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Question : 14 of 90
Marks:
+1,
-0
Solution:
Let
=+2+3=2+4+5=2+3+λ,=+4+5∴×=(15−4λ)−(10−λ)+52−1=+2+2∴ Shortest distance
=|‌| (15−4λ)−2(10−λ)+10 |
| √(15−4λ)2+(10−λ)2+25 |
|=‌⇒3(5−2λ)2=(15−4λ)2+(10−λ)2+25⇒5λ2−80λ+275=0 ∴ Sum of values of
λ=‌=16
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