Question

A vector of magnitude $2$ along a bisector of the angle between the two vectors $2i-2j+k$ and $i+2j-2k$ is

• A. $\frac{2}{\sqrt{10}}(3i-k)$
• B. $\frac{1}{\sqrt{26}}(i-4j+3k)$
• C. $\frac{2}{\sqrt{26}}(i-4j+3k)$
• D. None of these

Answer 1

Answer: (A), (C)

The correct options are
A $\frac{2}{\sqrt{10}}(3i-k)$
C $\frac{2}{\sqrt{26}}(i-4j+3k)$

Let $a=2i-2j+k$# and $b=i+2j-2k$

$\therefore \left|a\right|=3$ and $\left|b\right|=3$.

$\therefore$ a vector along bisectors

$=\frac{a}{\left|a\right|}\pm \frac{b}{\left|b\right|}=\frac{2i-2j+k}{3}\pm \frac{i+2j-2k}{3}$

$=i-\frac{1}{3}k,\frac{1}{3}i-\frac{4}{3}j+k.$

$\therefore$ The required vector

$=2.\frac{i-\frac{1}{3}k}{\sqrt{1^2+{(-\frac{1}{3})}^2}},2.\frac{\frac{1}{3}i-\frac{4}{3}j+k}{\sqrt{{\left(\frac{1}{3}\right)}^2+{(-\frac{4}{3})}^2+1^2}}$

$=\frac{2}{\sqrt{10}}(3i-k),\frac{2}{\sqrt{26}}(i-4j+3k)$