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Question : 61 of 160
Marks:
+1,
-0
Solution:
Let
A=(9,13,15)
B=(12,21,10) P=(5,7,3) and
Q=(x,y,z) Dr's
′S of
=(9−12,13−21,15−10) =(−3,−8,5)=(3,8,−5) Equation of line
overleftrightarrow‌AB is
‌=‌=‌=λ ‌=λ,‌=λ,‌=λ x=3λ+9,y=8λ+13,z=−5λ+15 ∴Q=(3λ+9,8λ+13,−5λ+15) Dr's of
PQ=(3λ+9−5,8λ+13−7,−5λ+15−3) Dr's of
PQ=(3λ+4,8λ+6,−5λ+12) Since,
overleftrightarrow‌AB‌⟂‌P‌Q ∴‌‌a1a2+b1b2+c1c2=0 3(3λ+4)+8(8λ+6)−5(−5λ+12)=0 9λ+12+64λ+48+25λ−60=0⇒λ=0 Put,
λ=0 in
Q ∴‌‌Q=(9,13,15) ∴ Foot of Perpendicular
Q=(9,13,15)
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