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← JEE Main 2022 · 28 Jul · Shift 1
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JEE Main 2022 · 28 Jul · Shift 1, Mathematics question 3

Let the vectors a⃗=(1+t)i^+(1−t)j^+k^\vec a=(1+t)\hat i+(1-t)\hat j+\hat k, b⃗=(1−t)i^+(1+t)j^+2k^\vec b=(1-t)\hat i+(1+t)\hat j+2\hat k and c⃗=ti^−tj^+k^\vec c=t\hat i-t\hat j+\hat k, t∈Rt\in\mathbf R be such that for α,β,γ∈R\alpha,\beta,\gamma\in\mathbf R, αa⃗+βb⃗+γc⃗=0⃗⇒α=β=γ=0\alpha\vec a+\beta\vec b+\gamma\vec c=\vec0\Rightarrow\alpha=\beta=\gamma=0. Then, the set of all values of t is :

Choose one option
Answer and worked solution

Official answer: C

Independence needs the determinant to be nonzero. Adding column 2 to column 1 and expanding gives 6t6t.

So t∈R−{0}t\in\mathbf R-\{0\}. (C)

Topics:Scalar and vector triple productsDeterminants and properties
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