Question

If the vector $-\hat{i}+\hat{j}-\hat{k}$ bisects the angle between the vector $\overrightarrow{c}$ and the vector $3\hat{i}+4\hat{j}$, then the unit vector in the 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 $x\hat{i}+y\hat{j}+z\hat{k}$ be the unit vector along $\overrightarrow{c}$. Since $-\hat{i}+\hat{j}-\hat{k}$ bisects the angle between $\overrightarrow{c}$ and $3\hat{i}+\hat{j}$.
Therefore,
$\lambda (-\hat{i}+\hat{j}-\hat{k})=(x\hat{i}+y\hat{j}+z\hat{k})+\frac{3\hat{i}+4\hat{j}}{5}$
$\Rightarrow x+\frac{3}{5}=-\lambda ,y+\frac{4}{5}=\lambda$ and $z=-\lambda$
Now, $x^2+y^2+z^2=1$ [$\because x\hat{i}+y\hat{j}+z\hat{k}$ is a unit vector]
But $\lambda \ne 0$. Because $\lambda =0$ implies that the given vectors are parallel.
$\therefore \lambda =\frac{2}{15}\Rightarrow x=-\frac{11}{15},y=-\frac{10}{15}$ and $z=-\frac{2}{15}$
Hence, $x\hat{i}+y\hat{j}+z\hat{k}=-\frac{1}{15}(11\hat{i}+10\hat{j}+2\hat{k})$