Question

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

• A. a non-empty finite set
• B. equal to $N$
• C. equal to $R -\{0\}$
• D. equal to $R$

Answer 1

Answer: (C)

By its given condition
$: \vec{a}, \vec{b}, \vec{c}$ are linearly independent vectors
$\Rightarrow[\overline{ a } \overline{ b } \overline{ c }] \neq 0$...(i)
Now, $[\bar{a} \bar{b} \bar{c}]$
$\ =\begin{vmatrix}1+ t & 1- t & 1 \\ 1- t & 1+ t & 2 \\ t & - t & 1\end{vmatrix} $
$ C _2 \rightarrow C _1+ C _2 $
$ \begin{vmatrix}1+ t & 2 & 1 \\ 1- t & 2 & 2 \\ t & 0 & 1\end{vmatrix} $
$=2 \begin{vmatrix}1+ t & 1 & 1 \\ 1- t & 1 & 2 \\ t & 0 & 1\end{vmatrix} $
$ =2[(1+ t )-(1- t )+ t ] $
$ =2[3 t ]=6 t $
$ {[\overline{ a } \overline{ b } \overline{ c }] \neq 0 \Rightarrow t \neq 0}$