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: Write the given vectors.
\[ \overrightarrow{OA}=-4\hat{i}+0\hat{j}+3\hat{k} \] \[ \overrightarrow{OB}=14\hat{i}+2\hat{j}-5\hat{k} \]

Step 2: Find magnitudes of the vectors.
\[ |\overrightarrow{OA}|=\sqrt{(-4)^2+0^2+3^2} \] \[ |\overrightarrow{OA}|=\sqrt{16+9}=5 \] Also, \[ |\overrightarrow{OB}|=\sqrt{14^2+2^2+(-5)^2} \] \[ |\overrightarrow{OB}|=\sqrt{196+4+25}=15 \]

Step 3: Find unit vectors along \(OA\) and \(OB\).
\[ \hat{a}=\frac{\overrightarrow{OA}}{|\overrightarrow{OA}|} =\frac{-4\hat{i}+3\hat{k}}{5} \] \[ \hat{a}=-\frac{4}{5}\hat{i}+\frac{3}{5}\hat{k} \] Similarly, \[ \hat{b}=\frac{\overrightarrow{OB}}{|\overrightarrow{OB}|} =\frac{14\hat{i}+2\hat{j}-5\hat{k}}{15} \] \[ \hat{b}=\frac{14}{15}\hat{i}+\frac{2}{15}\hat{j}-\frac{1}{3}\hat{k} \]

Step 4: Direction of angle bisector.
The direction of the internal angle bisector is along \[ \hat{a}+\hat{b} \] So, \[ \hat{a}+\hat{b} = -\frac{4}{5}\hat{i}+\frac{3}{5}\hat{k} + \frac{14}{15}\hat{i}+\frac{2}{15}\hat{j}-\frac{1}{3}\hat{k} \] \[ = \left(-\frac{12}{15}+\frac{14}{15}\right)\hat{i} +\frac{2}{15}\hat{j} +\left(\frac{9}{15}-\frac{5}{15}\right)\hat{k} \] \[ = \frac{2}{15}\hat{i} +\frac{2}{15}\hat{j} +\frac{4}{15}\hat{k} \] \[ = \frac{2}{15}(\hat{i}+\hat{j}+2\hat{k}) \] Thus, \(\overrightarrow{OD}\) is parallel to \[ \hat{i}+\hat{j}+2\hat{k} \]

Step 5: Use the given magnitude.
Let \[ \overrightarrow{OD}=\lambda(\hat{i}+\hat{j}+2\hat{k}) \] Then, \[ |\overrightarrow{OD}|=|\lambda|\sqrt{1^2+1^2+2^2} \] \[ |\overrightarrow{OD}|=|\lambda|\sqrt{6} \] Given, \[ |\overrightarrow{OD}|=\sqrt{6} \] So, \[ |\lambda|\sqrt{6}=\sqrt{6} \] \[ |\lambda|=1 \] Hence, \[ \lambda=\pm 1 \] Therefore, \[ \overrightarrow{OD}=\pm(\hat{i}+\hat{j}+2\hat{k}) \]

Step 6: Final conclusion.
Therefore, \[ \boxed{\pm(\hat{i}+\hat{j}+2\hat{k})} \]