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Question : 14 of 90
Marks:
+1,
-0
Solution:
A vector in the direction of the required line can be obtained by cross product of
|| =−9−9+9 Required line,
=(5−4+3)+λ′(−9−9+9) =(5−4+3)+λ(+−) Now distance of
(0,2,−2)
P.V. of P≡(5+λ)+(λ−4) +(3−λ) =(5+λ)+(λ−6)+(5−λ) ⋅(+−)=0 5+λ+λ−6−5+λ=0 λ=2 ||=√49+16+9 ||=√74
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