Let $\vec{a}=\hat{i}-\hat{k}, \vec{b}=x \hat{i}+\hat{j}+(1-x) \hat{k}$ and $\vec{c}=y \hat{i}+x…
Let $\vec{a}=\hat{i}-\hat{k}, \vec{b}=x \hat{i}+\hat{j}+(1-x) \hat{k}$ and $\vec{c}=y \hat{i}+x \hat{j}+(1+x-y) \hat{k}$. Then $[\vec{a}, \vec{b}, \vec{c}]$ depends on
only $y$
only $\mathrm{x}$
both $x$ and $y$
neither $x$ nor $y$
Solution
$\vec{a}=\hat{i}-\hat{k}, \vec{b}=x \hat{i}+\hat{j}+(1-x) \hat{k}$ and $\vec{c}=y \hat{i}+x \hat{j}+(1+x-y) \hat{k}$
$[\vec{a} \vec{b} \vec{c}]=\vec{a} \cdot(\vec{b} \times \vec{c})$
$
\begin{aligned}
& \vec{b} \times \vec{c}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
x & 1 & 1-x \\
y & x & 1+x-y
\end{array}\right|=\hat{i}\left(1+x-x-x^2\right)-\hat{j}\left(x+x^2-x y-y+x y\right)+\hat{k}\left(x^2-y\right) \\
& \vec{a} .(\vec{b} \times \vec{c})=1
\end{aligned}
$
which does not depend on $\mathrm{x}$ and $\mathrm{y}$