Three Dimensional Geometry Question 282
Question: The length and foot of the perpendicular from the point (7, 14, 5) to the plane $ 2x+4y-z=2, $ are
[AISSE 1987]
Options:
A) $ \sqrt{21},(1,,2,,8) $
B) $ 3\sqrt{21},(3,,2,,8) $
C) $ 21\sqrt{3},(1,,2,,8) $
D) $ 3\sqrt{21},(1,,2,,8) $
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Answer:
Correct Answer: D
Solution:
Let M be the foot of perpendicular from (7, 14, 5) to the given plane, then PM is normal to the plane. So, its d.r.’s are 2, 4, ?1. Since PM passes through $ P(7,14,5) $ and has d.r.’s 2, 4, ?1. Therefore, its equation is $ \frac{x-7}{2}=\frac{y-14}{4}=\frac{z-5}{-1}=r $ , (Say)
Þ $ x=2r+7 $ , $ y=4r+14 $ , $ z=-r+5 $ Let co-ordinates of M be $ (2r+7,,4r+14,,-r+5) $ Since M lies on the plane $ 2x+4y-z=2 $ , therefore $ 2(2r+7)+4(4r+14)-(-r+5)=2 $
Þ $ r=-3 $ So, co-ordinates of foot of perpependicular are $ M(1,,2,8) $ Now, PM = Length of perpendicular from P = $ \sqrt{{{(1-7)}^{2}}+{{(2-14)}^{2}}+{{(8-5)}^{2}}}=3\sqrt{21}. $
BETA
