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Equation of the perpendicular line from (3, –1, 11) to the line is
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a
x−31=y+1−6=z−114
b
x−32=y+15=z−117
c
x−31=y+111=z−113
d
x−31=y+16=z−114
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Correct option is A
Let P be the foot of the perpendicular from A (3, –1, 11) to the given line then P = (2r , 3r + 2, 4r + 3)D.r’s of A.P are (2r – 3, 3r + 3, 4r – 8)AP↔ is perpendicular to the given line⇒ 2(2r – 3) + 3(3r + 3) + 4(4r – 8)=0 ⇒ r = 1 P = (2, 5, 7)D.r’s of AP↔ are (1, –6, 4)
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The foot of the perpendicular from (a,b,c) on the line is the point where
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