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
$\Rightarrow [\overline{a}\overline{b}\overline{c}]\ne 0$...(i)
Now, $[\bar{a}\bar{b}\bar{c}]$
$=\left|\begin{array}{ccc}1+t& 1-t& 1\\ 1-t& 1+t& 2\\ t& -t& 1\end{array}\right|$
$C_2\to C_1+C_2$
$\left|\begin{array}{ccc}1+t& 2& 1\\ 1-t& 2& 2\\ t& 0& 1\end{array}\right|$
$=2\left|\begin{array}{ccc}1+t& 1& 1\\ 1-t& 1& 2\\ t& 0& 1\end{array}\right|$
$=2[(1+t)-(1-t)+t]$
$=2[3t]=6t$
$[\overline{a}\overline{b}\overline{c}]\ne 0\Rightarrow t\ne 0$