Question

If $x,y\mathrm{and}z$ are non-zero real numbers and $a=xi+2j,b=yj+3k\mathrm{and}c=xi+yj+zk$ are such that $a\times b=zi-3j+k$, then $[abc]$ is equal to:

• A. $3$
• B. $10$
• C. $9$
• D. $6$

Answer 1

The correct option is C

$9$

Explanation for the correct option.

Step 1: Find the value of $x,y\mathrm{and}z$.

We have $a=xi+2jandb=yj+3k$,so let us present it as

$\left[\begin{array}{ccc}i& j& k\\ x& 2& 0\\ 0& y& 3\end{array}\right]$

Now, $a\times b=6i-3xj+xyk$

It is given that $a\times b=zi-3j+k$

$\Rightarrow 6i-3xj+xyk=zi-3j+k$

$\Rightarrow z=6,x=1,y=1$

So, $c=i+j+6k$

Step 2: Find the value of $[abc]$.

$\begin{array}{rcl}[abc]& =& \left[a\times b\right]\times c\\ & =& \left(6i-3xj+xyk\right)\left(i+j+6k\right)\\ & =& 6-3+6\\ & =& 9\end{array}$

Hence, option C is correct.