NCERT Solutions
Class 11 Maths
Introduction to Three Dimensional Geometry
Ex.12.2 Q.1
Find the distance between the following pairs of points:
(1) (2, 3, 5) and (4, 3, 1)
(2) (–3, 7, 2) and (2, 4, –1)
(3) (–1, 3, –4) and (1, –3, 4)
(4) (2, –1, 3) and (–2, 1, 3)
The distance between points P (x1, y1, z1) and P (x2, y2, z2) is given by
PQ = √ {(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2}
(1) Distance between points (2, 3, 5) and (4, 3, 1)
= √ {(4 – 2)2 + (3 – 3)2 + (1 – 5)2}
= √ {22 + 0 + (-4)2}
= √ (4 + 16)
= √20
= 2√5
(2) Distance between points (–3, 7, 2) and (2, 4, –1)
= √ {(2 + 3)2 + (4 – 7)2 + (-1 – 2)2}
= √ {52 + (-3)2 + (-3)2}
= √ (25 + 9 + 9)
= √43
(3) Distance between points (–1, 3, –4) and (1, –3, 4)
= √ {(1 + 1)2 + (-3 – 3)2 + (4 + 4)2}
= √ {22 + (-6)2 + 82}
= √ (4 + 36 + 64)
= √104
= 2√26
(4) Distance between points (2, –1, 3) and (–2, 1, 3)
= √ {(-2 – 2)2 + (1 + 1)2 + (3 – 3)2}
= √ {(-4)2 + 22 + 0}
= √ (16 + 4)
= √20
= 2√5