Question

If$\mathit{a}\mathbf{.}\mathit{i}=4$, then $(\mathit{a}\mathbf{.}\mathit{j})\times (2\mathit{j}-3\mathit{k})=?$

• A. $12$
• B. $2$
• C. $0$
• D. $-12$

Answer 1

The correct option is D

$-12$

Explantion for the correct option:

Find out: $\mathbf{(}\mathit{a}\mathbf{.}\mathit{j}\mathbf{)}\mathbf{\times }\mathbf{(}\mathbf{2}\mathit{j}\mathbf{-}\mathbf{3}\mathit{k}\mathbf{)}\mathbf{=}\mathbf{?}$

Given

$a.i=4$

$\begin{array}{rcl}& & (a.j)\times (2j-3k)\\ & =& a.(j\times \left\{2j-3k\right\})\\ & =& a.(j\times 2j-3j\times k)\\ & =& a.(0-3i)\left\{j\times j=0\&j\times k=i\right\}\\ & =& a.\{-3(i\left)\right\}\\ & =& -3(a.i)\\ & =& -3\left(4\right)\mathbf{\because }\mathbf{a}\mathbf{.}\mathbf{i}\mathbf{}\mathbf{=}\mathbf{4}\\ & =& -12\end{array}$

Hence, the option (D) is correct.