JEE MAIN - Mathematics (2022 - 28th July Morning Shift - No. 3)
Let the vectors $$\vec{a}=(1+t) \hat{i}+(1-t) \hat{j}+\hat{k}, \vec{b}=(1-t) \hat{i}+(1+t) \hat{j}+2 \hat{k}$$ and $$\vec{c}=t \hat{i}-t \hat{j}+\hat{k}, t \in \mathbf{R}$$ be such that for $$\alpha, \beta, \gamma \in \mathbf{R}, \alpha \vec{a}+\beta \vec{b}+\gamma \vec{c}=\overrightarrow{0} \Rightarrow \alpha=\beta=\gamma=0$$. Then, the set of all values of $$t$$ is :
a non-empty finite set
equal to $$\mathbf{N}$$
equal to $$\mathbf{R}-\{0\}$$
equal to $$\mathbf{R}$$
Explanation
Clearly $$\overrightarrow a $$, $$\overrightarrow b $$, $$\overrightarrow c $$ are non-coplanar
$$\left| {\matrix{ {1 + t} & {1 - t} & 1 \cr {1 - t} & {1 + t} & 2 \cr t & { - t} & 1 \cr } } \right| \ne 0$$
$$ \Rightarrow (1 + t)(1 + t + 2t) - (1 - t)(1 - t - 2t) + 1({t^2} - t - t - {t^2}) \ne 0$$
$$ \Rightarrow (3{t^2} + 4t + 1) - (1 - t)(1 - 3t) - 2t \ne 0$$
$$ \Rightarrow (3{t^2} + 4t + 1) - (3{t^2} - 4t + 1) - 2t \ne 0$$
$$ \Rightarrow t \ne 0$$
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