Question

If x, y, z are non-zero real numbers, $\overline{a}=x\overline{i}+2\overline{j},\overline{b}=y\overline{j}+3\overline{k}$ and $\overline{c}=x\overline{i}+y\overline{j}+z\overline{k}$ are such that $\overline{a}\times \overline{b}=z\overline{i}-3\overline{j}+\overline{k}$ then $\left[\overline{a}\overline{b}\overline{c}\right]=$

• A. 3
• B. 10
• C. 9
• D. 6

Answer 1

The correct option is C 9

$\overline{a}\times \overline{b}=\left|\begin{array}{ccc}i& j& k\\ x& 2& 0\\ o& y& 3\end{array}\right|=6i-3x\hat{j}+xy\hat{k}$

and $\overline{a}\times \overline{b}=z\hat{i}-3\hat{j}+\hat{k}$ .....(given)

So, $z=6,x=1,y=1$

$\overline{c}=\hat{i}+\hat{j}+6\hat{k}$

$\therefore \left[abc\right]=(\overline{a}\times \overline{b})\cdot c=(6\hat{i}-3\hat{j}+\hat{k})(\hat{i}+\hat{j}+6\hat{k})$

$=6-3+6$

$=9$