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Question : 83 of 90
Marks:
+1,
-0
Solution:
P1=⋅(3‌−5‌+‌)=7P2=⋅(λ‌+‌−3‌)=9‌θ=sin‌−1(‌)‌⇒sin‌θ=‌∴cos‌θ=‌cos‌θ=‌⋅=‌| (3i−5j+K)(λi+j−3K) |
| √35⋅√λ2+10 |
‌=|‌|Square
⇒‌=‌⇒19λ2−120λ+125=0 ‌⇒19λ2−95λ−25λ+125=0‌⇒x=5,‌
Perpendicular distance of point
(38λ1,10λ2,2)≡(50,50,2) from plane
P1=‌| |3×50−5×50+2−7| |
| √35 |
=‌‌ Square ‌=‌=315
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