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Question : 8 of 90
Marks:
+1,
-0
Solution:
‌r1=(λ+4+3)+α(2+3+4))‌b=2+3+4‌r2=(2+4+7)+β(2+3+4)}‌a1+λ+4+3‌‌ Shortest dist. ‌=‌=‌‌‌| |(2+3+4)×((2−λ)+4)| |
| √29 |
=‌‌|−8−3(2−λ)+12+4(2−λ)|=13‌|12−4λ+(3λ−6)|=13‌144+16λ2+(3λ−6)2=169‌16λ2+(3λ−6)2=25=λ⇒=1‌
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