Question

If the vector $-\hat{i}+\hat{j}-\hat{k}$ bisect the angle between the vector $\overrightarrow{c}$ and the vector $3\hat{i}+4\hat{j}$ then the unit vector inthe direction of $\overrightarrow{c}$ is

• A. $\frac{1}{15}(11\hat{i}+10\hat{j}+2\hat{k})$
• B. $-\frac{1}{15}(11\hat{i}-10\hat{j}+2\hat{k})$
• C. $-\frac{1}{15}(11\hat{i}+10\hat{j}-2\hat{k})$
• D. $-\frac{1}{15}(11\hat{i}+10\hat{j}+2\hat{k})$

Answer 1

The correct option is D $-\frac{1}{15}(11\hat{i}+10\hat{j}+2\hat{k})$

Let $\hat{c}=x\hat{l}+y\hat{I}+Z\hat{k}$ where $x^2+y^2+z^2=1$

And unitvector along $=3\hat{I}+4\hat{I}=\frac{3\hat{I}+4\hat{I}}{\sqrt{5}}$

The biesectors of these two is given by

$r=t(\hat{a}+\hat{b})$

$\Rightarrow =t(x\hat{l}+y\hat{I}+z\hat{k}+\frac{3\hat{l}+4\hat{I}}{5})$

$\Rightarrow r=\frac{t}{5}(5x+3)\hat{l}(5y+4)\hat{I}+5Z\hat{k}$

But the bisector is given by $-\hat{l}+\hat{I}-\hat{k}$

Comparing$\left(2\right)$ and $\left(3\right),$ we get

$\frac{t}{5}(5x+3)=-1\Rightarrow x=\frac{-5+3t}{5t}$

$\frac{t}{5}(5y+4)=1\Rightarrow y=\frac{5-4t}{5t}$

$\frac{t}{5}\left(5z\right)=-1\Rightarrow z=\frac{-1}{t}$

But all the values is equation $\left(1\right),$we have

${\left(\frac{-5+3t}{5t}\right)}^2+{\left(\frac{5-4t}{5t}\right)}^2+{\left(\frac{-1}{t}\right)}^2=1$

$\Rightarrow \frac{25+9t^2+30t+25+16t^2-40t+25}{25t^2}=1$

$\Rightarrow 25t^2-10t+75=25t^2$

$\Rightarrow t=7.5$

Thus,

$x=-\frac{5+3t}{5t}\Rightarrow x=\frac{-11}{15}$

$y=\frac{5-4t}{5t}\Rightarrow y=\frac{-10}{15}$

$Z=\frac{-1}{t}\Rightarrow \frac{-2}{15}$

$\hat{e}=\frac{-11}{15}\hat{l}-\frac{10\hat{I}}{15}-\frac{2\hat{k}}{15}$