Let P(3,2,6) be a point in space and Q be a point on the line →r=(→i−→j+2→k)+μ(−3→i+→j+5→k) Then the value of μ for which the vector PQ is parallel to x−4y+3z=1 is
A
14
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B
−14
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C
18
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D
−18
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Solution
The correct option is B14 Let O be the origin, as Q lies on the given line, OQ=→i−→j+2→k+μ(−3→i+→j+5→k)
Also OP=3→i+2→j+6→k. As PQ is parallel to the plane x−4y+3z=1. PQ⋅(→i−4→j+3→k)=0 ⇒(OQ−OP)⋅(→i−4→j+3→k)=0 ⇒(1+4+6)+μ(−3−4+15)−(3−8+18)=0 ⇒11+8μ−13=0 ⇒μ=14 Ans: A