Question

Let \[ \overrightarrow{OA}=-4\hat{i}+3\hat{k},\quad \overrightarrow{OB}=14\hat{i}+2\hat{j}-5\hat{k}. \] \(\overrightarrow{OD}\) bisects \(\angle AOB\) and \[ |\overrightarrow{OD}|=\sqrt{6}, \] then \(\overrightarrow{OD}=\)

• A. \(\pm(\hat{i}+\hat{j}+2\hat{k})\)
• B. \(\pm(\hat{i}+2\hat{j}+\hat{k})\)
• C. \(\pm(2\hat{i}+\hat{j}+\hat{k})\)
• D. \(\pm \frac{1}{\sqrt{2}}(2\hat{i}+\hat{j}+\sqrt{7}\hat{k})\)

Hint:

The direction of the internal angle bisector of two vectors is obtained by adding their unit vectors: \[ \frac{\vec{a}}{|\vec{a}|}+\frac{\vec{b}}{|\vec{b}|}. \] Then adjust the magnitude according to the given condition.

Answer 1

The Correct Option is A

Step 1: Find magnitudes of $\overrightarrow{OA}$ and $\overrightarrow{OB}$.
$\overrightarrow{OA}=-4\hat{i}+3\hat{k}$: $|\overrightarrow{OA}|=\sqrt{16+0+9}=5$. $\overrightarrow{OB}=14\hat{i}+2\hat{j}-5\hat{k}$: $|\overrightarrow{OB}|=\sqrt{196+4+25}=15$.

Step 2: Find unit vectors along $OA$ and $OB$.
$\hat{a} = \frac{\overrightarrow{OA}}{5} = \frac{-4\hat{i}+3\hat{k}}{5}$ and $\hat{b} = \frac{\overrightarrow{OB}}{15} = \frac{14\hat{i}+2\hat{j}-5\hat{k}}{15}$.

Step 3: Find the direction of the angle bisector.
The internal angle bisector direction is $\hat{a}+\hat{b}$: \[ \hat{a}+\hat{b} = \left(-\frac{4}{5}+\frac{14}{15}\right)\hat{i} + \frac{2}{15}\hat{j} + \left(\frac{3}{5}-\frac{1}{3}\right)\hat{k} \] $= \left(-\frac{12}{15}+\frac{14}{15}\right)\hat{i}+\frac{2}{15}\hat{j}+\left(\frac{9}{15}-\frac{5}{15}\right)\hat{k} = \frac{2}{15}\hat{i}+\frac{2}{15}\hat{j}+\frac{4}{15}\hat{k} = \frac{2}{15}(\hat{i}+\hat{j}+2\hat{k})$

Step 4: Write $\overrightarrow{OD}$ in terms of the direction.
$\overrightarrow{OD} = \lambda(\hat{i}+\hat{j}+2\hat{k})$ for some scalar $\lambda$.

Step 5: Use the magnitude condition.
$|\overrightarrow{OD}|=\sqrt{6}$, so $|\lambda|\sqrt{1+1+4}=\sqrt{6} \Rightarrow |\lambda|\sqrt{6}=\sqrt{6} \Rightarrow |\lambda|=1$.

Step 6: State the answer.
\[ \boxed{\pm(\hat{i}+\hat{j}+2\hat{k})} \]