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← IIT-JEE 2010 · Paper 2
MathematicsSingle correct+5 / -2Question 23 of 57

IIT-JEE 2010 · Paper 2, Mathematics question 23

If the distance of the point P(1,−2,1)P(1,-2,1) from the plane x+2y−2z=αx+2y-2z=\alpha, where α>0\alpha>0, is 5, then the foot of the perpendicular from PP to the plane is

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Answer and worked solution

Official answer: A

Find α. ∣1−4−2−α∣3=5\dfrac{|1-4-2-\alpha|}3=5 gives ∣5+α∣=15|5+\alpha|=15, so α=10\alpha=10 (positive).

Foot of the perpendicular. Move from P along the normal (1,2,−2)(1,2,-2):
t=−1−4−2−109=53,P+t(1,2,−2)=(83,43,−73).t=-\frac{1-4-2-10}{9}=\frac53,\qquad P+t(1,2,-2)=\left(\frac83,\frac43,-\frac73\right).
(Check: 83+83+143=10\frac83+\frac83+\frac{14}3=10.)

(A)

Topics:Angles and distancesEquation of a plane
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