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Question : 29 of 90
Marks:
+1,
-0
Solution:
‌=−+(‌ DR's of ‌L1)‌=+−(DR′s‌ of ‌L2) ×=|| =0+2+2 (DR's of Line perpendicular to
L1 and
L2 )
DR of
AB line
‌=(0,2,2)=(3+µ−λ,3+µ+λ,3−µ−λ) ‌‌=‌=‌Solving above equation we get
µ=−‌ and
λ=‌ ‌‌ point ‌A=(‌,‌,‌)‌B=(‌,‌,‌)‌‌ Point of ‌AB=(‌,2,6)=(α,β,γ)‌2(α+β+γ)=5+4+12=21
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