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NCERT Class XI Mathematics - Introduction to Three Dimensional Geometry - Solutions

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Question : 5 of 20
Marks: +1, -0
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 = (4−2)2+(3−3)2+(1−5)2\sqrt{(4-2)^2+(3-3)^2+(1-5)^2} = 4+0+16\sqrt{4+0+16} = 20\sqrt{20} = 252\sqrt{5} units.
(ii) The distance PQ between the points P(–3, 7, 2) and Q(2, 4, –1) is
PQ =
(2−(−3))2+(4−7)2+(1−12)2\sqrt{(2-(-3))^2+(4-7)^2+(1-12)^2}
=
(2+3)2+(4−7)2+(−1−2)2\sqrt{(2+3)^2+(4-7)^2+(-1-2)^2}
= 25+9+9\sqrt{25+9+9} = 43\sqrt{43} units
(iii) The distance PQ between the points P(–1, 3, –4) and Q(1, –3, 4) is
PQ =
(1−(−1))2+(−3−3)2+(4−(−4))2\sqrt{(1-(-1))^2+(-3-3)^2+(4-(-4))^2}
= 4+36+64\sqrt{4+36+64} = 104\sqrt{104} = 2262\sqrt{26} units
(iv) The distance PQ between the points P(2, –1, 3) and Q(–2, 1, 3) is
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