Question

If a vector $r$ of magnitude $3\sqrt{6}$ is directed along the bisector of the angle between the vectors $a=7i-4j-4k$ and $b=-2i-j+2k$, then $r$ is equal to

• A. $-i-7j-2k$
• B. $i-7j-2k$
• C. $i+7j-2k$
• D. $\pm (i-7j+2k)$

Answer 1

The correct option is D $\pm (i-7j+2k)$

The required vector $r=t(a+b),t$ is a scalar

$\Rightarrow r=t[\frac{1}{9}(7i-4j-4k)+\frac{1}{3}(-2i-j+2k)]$

$=\frac{t}{9}(i-7j+2k)$

Since $r=3\sqrt{6}\Rightarrow {\left|r\right|}^2=54$

$\Rightarrow \frac{t^2}{81}(1+49+4)=54\Rightarrow t^2=81\Rightarrow t=\pm 9$

Hence, $r=\pm (i-7j+2k)$