Question

If $a\overline{i}+b\overline{j}+c\overline{k}$ is a bisector of angle between $\overline{i}-\overline{j}$ and $\overline{j}-\overline{k}$, then

• A. $a=b$
• B. $a=c$
• C. $a+c=0=b$
• D. $b=c$

Answer 1

The correct option is C $a+c=0=b$

$a\overline{i}+\overline{b}j+c\overline{k}=t(\frac{\overline{i}-\overline{j}}{\sqrt{2}}+\frac{\overline{j}-\overline{k}}{\sqrt{2}})$

$\Rightarrow a\overline{i}+\overline{b}j+c\overline{k}=\frac{t}{\sqrt{2}}(\overline{i}-\overline{k})$
$\Rightarrow a=\frac{t}{\sqrt{2}},b=0,c=-\frac{t}{\sqrt{2}}$
$\Rightarrow a=-c,b=0\Rightarrow a+c=0,b=0$