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Question : 5 of 90
Marks:
+1,
-0
Solution:
‌+=‌⟨4,3,δ⟩⋅+⟨−2,1,3⟩=‌⇒=⟨2,4,3+δ⟩ ‌×=||‌=(9−δ)−(2δ+12)+k(10)‌|AB×AC|2=(9−δ)2+(2δ+12)2+(10)2‌=5δ2+30δ+325‌‌ Area of ‌∆ABC=5√6‌⇒‌|×|=5√6‌⇒|×|2=600‌⇒5δ2+30δ−275=0‌⇒S2+6δ−55=0‌⇒(δ+11)(δ−5)=0‌δ=5‌=<2,3,8>‌⋅=<2,4,8>.<4,3,5>‌=8+12+40=60
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