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

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Question : 18 of 20
Marks: +1, -0
Find the coordinates of a point on y-axis which are at a distance of 5 2\sqrt{2} from the point P(3, –2, 5).
Solution:  
Let Q(0, y, 0) be any point on y-axis.
Then PQ = (0−3)2+(y+2)2+(0−5)2\sqrt{(0-3)^2+(y+2)^2+(0-5)^2}
= 9+y2+4+4y+25\sqrt{9+y^2+4+4y+25} = y2+4y+38\sqrt{y^2+4y+38}
But y2+4y+38\sqrt{y^2+4y+38} = 525\sqrt{2}
on squaring both sides, we get y2y^2 + 4y + 38 = 50 ⇒ y2y^2 + 4y – 12 = 0
⇒ (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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