If $\vec{a}=\hat{i}-\hat{k}, \vec{b}=x \hat{i}+\hat{j}+(1-x) \hat{k} \quad$ and $\quad \vec{c}=y \hat{i}+x…
If $\vec{a}=\hat{i}-\hat{k}, \vec{b}=x \hat{i}+\hat{j}+(1-x) \hat{k} \quad$ and $\quad \vec{c}=y \hat{i}+x \hat{j}+(1+x-y) \hat{k}$ then $[\vec{a} \vec{b} \vec{c}]$ depends on
neither $x$ nor $y$
only $x$
only $y$
both $x$ and $y$
Solution
$\begin{aligned} & {[\vec{a} \vec{b} \vec{c}]=\left|\begin{array}{ccc}1 & 0 & -1 \\ x & 1 & 1-x \\ y & x & 1+x-y\end{array}\right|=\left|\begin{array}{ccc}0 & 0 & -1 \\ 1 & 1 & 1-x \\ 1+x & x & 1+x-y\end{array}\right|} \\ & =-1(x-1-x) \\ & =1 \text { which is independent of both } x \text { and } y\end{aligned}$