# Chapter 12 Introduction to Three Dimensional Geometry Ex – 12.2

Question 1.

Find the distance between the following pairs of points:

(i) (2, 3, 5) and (4, 3, 1)
(ii) (-3, 7, 2) and (2, 4, -1)
(iii) (-1, 3, -4) and (1, -3, 4)
(iv) (2, -1, 3) and (-2, 1, 3)

Solution:

(i) The distance PQ between the points P(2, 3, 5) and Q(4, 3, 1) is
$PQ=\sqrt { \left( 4-2 \right) ^{ 2 }+\left( 3-3 \right) ^{ 2 }\left( 1-5 \right) ^{ 2 } }$
$\sqrt { 4+0+16= } \sqrt { 20 } =2\sqrt { 5 } units.$

(ii) The distance PQ between the points P(-3, 7, 2) and Q(2, 4, -1) is
$PQ=\sqrt { \left[ 2-\left( -3 \right) \right] ^{ 2 }+\left( 4-7 \right) ^{ 2 }\left( -1-2 \right) ^{ 2 } }$
$=\sqrt { \left( 2+3 \right) ^{ 2 }+\left( 4-7 \right) ^{ 2 }+\left( -1-2 \right) ^{ 2 } }$
$=\sqrt { 25+9+9 } =\sqrt { 43 } units$

(iii) The distance PQ between the points P(-1, 3, -4) and Q(1, -3, 4) is
$PQ=\sqrt { \left[ 1-\left( -1 \right) \right] ^{ 2 }+\left( -3-3 \right) ^{ 2 }\left[ 4-\left( -4 \right) \right] ^{ 2 } }$
$=\sqrt { 4+36+64 } =\sqrt { 104 } =2\sqrt { 26 } units$

(iv) The distance PQ between the points P(2, -1, 3) and Q(-2, 1, 3) is
$PQ=\sqrt { \left( -2-2 \right) ^{ 2 }+\left[ 1-\left( -1 \right) \right] ^{ 2 }+\left( 3-3 \right) ^{ 2 } }$
$=\sqrt { 16+4+0 } =\sqrt { 20 } =2\sqrt { 5 } units$

Question 2.

Show that the points (-2, 3, 5), (1, 2, 3) and (7, 0, -1) are collinear.

Solution:

Let A(-2, 3, 5), B(1, 2, 3) and C(7, 0, -1) be three given points.

Now AC = AB + BC

Thus, points A, B and C are collinear.

Question 3.

Verify the following:

(i) (0, 7, -10), (1, 6, -6) and (4, 9, -6) are the vertices of an isosceles triangle.

(ii) (0, 7, 10), (-1, 6, 6) and (-4, 9, 6) are the vertices of a right angled triangle.

(iii) (-1, 2, 1), (1, -2, 5), (4, -7,8) and (2, -3,4) are the vertices of a parallelogram.

Solution:

(i) Let A(0, 7, -10), B(l, 6, -6) and C(4, 9, -6) be three vertices of triangle ABC. Then

Now, AB = BC

Thus, ABC is an isosceles triangle.

(ii) Let A(0, 7,10), B(-l, 6, 6) and C(-A, 9, 6) be three vertices of triangle ABC. Then

Now, AC2 = AB2 + BC2
Thus, ABC is a right angled triangle.

(iii) Let A(-1, 2, 1), B(1, -2, 5) and C(4, -7, 8) and D(2, -3,4) be four vertices of quadrilateral ABCD. Then

Now AB = CD, BC = AD and AC ≠ BD
Thus A, B, C and D are vertices of a parallelogram ABCD.

Question 4.

Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, -1).

Solution:

Let A(x, y, z) be any point which is equidistant from points B(1, 2, 3) and C(3, 2, -1).

Question 5.

Find the equation of the set of points P, the sum of whose distances from A(4, 0, 0) and B(-4,0,0) is equal to 10.

Solution:

Let P(x, y, z) be any point.