Let the vectors a=(1+t) i+(1-t) j+k, b=(1-t) i+(1+t) j+2 k and c=t i-t j+k, t R be such that for α, β, γ R, α a+β b+γ c=0 ⇒ α=β=γ=0. Then, the set of all values of t is:

Vector Algebra: JEE Main 2022 Mathematics question

JEE Main 2022 · 28 Jul, Morning shift · Q5696

JEE Main 202228 Jul, Morning shiftMathematicsMediumVector Algebra184× since 2002

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 :

JEE Main 2022 · 28 Jul, Morning shift · Q5696

JEE Main 202228 Jul, Morning shiftMathematicsMediumVector Algebra184× since 2002

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 :