Let the vectors $\vec{a}=(1+t) \hat{i}+(1-t) \hat{j}+\hat{k}$, $\overrightarrow{\mathrm{b}}=(1-\mathrm{t}) \hat{\mathrm{i}}+(1+\mathrm{t}) \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{c}}=\mathrm{t} \hat{\mathrm{i}}-\mathrm{t} \hat{\mathrm{j}}+\hat{\mathrm{k}}, \quad \mathrm{t} \in \mathrm{R}$ be such that for $\alpha, \beta, \gamma \in \mathrm{R}, \alpha \overrightarrow{\mathrm{a}}+\beta \overrightarrow{\mathrm{b}}+\gamma \overrightarrow{\mathrm{c}}=\overrightarrow{0}$ $\Rightarrow \alpha=\beta=\gamma=0$. Then, the set of all values of t is :

(C) equal to $\mathrm{R}-\{0\}$

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Vector Algebra

Let the vectors

a = (1 + t) \hat{i} + (1 - t) \hat{j} + \hat{k}

b = (1 - t) \hat{i} + (1 + t) \hat{j} + 2 \hat{k}

and

c = t \hat{i} - t \hat{j} + \hat{k}, t \in R

be such that for

α, β, γ \in R, α a + β b + γ c = 0

\Rightarrow α = β = γ = 0

. Then, the set of all values of t is :

Answer: (C)

The given condition means

a, b, c

are linearly independent, which requires

[a b c] \neq = 0

. Computing the determinant:

1 + t 1 - t t 1 - t 1 + t - t 121

. Using column operations and simplification, this determinant equals

6 t

. For linear independence,

6 t \neq = 0

, so

t \neq = 0

. Therefore, the set of all values is

R - {0}

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Single correct option