QJEEM22Ma9d1b: JEE Main 2022 Physics - Class 11 JEEM22Ma9d1b Solution: Let the vectors a → = ( 1 + t ) i ^ + ( 1 − t ) j ^ + k ^ , b → …

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QJEEM22Ma9d1b: JEE Main 2022 Physics - Class 11 JEEM22Ma9d1b Solution: Let the vectors a → = ( 1 + t ) i ^ + ( 1 − t ) j ^ + k ^ , b → …

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 $\vec{c} = t \hat{i} - t \hat{j} + \hat{k}, t \in R$ be such that for $α, β, γ \in R, α \vec{a} + β \vec{b} + γ \vec{c} = \vec{0} \Rightarrow α = β = γ = 0$ . Then, the set of all values of $t$ is :

Multiple Choice Options

A a non-empty finite set
B equal to N
C equal to R−{0}
D equal to R

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Accepted Answer

Clearly $\vec{a}$ , $\vec{b}$ , $\vec{c}$ are non-coplanar

$| \begin{matrix} 1 + t & 1 - t & 1 \\ 1 - t & 1 + t & 2 \\ t & - t & 1 \end{matrix} | \neq 0$

$\Rightarrow (1 + t) (1 + t + 2 t) - (1 - t) (1 - t - 2 t) + 1 (t^{2} - t - t - t^{2}) \neq 0$

$\Rightarrow (3 t^{2} + 4 t + 1) - (1 - t) (1 - 3 t) - 2 t \neq 0$

$\Rightarrow (3 t^{2} + 4 t + 1) - (3 t^{2} - 4 t + 1) - 2 t \neq 0$

$\Rightarrow t \neq 0$

QJEEM22Ma9d1b: JEE Main 2022 Physics - Class 11 JEEM22Ma9d1b Solution: Let the vectors a → = ( 1 + t ) i ^ + ( 1 − t ) j ^ + k ^ , b → …

Multiple Choice Options

Step-by-Step Verified Solution

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