NCERT Class XI Mathematics - Introduction to Three Dimensional Geometry - Solutions
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Question : 18
Total: 20
Find the coordinates of a point on y-axis which are at a distance of 5 √ 2
Solution:
Let Q(0, y, 0) be any point on y-axis.
Then PQ =√ ( 0 − 3 ) 2 + ( y + 2 ) 2 + ( 0 − 5 ) 2
=√ 9 + y 2 + 4 + 4 y + 25 √ y 2 + 4 y + 38
But√ y 2 + 4 y + 38 5 √ 2
on squaring both sides, we gety 2 y 2
⇒ (y – 2)(y + 6) = 0 ⇒ y = 2, –6
Thus coordinates of point Q are (0, 2, 0) and (0, –6, 0).
Then PQ =
=
But
on squaring both sides, we get
⇒ (y – 2)(y + 6) = 0 ⇒ y = 2, –6
Thus coordinates of point Q are (0, 2, 0) and (0, –6, 0).
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