CBSE Class 11 Maths Revision Notes Chapter 12

Q.1 Show that the points A (1, 2, 3), B (–1, –2, –1), C (2, 3, 2) and D (4, 7, 6) are the vertices of a parallelogram ABCD, but it is not a rectangle.

Ans

To show ABCD is a parallelogram we need to show that opposite sides are equal.

AB = √{(– 1 – 1)² + (– 2 – 2)² + (– 1 – 3)²} = √(4 + 16 + 16) = 6
BC = √{(2 + 1)² + (3 + 2)² + (2 + 1)²} = √(9 + 25 + 9) = √43
CD = √{(4 – 2)² + (7 – 3)² + (6 – 2)²} = √(4 + 16 + 16) = 6
AD = √{(1 – 4)² + (2 – 7)² + (3 – 6)²} = √(9 + 25 + 9) = √43
Since AB = CD and BC = DA, ABCD is a parallelogram.
AC = √{(2 – 1)² + (3 – 2)² + (2 – 3)²} = √(1 + 1 + 1) = √3
BD = √{(4 + 1)² + (7 + 2)² + (6 + 1)²} = √(25 + 81 + 49) = √155
Since AC ≠ BD, ABCD is not a rectangle.

Q.2 Find the ratio in which the line joining the points (12, – 4, 8) and (27, – 9, 18) is divided by the plane 2x + 2y – z = 9. Also find the coordinates of the point of division.

Ans

Let P be the point which divides the line joining the points in ratio k : 1.
Then coordinates of P are
{(27k + 12)/(k + 1), (– 9k – 4)/(k + 1), (18k + 8)/(k + 1)} …(1)
Since, P lies on the plane 2x + 2y – z = 9, therefore coordinates of P must satisfy the equation of the plane
i.e., {2(27k + 12)/(k + 1) + 2(– 9k – 4)/(k + 1) – (18k + 8)/(k + 1)} = 9
⇒ k = 1/9.
So, the required ratios is 1 : 9
Putting k = 1/9 in (1), we get the coordinates of P (135/10, – 9/2, 9)

Q.3 Find the centroid of a triangle, mid-points of whose sides are (1, 2, – 3), (2, 0, 1) and (– 1, 1, – 4).

Ans

Let A(x₁, y₁, z₁), B(x₂, y₂, z₂) and C(x₃, y₃, z₃) are vertices of triangle ABC.
P(1, 2, –3), Q(3, 0, 1) and R(– 1, 1, – 4) are mid-points of sides AB, BC and AC respectively.

P is the mid-point of AB, therefore
(x₁ + x₂)/2 = 1 …(1)
(y₁ + y₂)/2 = 2 …(2)
(z₁ + z₂)/2 = – 3 …(3)
Q is the mid-point of BC, therefore
(x₂ + x₃)/2 = 3 …(4)
(y₂ + y₃)/2 = 0 …(5)
(z₂ + z₃)/2 = 1 …(6)
R is the mid-point of AC, therefore
(x₁ + x₃)/2 = – 1 …(7)
(y₁ + y₃)/2 = 1 …(8)
(z₁ + z₃)/2 = – 4 …(9)
Adding (1), (4) and (7), we get
(x₁ + x₂)/2 + (x₂ + x₃)/2 + (x₁ + x₃)/2 = 1 + 3 – 1
⇒ (x₁ + x₂ + x₃) = 3
Adding (2), (5) and (8), we get
(y₁ + y₂)/2 + (y₂ + y₃)/2 + (y₁ + y₃)/2 = 2 + 0 + 1
⇒ (y₁ + y₂ + y₃) = 3
Adding (3), (6) and (9), we get
(z₁ + z₂)/2 + (z₂ + z₃)/2 + (z₁ + z₃)/2 = – 3 + 1 – 4
⇒ (z₁ + z₂ + z₃) = – 6
We have centroid O of triangle is given by
{(x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3, (z₁ + z₂ + z₃)/3} = {3/3, 3/3, –6/3}
= (1, 1, – 2)
Therefore, centroid is (1, 1, – 2).

Q.4 A point is on the x-axis. What is its y-coordinate and z-coordinate?

Ans

y and z- coordinate both are zero.

Q.5 Name the octants in which the following points lie: (2, 3, 4), (1, –2, 6).

Ans

I, IV

Q.6 Fill in the blanks:

(i) The x- axis and z-axis taken together determine a plane known as_______.
(ii) The coordinates of points in the XZ-plane are of the form _______.

Ans

(i) XZ – plane
(ii) (x, 0, z)

Q.7 Find the distance between the points P(1, 0, 4) and Q (– 4, 1, 0).

Ans

PQ = √{(– 4 – 1)² + (1 – 0)² + (0 – 4)²}
= √(25 + 1 + 16)
= √42 units.

Q.8 Find the coordinates of the point which divides externally the line segment joining the points (1, –2, 3) and (3, 4, –5) in the ratio 2 : 3.

Ans

Let P (x, y, z) be the point which divides segment joining A (1, – 2, 3) and B (3, 4, – 5) externally in the ratio 2 : 3. Then
x = {2(3) + (– 3)(1)}/(2 – 3) = – 3
y = {2(4) + (– 3) (- 2)}/(2 – 3) = – 14
z = {2(- 5) + (– 3) (3)}/(2 – 3) = 19
Therefore, the required point is (– 3, –14, 19).

Q.9 Write the coordinates of the mid-point of the line segment joining two points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂).

Ans

$\text{Coordinates of midpoints :}\left(\frac{\left({\mathrm{x}}_1+{\mathrm{x}}_2\right)}{2},\frac{\left({\mathrm{y}}_1+{\mathrm{y}}_2\right)}{2},\frac{\left({\mathrm{z}}_1+{\mathrm{z}}_2\right)}{2}\right)\text{.}$

Q.10 Write the coordinates of the centroid of the triangle, whose vertices are P(x₁, y₁, z₁), Q(x₂, y₂, z₂) and R (x₃, y₃, z₃).

Ans

$\text{Required coordinates are}\left(\frac{\left({\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3\right)}{3},\frac{\left({\mathrm{y}}_1+{\mathrm{y}}_2+{\mathrm{y}}_3\right)}{3},\frac{\left({\mathrm{z}}_1+{\mathrm{z}}_2+{\mathrm{z}}_3\right)}{3}\right)\text{.}$

Q.11 A point is in the YZ-plane. What can you say about its x-coordinate?

Ans

Its x- coordinate is zero.

Q.12 Three coordinate planes divide the space into eight parts knowns as____.

Ans

octants

Q.13 Find the image of (–2,3,4) in the yz- plane.

Ans

The image of the point (–2,3,4) in the yz- plane is given by (2,3,4).

Q.14 Find the distance of the point P(–4, 3, 5) from the x- axis.

Ans

Let point on X- axis be (– 4, 0, 0) and given point is (– 4, 3, 5).
Required distance = √{(– 4 + 4)² +(3 – 0)² +(5 – 0)² }
= √(9 + 25)
= √34.

Q.15 Prove that the points P(1, 2, 3), Q(–1, –1, –1) and R(3, 5, 7) are collinear.

Ans

We have
PQ = √{(– 1 – 1)² + (– 1 – 2)² + (– 1 – 3)²} = √(4 + 9 + 16) = √29.
QR = √{(3 + 1)² + (5 + 1)² + (7 + 1)²} = √(16 + 36 + 64) = 2√29.
PR = √{(3 – 1)² + (5 – 2)² + (7 – 3)²} = √(4 + 9 + 16) = √29.
Therefore, QR = PQ + PR.
Hence, the given points are collinear.

Q.16 Find the point in XY-plane which is equidistant from three points A(2,0,3), B(0,3,2) and C(0,0,1).

Ans

The z-coordinate = 0 on xy-plane.
Let P (x, y, 0) be a point on xy-plane such that PA = PB = PC
Now, PA = PB
⇒ PA² = PB²
⇒ (x – 2)² + (y – 0)² + (0 – 3)² = (x – 0)² + (y – 3)² + (0 – 2)²
⇒ 4x – 6y = 0 or 2x – 3y = 0 …(1)
PB = PC
PB² = PC²
⇒ (x – 0)² + (y – 3)² + (0 – 2)² = (x – 0)² + (y – 0)² + (0 – 1)²
⇒ – 6y + 12 = 0 or y = 2 …(2)
By (1), we get x = 3
Hence, the required point is (3, 2, 0).

Q.17 Find the locus of the point which is equidistant from the points A(0,2,3) and B(2,–2,1).

Ans

Let P(x, y, z) be the point equidistant from A and B.
Therefore,
PA = PB
⇒ PA² = PB²
⇒ (x – 0)² + (y – 2)² + (z – 3)² = (x – 2)² + (y + 2)² + (z – 1)²
⇒ x² + y² – 4y + 4 + z² – 6z + 9 = x² – 4x + 4 + y² + 4y + 4 +z² – 2z + 1
⇒ 4x – 8y – 4z + 4 = 0 or x – 4y – z + 1 = 0
Hence, the required locus is x – 4y – z + 1 = 0.

Q.18 Find the coordinates of a point equidistant from the four points O(0, 0, 0), A(a, 0, 0), B(0, b, 0) and C(0, 0, c).

Ans

Let P (x, y, z) be the point equidistant from O, A, B and C.
Then OP = PA = PB = PC
OP = PA
⇒ OP² = PA²
⇒ x² + y² + z² = (x – a)² + (y – 0)² + (z – 0)²
⇒ 0 = –2ax + a²
⇒ x = a/2.
OP = PB
⇒ OP² = PB²
⇒ x² + y² + z² = (x – 0)² + (y – b)² + (z – 0)²
⇒ 0 = –2by + b²
⇒ y = b/2.
OP = PC
⇒ OP² = PC²
⇒ x² + y² + z² = (x – 0)² + (y – 0)² + (z – c)²
⇒ 0 = –2cz + c²
⇒ z = c/2.
Hence, the required point is (a/2, b/2, c/2).

Q.19 Find the ratio in which the line joining the points (1, 2, 3) and (– 3, 4, – 5) is divided by the xy-plane. Also, find the coordinates of the point of division.

Ans

Let P be the point which divides the line joining the given points in raio k :1.
Then the coordinates of P are {(–3k + 1)/(k + 1), (4k + 2)/(k + 1), (–5k + 3)/(k + 1)
P lies on xy-plane i.e. z = 0
(– 5k + 3)/(k + 1) = 0
⇒ k = 3/5.
So required ratio is 3 : 5.
Coordinates of point P = {(1 – 3 × 3/5)/(3/5 + 1), (2 + 4 × 3/5)/ (3/5 + 1), 0}
= (– 1/2, 11/4, 0)
and coordinates of P are (– 1/2, 11/4, 0).

Q.20 Find the equation of set of points P such that PA² + PB² = 2k², where A and B are the points (1, 2, 3) and (1, 0, 0), respectively.

Ans

Let the coordinates of point P be (x, y, z).
PA² = (x – 1)² + (y – 2)² + (z – 3)²
PB² = (x – 1)² + (y – 0)² + (z – 0)²
By the given condition, PA² + PB² = 2k², we have
(x – 1)² + (y – 2)² + (z – 3)² + (x – 1)² + (y – 0)² + (z – 0)² = 2k²
⇒ 2x² + 2y² + 2z² – 4x – 4y – 6z = 2k² – 15

Q.21 Using section formula, prove that the three points (– 2, 3, 5), (1, 2, 3) and (7, 0, –1) are collinear.

Ans

Let A (– 2, 3, 5), B(1, 2, 3) and C (7, 0, –1) be the given points.
Let the point C divides AB in the ratio k : 1.
Then coordinates of the point C are
{(k – 2)/(k + 1), (2k + 3)/(k +1), (3k + 5)/(k +1)
But the cordinate of C are (7, 0, –1),
So, (k – 2)/(k +1) = 7 ⇒ k – 2 = 7k + 7 ⇒ k = – 3/2
(2k + 3)/(k +1) = 0 ⇒ 2k + 3 = 0 ⇒ k = – 3/2
(3k + 5)/(k +1) = – 1 ⇒ 3k + 5 = – k – 1
⇒ 4k = – 6 ⇒ k = –3/2
From each of these equations we get k = – 3/2
Therefore, C (7, 0, –1) is a point which divides AB externally in the ratio 3 : 2 .
Hence A, B, C are collinear.

Q.22 A is a point (1, 3, 4) and B is the point (1, –2, –1). A point P moves so that 3PA = 2PB. Find the locus of P.

Ans

$\begin{array}{l}\text{We have}\\ 3\mathrm{PA}=2\mathrm{PB}\Rightarrow 9{\mathrm{PA}}^2=4{\mathrm{PB}}^2\\ 9\left\{{(\mathrm{x}-1)}^2+{(\mathrm{y}-3)}^2+{(\mathrm{z}-4)}^2\right\}\\ =4\left\{{(\mathrm{x}-1)}^2+{(\mathrm{y}+2)}^2+{(\mathrm{z}+1)}^2\right\}\\ 9\left\{{\mathrm{x}}^2-2\mathrm{x}+1+{\mathrm{y}}^2-6\mathrm{y}+9+{\mathrm{z}}^2-8\mathrm{z}+16\right\}\\ =4\left\{{\mathrm{x}}^2-2\mathrm{x}+1+{\mathrm{y}}^2+4\mathrm{y}+4+{\mathrm{z}}^2+2\mathrm{z}+1\right\}\\ 9{\mathrm{x}}^2-18\mathrm{x}+9+9{\mathrm{y}}^2-54\mathrm{y}+81+9{\mathrm{z}}^2-72\mathrm{z}+144\\ =4{\mathrm{x}}^2-8\mathrm{x}+4+4{\mathrm{y}}^2+16\mathrm{y}+16+4{\mathrm{z}}^2+8\mathrm{z}+4\\ 5{\mathrm{x}}^2+5{\mathrm{y}}^2+5{\mathrm{z}}^2-10\mathrm{x}-70\mathrm{y}-80\mathrm{z}-210=0\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2-2\mathrm{x}-14\mathrm{y}-16\mathrm{z}-42=0\\ \text{Which is locus of point P.}\end{array}$

Q.23 Three vertices of a parallelogram ABCD are A (4, 0, 3), B (3, 4, – 2) and C (– 2, 0, 1). Find the coordinates of the fourth vertex.

Ans

Let the fourth vertex be D(x, y, z)

O is mid-point of line AC
Therefore, coordinates of O are {(4 – 2)/2, (0 + 0)/2, (3 + 1)/2} = {1, 0, 2}
Now O is also mid-point of BD, therefore
(x + 3)/2 = 1 ⇒ x = – 1
(y + 4)/2 = 0 ⇒ y = – 4
(z – 2)/2 = 2 ⇒ z = 6.
Thus, fourth vertex be D(– 1, – 4, 6)

Q.24 Find the image of (– 4,0,0) in the xy-plane and (– 5,0,3) in the xz-plane.

Ans

The image of (– 4,0,0) in the xy-plane is (– 4,0,0) and the image of (– 5,0,3) in the xz-plane is (– 5,0,3).

Q.25 Name the octants in which the following points lie: (7,4,–3) and (–5,-3,–2).

Ans

Points (7,4,–3) and (–5,-3,–2) lie in V and VII Octants respectively.

Q.26 Find the distance between (1,2,3) and (–4, 2, –5).

Ans

$\begin{array}{l}\text{We\hspace{0.33em}have}(1,2,3)\text{\hspace{0.33em}and\hspace{0.33em}}(-4,2,-5)\text{\hspace{0.33em}Distance between two points}\\ =\sqrt{{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_2\right)}^2+{\left({\mathrm{z}}_2-{\mathrm{z}}_1\right)}^2}\\ =\sqrt{{(-4-1)}^2+{(2-2)}^2+{(-5-3)}^2}\\ =\sqrt{{(-5)}^2+{\left(0\right)}^2+{(-8)}^2}\\ =\sqrt{25+0+64}\\ =\sqrt{89}\mathrm{units}\end{array}$

Q.27 Find the coordinate of the point which divides the line segment joining the points (1,–2,2) and (3,4,–5) in the ratio of 1:3 externally.

Ans

$\begin{array}{l}\text{For external intersection point,}\\ \mathrm{x}=\frac{{\mathrm{m}}_2{\mathrm{x}}_1-{\mathrm{m}}_1{\mathrm{x}}_2}{{\mathrm{m}}_1-{\mathrm{m}}_2},\mathrm{y}=\frac{{\mathrm{m}}_2{\mathrm{y}}_1-{\mathrm{m}}_1{\mathrm{y}}_2}{{\mathrm{m}}_1-{\mathrm{m}}_2},\mathrm{z}=\frac{{\mathrm{m}}_2{\mathrm{z}}_1-{\mathrm{m}}_1{\mathrm{z}}_2}{{\mathrm{m}}_1-{\mathrm{m}}_2}\\ \mathrm{x}=\frac{3\left(1\right)-1\left(3\right)}{1-3},\mathrm{y}=\frac{3(-2)-1\left(4\right)}{1-3},\mathrm{z}=\frac{3\left(2\right)-1(-5)}{1-3}\\ \mathrm{x}=\frac{0}{-2},\mathrm{y}=\frac{-10}{-2},\mathrm{z}=\frac{11}{-2}\\ \mathrm{x}=0,\mathrm{y}=5,\mathrm{z}=\frac{11}{-2}\\ \text{Thus, the coordinates of intesection point is\hspace{0.33em}}\left(0,5,\frac{11}{-2}\right)\text{.}\end{array}$

Q.28 Find the ratio in which the line segment joining the points (4, 8, 10) and (6, 10, -8) is divided by the XZ-plane.

Ans

$\begin{array}{l}\text{Let}\mathrm{XZ}\text{-plane divides the segment joining}\mathrm{A}(4,8,10)\text{\hspace{0.33em}and}\mathrm{B}(6,10,-8)\text{at}\mathrm{P}(\mathrm{x},\mathrm{y},\mathrm{z})\text{in the ratio}\mathrm{k}:1.\\ \text{Then the coordinates of P are}\\ \left(\frac{1\left(4\right)+\mathrm{k}\left(6\right)}{\mathrm{k}+1},\frac{1\left(8\right)+\mathrm{k}\left(10\right)}{\mathrm{k}+1},\frac{1\left(10\right)-\mathrm{k}\left(8\right)}{\mathrm{k}+1}\right)\\ \left(\frac{4+6\mathrm{k}}{\mathrm{k}+1},\frac{8+10\mathrm{k}}{\mathrm{k}+1},\frac{10-8\mathrm{k}}{\mathrm{k}+1}\right)\\ \text{Since, P lies on XZ-plane}\\ \text{So, its Y-coordinate is zero, i.e.,}\frac{8+10\mathrm{k}}{\mathrm{k}+1}=0\\ \Rightarrow 8+10\mathrm{k}=0\\ \Rightarrow 10\mathrm{k}=-8\Rightarrow \mathrm{k}=\frac{-8}{10}\Rightarrow \mathrm{k}=-\frac{4}{5}\\ \text{Therefore, XZ-plane divides AB externally\hspace{0.33em}in the ratio}4:5.\end{array}$

Q.29 Check whether (1,2,3), (–1,–2,–1) and (2,3,2) are collinear or not.

Ans

$\begin{array}{l}\text{Let}\mathrm{A}(1,2,3),\mathrm{B}(-1,-2,-1),\mathrm{C}(2,3,2)\text{are collinear\hspace{0.33em}and B divides AC in k:1.}\\ \text{then,}\\ \text{coordinates of}\mathrm{B}=\left(\frac{1\left(2\right)+\mathrm{k}\left(1\right)}{\mathrm{k}+1},\frac{1\left(3\right)+\mathrm{k}\left(2\right)}{\mathrm{k}+1},\frac{1\left(2\right)+\mathrm{k}\left(3\right)}{\mathrm{k}+1}\right)\\ (-1,-2,-1)=\left(\frac{2+\mathrm{k}}{\mathrm{k}+1},\frac{3+2\mathrm{k}}{\mathrm{k}+1},\frac{2+3\mathrm{k}}{\mathrm{k}+1}\right)\\ \Rightarrow -1=\frac{2+\mathrm{k}}{\mathrm{k}+1},-2=\frac{3+2\mathrm{k}}{\mathrm{k}+1}\text{and\hspace{0.33em}}-1=\frac{2+3\mathrm{k}}{\mathrm{k}+1}\\ \Rightarrow -\mathrm{k}-1=2+{\mathrm{k}}_{\mathrm{t}}-2\mathrm{k}-2=3+2\mathrm{k}\text{and}-\mathrm{k}-1=2+3\mathrm{k}\\ \Rightarrow \mathrm{k}=-\frac{3}{2},\mathrm{k}=-\frac{5}{4}\text{and}\mathrm{k}=-\frac{3}{4}.\\ \text{Since, ratio is not same so}\mathrm{A},\mathrm{B}\text{and}\mathrm{C}\text{are not collinear.}\end{array}$

Q.30 Check whether (–2,3,5), (1,2,3) and (7,0,–1) are collinear or not.

Ans

$\begin{array}{l}\text{Let}\mathrm{A}(-2,3,5),\mathrm{B}(1,2,3),\mathrm{C}(7,0,-1)\text{are collinear\hspace{0.33em}and B divides AC in k:1.}\\ \text{then,}\\ \text{coordinates of}\mathrm{B}=\left(\frac{1\left(7\right)+\mathrm{k}(-2)}{\mathrm{k}+1},\frac{1\left(0\right)+\mathrm{k}\left(3\right)}{\mathrm{k}+1},\frac{1(-1)+\mathrm{k}\left(5\right)}{\mathrm{k}+1}\right)\\ (1,2,3)=\left(\frac{7-2\mathrm{k}}{\mathrm{k}+1},\frac{0+3\mathrm{k}}{\mathrm{k}+1},\frac{-1+5\mathrm{k}}{\mathrm{k}+1}\right)\\ \Rightarrow 1=\frac{7-2\mathrm{k}}{\mathrm{k}+1},2=\frac{0+3\mathrm{k}}{\mathrm{k}+1}\text{and}3=\frac{-1+5\mathrm{k}}{\mathrm{k}+1}\\ \Rightarrow \mathrm{k}+1=7-2{\mathrm{k}}_{1}2\mathrm{k}+2=3\mathrm{k}\text{and}3\mathrm{k}+3=-1+5\mathrm{k}\\ \Rightarrow \mathrm{k}=\frac{6}{3}=2,\mathrm{k}=2\text{and}\mathrm{k}=2\\ \text{Sinceratio\hspace{0.33em}is\hspace{0.33em}not\hspace{0.33em}same, so}{\mathrm{A}}_{\mathrm{t}}\mathrm{B}\text{and}\mathrm{C}\text{are not collinear.}\end{array}$

Q.31 Find the centroid of triangle ABC whose vertices are (3,–5,7), (–1,7,-6) and (1,1,2).

Ans

$\begin{array}{l}\text{Centroid\hspace{0.33em}of\hspace{0.33em}triangle\hspace{0.33em}}=\left(\frac{{\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3}{3},\frac{{\mathrm{y}}_1+{\mathrm{y}}_2+{\mathrm{y}}_3}{3},\frac{{\mathrm{z}}_1+{\mathrm{z}}_2+{\mathrm{z}}_3}{3}\right)\\ =\left(\frac{3-1+1}{3},\frac{-5+7+1}{3},\frac{7-6+2}{3}\right)\\ =(1,1,1)\end{array}$

Q.32 Find the distance between (2, –1, 3) and (–2, 1, 3).

Ans

$\begin{array}{l}\text{We\hspace{0.33em}have\hspace{0.33em}}(2,-1,3)\mathrm{and}(-2,1,3)\text{\hspace{0.33em}Distance\hspace{0.33em}between\hspace{0.33em}two\hspace{0.33em}points}\\ =\sqrt{{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_2\right)}^2+{\left({\mathrm{z}}_2-{\mathrm{z}}_1\right)}^2}\\ =\sqrt{{(-2-2)}^2+{(1+1)}^2+{(3-3)}^2}\\ =\sqrt{{(-4)}^2+{\left(2\right)}^2+{\left(0\right)}^2}\\ =\sqrt{16+4+0}\\ =\sqrt{20}\mathrm{units}\\ =2\sqrt{5}\mathrm{units}\end{array}$

Q.33 Find the equation of the set of the points P such that its distances from the points A(2,3,4) and B(–2,1,5) are equal.

Ans

$\begin{array}{l}\text{Let}\mathrm{P}(\mathrm{x},\mathrm{y},\mathrm{z})\text{is equidistant from}\mathrm{A}(2,3,4)\text{and}\mathrm{B}(-2,1,5)\text{,}\\ \text{Then,\hspace{0.33em}}\mathrm{PA}=\mathrm{PB}\Rightarrow {\mathrm{PA}}^2={\mathrm{PB}}^2\\ {\left(2-\mathrm{x}\right)}^2+{\left(3-\mathrm{y}\right)}^2+{\left(4-\mathrm{z}\right)}^2={\left(-2-\mathrm{x}\right)}^2+{\left(1-\mathrm{y}\right)}^2+{\left(5-\mathrm{z}\right)}^2\\ \sout{4}-4\mathrm{x}+\sout{{\mathrm{x}}^2}+9-6\mathrm{y}+\sout{{\mathrm{y}}^2}+16-8\mathrm{z}+\sout{{\mathrm{z}}^2}\\ =\sout{4}+4\mathrm{x}+\sout{{\mathrm{x}}^2}+1-2\mathrm{y}+\sout{{\mathrm{y}}^2}+25-50\mathrm{z}+\sout{{\mathrm{z}}^2}\\ -4\mathrm{x}+25-6\mathrm{y}-8\mathrm{z}=4\mathrm{x}-2\mathrm{y}-50\mathrm{z}+26\\ -8\mathrm{x}-4\mathrm{y}+42\mathrm{z}-1=0\\ \Rightarrow 8\mathrm{x}+4\mathrm{y}-42\mathrm{z}+1=0\end{array}$

Q.34 Verify that (0,7,–10),(1,6,–6) and (4,9,–6) are the vertices of an isosceles triangle.

Ans

$\begin{array}{l}\text{Let}\mathrm{A}(0,7,-10),\mathrm{B}(1,6,-6)\text{and}\mathrm{C}(4,9,-6)\\ \text{are the vertices of triangle}\mathrm{ABC}\text{.}\\ 50,\mathrm{AB}=\sqrt{{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^2+{\left({\mathrm{z}}_2-{\mathrm{z}}_1\right)}^2}\\ =\sqrt{{(1-0)}^2+{(6-7)}^2+{(-6+10)}^2}\\ =\sqrt{1+1+16}\\ =\sqrt{18}\text{units}\\ \mathrm{BC}=\sqrt{{(4-1)}^2+{(9-6)}^2+{(-6+6)}^2}\\ =\sqrt{9+9+0}\\ =\sqrt{18}\text{units}\\ \mathrm{CA}=\sqrt{{(4-0)}^2+{(9-7)}^2+{(-6+10)}^2}\\ =\sqrt{16+4+16}\\ =\sqrt{18}\text{units}\\ \text{Since,}\mathrm{AB}=\mathrm{BC}=\mathrm{CA}\\ \text{So,}\mathrm{\Delta ABC}\text{is equilateral triangle.}\end{array}$

Q.35 Find the lengths of the medians in triangle with vertices A(0,0,6),B(0,4,0) and (0,4,0).

Ans

$\begin{array}{l}\text{For median AD,}\\ \text{Coordinates of mid point}\mathrm{D}\text{of}\mathrm{BC}=\left(\frac{0+0}{2},\frac{4+4}{2},\frac{0+0}{2}\right)\\ =(0,4,0)\\ \text{So, length of median}\mathrm{AD}=\sqrt{{(0-0)}^2+{(4-0)}^2+{(0-6)}^2}\\ =\sqrt{0+16+36}\\ =\sqrt{52}\text{units}\\ \text{For median CF,}\\ \text{Coordinates of mid point}\mathrm{F}\text{of}\mathrm{AB}=\left(\frac{0+0}{2},\frac{0+4}{2},\frac{6+0}{2}\right)\\ =(0,2,3)\\ \text{So, length of median}\mathrm{CF}=\sqrt{{(0-0)}^2+{(4-2)}^2+{(0-3)}^2}\\ =\sqrt{0+4+9}\\ =\sqrt{13}\text{units}\\ \text{For median BE,}\\ \text{Coordinates of mid point}\mathrm{E}\text{of}\mathrm{AC}=\left(\frac{0+0}{2},\frac{0+4}{2},\frac{6+0}{2}\right)\\ =(0,2,3)\\ \text{So, length of median BE}=\sqrt{{(0-0)}^2+{(2-4)}^2+{(3-0)}^2}\\ =\sqrt{0+4+9}\\ =\sqrt{13}\text{units}\end{array}$

Q.36 Find the equation of the set of points which are equidistant from the points (–1, 2, 1) and (1, –2, 5).

Ans

$\begin{array}{l}\text{Let}\mathrm{P}(\mathrm{x},\mathrm{y},\mathrm{z})\text{be equidistant from point}\mathrm{A}\left(-1,2,1\right)\text{and}\mathrm{B}\left(1,-2,5\right)\\ \text{Then,}\mathrm{PA}=\mathrm{PB}\Rightarrow {\mathrm{PA}}^2={\mathrm{PB}}^2\\ \Rightarrow {\left(-1-\mathrm{x}\right)}^2+{\left(2-\mathrm{y}\right)}^2+{\left(1-\mathrm{z}\right)}^2={\left(1-\mathrm{x}\right)}^2+{\left(-2-\mathrm{y}\right)}^2+{\left(5-\mathrm{z}\right)}^2\\ \Rightarrow 1+2\mathrm{x}+\sout{{\mathrm{x}}^2}+4-4\mathrm{y}+\sout{{\mathrm{y}}^2}+1-2\mathrm{z}+\sout{{\mathrm{z}}^2}\\ =1-2\mathrm{x}+\sout{{\mathrm{x}}^2}+4+4\mathrm{y}+\sout{{\mathrm{y}}^2}+25-10\mathrm{z}+\sout{{\mathrm{z}}^2}\\ 2\mathrm{x}-4\mathrm{y}++1-2\mathrm{z}=-2\mathrm{x}+4\mathrm{y}+25-10\mathrm{z}\\ 2\mathrm{x}+2\mathrm{x}-4\mathrm{y}-4\mathrm{y}+10\mathrm{z}-2\mathrm{z}+1-25=0\\ 4\mathrm{x}-8\mathrm{y}+8\mathrm{z}-24=0\\ \Rightarrow \mathrm{x}-2\mathrm{y}+2\mathrm{z}-6=0\end{array}$

Q.37 Find the coordinates of a point on x-axis which are at a distance of 5√2 from the point P(3,–2,5).

Ans

$\begin{array}{l}\text{Let coordinates of a point on}\mathrm{x}\text{-axis be}\mathrm{A}(\mathrm{x},0,0)\text{.}\\ \text{Distance of point}\mathrm{A}\text{from point}\mathrm{P}(3,-2,5)=5\sqrt{2}\\ \Rightarrow \mathrm{AP}=5\sqrt{2}\Rightarrow {\mathrm{AP}}^2=50\\ {\left(3-\mathrm{x}\right)}^2+{\left(-2-0\right)}^2+{\left(5-0\right)}^2=50\\ 9-6\mathrm{x}+{\mathrm{x}}^2+4+25=50\\ {\mathrm{x}}^2-6\mathrm{x}+38-50=0\\ {\mathrm{x}}^2-6\mathrm{x}-12=0\\ \mathrm{x}=\frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}}\\ =\frac{-(-6)\pm \sqrt{{(-6)}^2-4\left(1\right)(-12)}}{2\left(1\right)}\\ =\frac{6\pm \sqrt{36+48}}{2\left(1\right)}\\ =\frac{6\pm \sqrt{84}}{2}\\ =\frac{6\pm 2\sqrt{21}}{2}\\ \mathrm{x}=(3+\sqrt{21}),(3-\sqrt{21})\\ \text{So, the coordinates of point}\mathrm{A}\text{on}\mathrm{x}\text{-axis are}\\ (3+\sqrt{21},0,0)\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}or\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}(3-\sqrt{21},0,0).\end{array}$

Q.38 A point R with z-coordinate 8 lies on the line segment joining the points P(2,-3,4) and Q(8,0,10). Find the ratio in which the line PQ is divided by the point R.

Ans

$\begin{array}{l}\text{Let}\mathrm{R}\text{divides PQ in ratio}\mathrm{k}:1.\\ \text{then coordinates of}\mathrm{R}=\left(\frac{1\left(8\right)+\mathrm{k}\left(2\right)}{\mathrm{k}+1},\frac{1\left(0\right)+\mathrm{k}(-3)}{\mathrm{k}+1},\frac{1\left(10\right)+\mathrm{k}\left(4\right)}{\mathrm{k}+1}\right)\\ (0,0,8)=\left(\frac{8+2\mathrm{k}}{\mathrm{k}+1},\frac{-3\mathrm{k}}{\mathrm{k}+1},\frac{10+4\mathrm{k}}{\mathrm{k}+1}\right)\\ \Rightarrow \frac{10+4\mathrm{k}}{\mathrm{k}+1}=8\Rightarrow 10+4\mathrm{k}=8\mathrm{k}+8\\ \Rightarrow 10-8=8\mathrm{k}-4\mathrm{k}\\ \Rightarrow 2=4\mathrm{k}\Rightarrow \mathrm{k}=\frac{2}{4}=\frac{1}{2}\\ \text{So, point}\mathrm{R}\text{divides}\mathrm{PQ}\text{in the ratio}\frac{1}{2}:1\\ \text{i.e.,}1:2\end{array}$

Q.39 By using distance formula, prove that the three points ( – 4,6,10), (2,4,6) and (14, 0, – 2) are collinear.

Ans

$\begin{array}{l}\text{Let points}\mathrm{A}(-4,6,10),\mathrm{B}(2,4,6)\text{and C}(14,0,-2)\text{are\hspace{0.33em}given.}\\ \mathrm{AB}=\sqrt{{(2+4)}^2+{(4-6)}^2+{(6-10)}^2}\\ =\sqrt{36+4+16}\\ =\sqrt{56}\\ =2\sqrt{14}\\ \mathrm{BC}=\sqrt{{(14-2)}^2+{(0-4)}^2+{(-2-6)}^2}\\ =\sqrt{144+16+64}\\ =\sqrt{224}\\ =4\sqrt{14}\\ \mathrm{AC}=\sqrt{{(14+4)}^2+{(0-6)}^2+{(-2-10)}^2}\\ =\sqrt{324+36+144}\\ =\sqrt{504}\\ =6\sqrt{14}\\ \mathrm{AB}+\mathrm{BC}=2\sqrt{14}+4\sqrt{14}\\ =6\sqrt{14}=\mathrm{AC}\\ \text{Thus,}{\mathrm{A}}_{\mathrm{t}}\text{B andC are collinear.}\end{array}$