JEE Main 2022 (28 Jul, Shift 1) Maths Q3: Vector Algebra

Question

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}=\vec{0}\Rightarrow\alpha=\beta=\gamma=0

. Then, the set of all values of

t

is :

Official answer

Option 3

NTA final key (2022 Session 2).

Accepted Answer

Option 3: equal to $\mathbf{R}-\{0\}$

Answer

a non-empty finite set

Answer

equal to $\mathbf{N}$

Answer

equal to $\mathbf{R}-\{0\}$

Answer

equal to $\mathbf{R}$

JEE Main 28 July 2022, Shift 1, Maths Q3