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Question : 73 of 90
Marks:
+1,
-0
Solution:
Let
1=λNow
2=−1=(+2−4)−λ(4+3+5)=(1−4λ)+(2−3λ)−(5λ+4)2⋅=0⇒4(1−4λ)+3(2−3λ)−5(5λ+4)=0⇒4−16α+6−9λ−25λ−20=0⇒50λ=−10⇒λ=‌2=(1+‌)‌+(2+‌)‌−(−1+4)‌2=‌+‌−352=9‌+13‌−15‌52⋅(++)=9+13−15=7
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