Question

If the vector $-\hat{i}+\hat{j}-\hat{k}$ bisects the angles between the vector $\hat{c}$ and the vector $3\hat{i}+4\hat{j}$, then the unit vector in the direction of $\hat{c}$

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

consider the problem

Let unit vector

$\hat{c}=x\hat{i}+y\hat{j}+z\hat{k}$

therefore,

$x^2+y^2+z^2=1$ ---- $\left(1\right)$ (since $\hat{c}$ is unit vector)

And unit vector along $3\hat{i}+4\hat{j}$

$\begin{array}{c}3\hat{i}+4\hat{j}=\frac{3\hat{i}+4\hat{j}}{\sqrt{3^2+4^2}}\\ =\frac{3\hat{i}+4\hat{j}}{\sqrt{9+16}}\\ =\frac{3\hat{i}+4\hat{j}}{\sqrt{25}}\\ =\frac{3\hat{i}+4\hat{j}}{5}\end{array}$

And bisectors of them two is

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

$r=t(x\hat{i}+y\hat{j}+z\hat{k}+\frac{3\hat{i}+4\hat{j}}{5})$

$r=\frac{t}{5}\left(\left(5x+3\right)\hat{i}+\left(5y+4\right)\hat{j}+5z\hat{k}\right)$ ----- $\left(2\right)$

But the bisector is given by $-\hat{i}-\hat{k}-\hat{z}$ ---- $\left(3\right)$

On comparing $\left(2\right)$ and $\left(3\right)$.

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

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

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

Putting the values in first equation

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

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

$t=7.5$

Thus,

$x=-\frac{5+3t}{5t}\Rightarrow =-\frac{5+3\times 7.5}{5\times 7.5}=-\frac{11}{15}$

$y=\frac{5-4t}{5t}\Rightarrow y=\frac{5-4\times 7.5}{5\times 7.5}$

$z=-\frac{1}{t}=-\frac{1}{7.5}=-\frac{2}{15}$

Hence unit vector

$\begin{array}{c}\hat{c}=-\frac{11}{15}\hat{i}-\frac{10}{15}\hat{j}-\frac{2}{15}\hat{k}\\ \hat{c}=-\frac{1}{15}\left(11\hat{i}+10\hat{j}+2\hat{k}\right)\end{array}$

Hence, Option $C$ is the correct answer.