} + 5 \hat{j} + 2 \hat{k}) and \\ \vec{r} = (3 \hat{i} + 2 \hat{j} + 4 \hat{k}) + μ (3 \hat{i} + 4 \hat{j} + 2 \hat{k}) \\ Here, \vec{a_{1}} = \hat{i} + \hat{j} + \hat{k}, \\ {\vec{a}}_{2} = 3 \hat{i} + 2 \hat{j} + 4 \hat{k} \\ \Rightarrow {\vec{a}}_{2} - \vec{a_{1}} = 2 \hat{i} + \hat{j} + 3 \hat{k} \\ a n d {\vec{b}}_{1} = 3 \hat{i} + 5 \hat{j} + 2 \hat{k}, \\ {\vec{b}}_{2} = 3 \hat{i} + 4 \hat{j} + 2 \hat{k} \\ ∴ {\vec{b}}_{1} \times {\vec{b}}_{2} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 5 & 2 \\ 3 & 4 & 2 \end{matrix}| \\ = \hat{i} (10 - 8) - \hat{j} (6 - 6) + \hat{k} (12 - 15) \\ = \hat{i} (2) - \hat{j} (0) + \hat{k} (- 3) \\ = 2 \hat{i} - 3 \hat{k} \\ |{\vec{b}}_{1} \times {\vec{b}}_{2}| = \sqrt{4 + 0 + 9} \\ = \sqrt{13} \\ ({\vec{a}}_{2} - \vec{a_{1}}) . ({\vec{b}}_{1} \times {\vec{b}}_{2}) = (2 \hat{i} + \hat{j} + 3 \hat{k}) . (2 \hat{i} - 3 \hat{k}) \\ =4+0-9 \\ =-5 \end{array}$ $\begin{array}{l} The distance between two skew lines is given by, \\ d = \frac{|({\vec{a}}_{2} - \vec{a_{1}}) . ({\vec{b}}_{1} \times {\vec{b}}_{2})|}{|({\vec{b}}_{1} \times {\vec{b}}_{2})|} \\ = \frac{|- 5|}{13} \\ = \frac{5}{13} \end{array}$

Q.50

Ans

Q.51

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

Ans

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

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

Ans

$\begin{array}{l} The equation of planes are 3 x -6 y +2 z -7 = 0 and \\ 6 x +3 y -2 z = 5. \\ Here a_{1} = 3, b_{1} = - 6, c_{1} = 2 \\ a_{1} = 6, b_{1} = 3, c_{1} = - 2 \\ Let θ be the angle between the planes . \\ cosθ = |\frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt{{(a_{1})}^{2} + {(b_{1})}^{2} + {(c_{1})}^{2}} \sqrt{} {(a_{2})}^{2} + {(b_{2})}^{2} + {(c_{3})}^{2}}| \\ = |\frac{3 \times 6 - 6 \times 3 + 2 \times - 2}{\sqrt{{(3)}^{2} + {(- 6)}^{2} + {(2)}^{2}} \sqrt{{(6)}^{2} + {(3)}^{2} + {(- 2)}^{2}}}| \\ = |\frac{18 - 18 - 4}{\sqrt{{(3)}^{2} + {(- 6)}^{2} + {(2)}^{2}} \sqrt{{(6)}^{2} + {(3)}^{2} + {(- 2)}^{2}}}| \\ = |\frac{- 4}{\sqrt{9 + 36 + 4} \sqrt{9 + 36 + 4}}| \\ = |\frac{4}{\sqrt{49} \sqrt{49}}| \\ = |\frac{4}{7 \times 7}| \\ = |\frac{4}{49}| \\ cosθ = \frac{4}{49} \\ \Rightarrow θ = \cos^{- 1} (\frac{4}{49}) \end{array}$

Q.53

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

Ans

Q.54

Ans

Q.55

$Find the image of the point (3, - 2, 1) in the plane 3 x - y +4 z = 2.$

Ans

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

Q.56

Ans

$\begin{array}{l} We have the vector equation of lines \\ \vec{r} = (\hat{i} + \hat{j} + \hat{k}) + λ (3 \hat{i} + 5 \hat{j} + 2 \hat{k}) and \\ \vec{r} = (3 \hat{i} + 2 \hat{j} + 4 \hat{k}) + μ (3 \hat{i} + 4 \hat{j} + 2 \hat{k}) \\ Here, \vec{a_{1}} = \hat{i} + \hat{j} + \hat{k}, \\ {\vec{a}}_{2} = 3 \hat{i} + 2 \hat{j} + 4 \hat{k} \\ \Rightarrow {\vec{a}}_{2} - \vec{a_{1}} = 2 \hat{i} + \hat{j} + 3 \hat{k} \\ and {\vec{b}}_{1} = 3 \hat{i} + 5 \hat{j} + 2 \hat{k}, \\ {\vec{b}}_{2} = 3 \hat{i} + 4 \hat{j} + 2 \hat{k} \\ ∴ {\vec{b}}_{1} \times {\vec{b}}_{2} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 5 & 2 \\ 3 & 4 & 2 \end{matrix}| \\ = \hat{i} (10 - 8) - \hat{j} (6 - 6) + \hat{k} (12 - 15) \\ = \hat{i} (2) - \hat{j} (0) + \hat{k} (- 3) \\ = 2 \hat{i} - 3 \hat{k} \\ |{\vec{b}}_{1} \times {\vec{b}}_{2}| = \sqrt{4 + 0 + 9} \\ = \sqrt{13} \\ ({\vec{a}}_{2} - \vec{a_{1}}) . ({\vec{b}}_{1} \times {\vec{b}}_{2}) = (2 \hat{i} + \hat{j} + 3 \hat{k}) . (2 \hat{i} - 3 \hat{k}) \\ =4+0-9 \\ =-5 \end{array}$ $\begin{array}{l} T h e d i s \tan c e between two skew lines is given by, \\ d= \frac{|({\vec{a}}_{2} - \vec{a_{1}}) . ({\vec{b}}_{1} \times {\vec{b}}_{2})|}{|({\vec{b}}_{1} \times {\vec{b}}_{2})|} \\ = \frac{|- 5|}{13} \\ = \frac{5}{13} \end{array}$

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Certainly! The Class 12 Mathematics Chapter 11 notes provide all kinds of objective subjective and CBSE extra questions. Solutions for the same are also available to students for quick reference.

The CBSE revision notes include unlimited practice questions and their solutions. This step-by-step method helps students to learn several shortcut strategies to solve complex problems. Each question will be based on the different concepts included in the Chapter. By using the Class 12 Mathematics Chapter 11 notes, students will learn how to approach a problem and solve it.

CBSE Class 12 Maths Revision Notes Chapter 11

Class 12 Mathematics Chapter 11 Notes

NCERT Class 12 Mathematics Chapter 11 Notes: Key Topics

Introduction:

Representation of a point in Cartesian and Vector form

Direction Cosines of a Line

Direction Ratios of a Line

Relationship between the Direction Ratios and Direction Cosines

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

Equation of a line in Space

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

Distance between two lines:

PLANE:

Equation in normal form:

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

Equation of a plane through three non-collinear points

Intercept form of an equation:

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

Coplanarity of Two Lines:

Angle between Two Planes

Angle between line and plane:

Class 12 Mathematics Chapter 11 Notes: Exercise & Solutions

NCERT Class 12 Mathematics Chapter 11 Notes: Key Features

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CBSE Class 12 Maths Revision Notes Chapter 11

Class 12 Mathematics Chapter 11 Notes

NCERT Class 12 Mathematics Chapter 11 Notes: Key Topics

Introduction:

Representation of a point in Cartesian and Vector form

Direction Cosines of a Line

Direction Ratios of a Line

Relationship between the Direction Ratios and Direction Cosines

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

Equation of a line in Space

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

Distance between two lines:

PLANE:

Equation in normal form:

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

Equation of a plane through three non-collinear points

Intercept form of an equation:

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

Coplanarity of Two Lines:

Angle between Two Planes

Angle between line and plane:

Class 12 Mathematics Chapter 11 Notes: Exercise & Solutions

NCERT Class 12 Mathematics Chapter 11 Notes: Key Features

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