Question

Let \( \vec{OA} = -4\hat{i} + 3\hat{k} \), \( \vec{OB} = 14\hat{i} + 2\hat{j} - 5\hat{k} \), and vector \( \vec{OD} \) bisects \( \angle AOB \), with \( |\vec{OD}| = \sqrt{6} \). Then: \[ \vec{OD} = ? \]

• 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:

To find bisecting vectors, use normalized direction vectors or combine unit vectors with proportional scaling.

Answer 1

The Correct Option is A

Use the angle bisector theorem for vectors: \[ \vec{OD} = \text{unit vector in the direction of } \vec{OA} + \vec{OB} \Rightarrow \vec{A} = -4\hat{i} + 0\hat{j} + 3\hat{k},\quad \vec{B} = 14\hat{i} + 2\hat{j} - 5\hat{k} \Rightarrow \vec{A} + \vec{B} = 10\hat{i} + 2\hat{j} - 2\hat{k} \] Direction vector of angle bisector is proportional to: \[ \vec{v} = \frac{\vec{OA}}{|\vec{OA}|} + \frac{\vec{OB}}{|\vec{OB}|} \Rightarrow |\vec{OA}| = \sqrt{(-4)^2 + 0^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \] \[ |\vec{OB}| = \sqrt{14^2 + 2^2 + (-5)^2} = \sqrt{196 + 4 + 25} = \sqrt{225} = 15 \] So: \[ \begin{align} \vec{OD} = \frac{1}{5}(-4\hat{i} + 3\hat{k}) + \frac{1}{15}(14\hat{i} + 2\hat{j} - 5\hat{k})
= \left(-\frac{4}{5} + \frac{14}{15}\right)\hat{i} + \frac{2}{15}\hat{j} + \left(\frac{3}{5} - \frac{5}{15}\right)\hat{k} = \hat{i} + \hat{j} + 2\hat{k} \quad (\text{after simplification and normalizing}) \] Since the magnitude is given as \( \sqrt{6} \), vector must be: \[ \vec{OD} = \pm(\hat{i} + \hat{j} + 2\hat{k}) \]