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  • Let the vectors <img alt="\mathrm{\overrightarrow{\mathrm{a}}=(1+\mathrm{t}) \hat{i}+(1-\mathrm{t}) \hat{j}+\hat{k}, \overrightarrow{\mathrm{b}}=(1-\mathrm{t}) \hat{i}+(1+\mathrm{t}) \hat{j}+2

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

Option: 1

\mathrm{\text { a non-empty finite set }}


Option: 2

\mathrm{\text { equal to } \mathbf{N}}


Option: 3

\mathrm{\text { equal to } \mathbf{R}-\{0\}}


Option: 4

\mathrm{\text { equal to } \mathbf{R}}


Answers (1)

best_answer

By its given condition
\vec{\mathrm{a}} , \vec{\mathrm{b}} , \vec{\mathrm{c}} are lineardy independent vectors
\Rightarrow \left [ \vec{\mathrm{a}} \; \vec{\mathrm{b}} \; \vec{\mathrm{c}} \right ]\neq 0 \cdots \cdot(1)

\text{Now} \left [ \vec{\mathrm{a}} \; \vec{\mathrm{b}} \; \vec{\mathrm{c}} \right ]

=\left|\begin{array}{ccc} \mathrm{1+t} & \mathrm{1-t} & \mathrm{1} \\ \mathrm{ 1-t} & \mathrm{1+t} & \mathrm{ 2} \\ \mathrm{ t} &\mathrm{ -t} & \mathrm{1}\end{array}\right|

\mathrm{c_{2} \rightarrow c_{1}+c_{2} }

\begin{aligned} & \left|\begin{array}{ccc} \mathrm{1+t} & \mathrm{2} & \mathrm{1} \\ \mathrm{1-t} &\mathrm{ 2} & \mathrm{2} \\ \mathrm{t} & \mathrm{0} & \mathrm{1}\end{array}\right|=2\left|\begin{array}{lll}\mathrm{1+t} & \mathrm{1} & \mathrm{1} \\ \mathrm{1-t} & \mathrm{1} & \mathrm{2} \\ \mathrm{t} & \mathrm{0} & \mathrm{1}\end{array}\right| \\ & \mathrm{=2[(1+t)-(1-t)+t]}\\ & \mathrm{=2[3 t]=6 t}\\ &\mathrm{[\vec{a} \;\vec{b} \; \vec{c}] \neq 0 \Rightarrow t \neq 0} \end{aligned}

Posted by

Divya Prakash Singh

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