Question

Let $\overrightarrow{a}=\hat{i}-\hat{k},\overrightarrow{b}=x\hat{i}+\hat{j}+(1-x)\hat{k}$ and $\overrightarrow{c}=y\hat{i}+x\hat{j}+(1+x-y)\hat{k},$ then $\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]$ depends on

• A. Only $x$
• B. only $y$
• C. neither $x$ nor $y$
• D. both $x$ and $y$.

Answer 1

The correct option is C neither $x$ nor $y$

$\left[\overline{a}\overline{b}\overline{c}\right]=\overline{a}\cdot (\overline{b}\times \overline{c})=\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|$ $\left|\begin{array}{ccc}1& 0& -1\\ x& 1& 1-x\\ y& x& 1+x-y\end{array}\right|$
$C_3\to C_3+C_1$
$\left|\begin{array}{ccc}1& 0& 0\\ x& 1& 1\\ y& x& 1+x\end{array}\right|=1\left(1\right)=1$
which is independent of x and y.