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

The unit vector in direction of angle bisector is the sum or difference of unit vectors of both the angle arms.

Thus, $\hat{c}\pm \frac{3\overrightarrow{i}+4\overrightarrow{j}}{5}$ = $\lambda (-\overrightarrow{i}+\overrightarrow{j}-\overrightarrow{k})$

We need to find $\lambda$ and so we can rather solve the question by seeing at the options

The denominator is $15$ and so we make the denominator of $\frac{3\overrightarrow{i}+4\overrightarrow{j}}{5}$ also as $15$, and so numerator becomes $9\overrightarrow{i}+12\overrightarrow{j}$.

We need some multiple of $-\overrightarrow{i}+\overrightarrow{j}-\overrightarrow{k}$ as the RHS.

So, $-11\overrightarrow{i}-10\overrightarrow{j}$ will be approriate

The third component of $\overrightarrow{k}$ can then be selected appropriately to get option $D$ as the answer.