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Question : 4 of 38
Marks:
+1,
-0
Solution:
×=‌ and ‌×=⇒, and
are mutually perpendicular.
‌(×)×=‌⇒‌‌(⋅)−(⋅)=‌⇒‌‌(||2−1)−(⋅)= ‌⇒‌‌||2=1‌ and ‌⋅=0‌||||=|ω|⇒||=√6‌⋅=0⇒α+β−2γ=0‌‌ and ‌−tα+β+γ=0‌α−tβ+γ=0‌α+β−tγ=0(ii) - (i)
⇒‌‌α(1+t)=(t+1)β(iii) - (ii)
⇒‌‌β(1+t)=(1+t)γ⇒α(1+t)=β(1+t)=γ(1+t)either
t=−1 or
α=β=γ⇒√α2+α2+α2=√6⇒‌‌α=√2,−tα=−α−α⇒t=2Since
α=√3⇒‌‌t=−1⇒α+β+γ=0α+β−2γ=0⇒γ=0(Q)
⟶1α+β=0⇒(β+γ)=−α⇒(β+γ)2=α2=3,(R)⟶4⇒||2=1⟶(P)⟶2If
α=√3⇒γ2=0⟶(Q)⟶1If
α=√3⇒(β+γ)2=(√3)2=3,(R)⟶4If
α=√2,t+3=(2)+3=5, (S)
⟶5
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