Question

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

Answer 1

Let $x\hat{i}+y\hat{j}+z\hat{k}$ be the unit vector along $\overrightarrow{c}$

Since, $-\hat{i}+\hat{j}-\hat{k}$ bisect the angle between $\overrightarrow{c}$ & $3\hat{i}+4\hat{j}$

$\therefore \lambda (-\hat{i}+\hat{j}-\hat{k})=x\hat{i}+y\hat{j}+z\hat{k}+\frac{3\hat{i}+4\hat{j}}{5}$

$\Rightarrow \frac{x+3}{5}=-\lambda ,\frac{y+4}{5}=\lambda ,z=-\lambda$

Now, $x^2+y^2+z^2=1$

$\Rightarrow {(-\lambda -\frac{3}{5})}^2+{(\lambda -\frac{4}{5})}^2+{(-\lambda )}^2=1$

$\therefore \lambda =0$, $\lambda =\frac{2}{15}$

But $\lambda \ne 0$, because $\lambda =0$ implies that the given vectors are parallel.

$\therefore \lambda =\frac{2}{15}$

$x=\frac{-11}{5},y=\frac{-10}{15},z=\frac{-2}{15}$

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