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Question : 9 of 100
Marks:
+1,
-0
Solution:
a=⟨α,β,3⟩ b=⟨−β,−α,−1⟩ c=⟨1,−2,−1⟩ a⋅b=1 −αβ−αβ−3=1 αβ=−2 b⋅c=−3 −β+2α+1=−3 β−2α=4 ⇒β−2()=4 ⇒β2+4=4β⇒β2−4β+4=0 ⇒(β−2)2=0⇒β=2 αβ=−2⇒α⋅2=−2 ⇒α=−1 Hence,
[(a×b)⋅c] a=⟨−1,2,3⟩ b=⟨−2,1,−1⟩ c=⟨1,−2,−1⟩ =−5+14−3=6 ∴[(a×b)⋅c]=×6=2
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