Question

Let O be the origin and the position vector of A and B be $2\hat{i}+2\hat{j}+\hat{k}$ and $2\hat{i}+4\hat{j}+4\hat{k}$ respectively. If the internal bisector of$\angle AOB$ meets the line AB at C, then the length of OC is

• A. $\frac{2}{3}\sqrt31$
• B. $\frac{2}{3}\sqrt34$
• C. $\frac{3}{4}\sqrt34$
• D. $\frac{3}{2}\sqrt31$

Answer 1

Answer: (B)

$\frac{2}{3}\sqrt34$

Answer 2

Step 1: Find the Coordinates of Points A and B

A = (2, 2, 1) and B = (2, 4, 4)

Step 2: Use the Internal Division Formula

The internal bisector of ∠AOB divides AB in the ratio OA : OB = 1 : 2. Using the section formula, the coordinates of C are:

$C = \frac{1 \cdot B + 2 \cdot A}{1 + 2} = \frac{1 \cdot (2, 4, 4) + 2 \cdot (2, 2, 1)}{3} = \left(2, \frac{8}{3}, 2\right)$

Step 3: Calculate the Length of OC

The vector OC has coordinates (2, $\frac{8}{3}$, 2). Using the distance formula:

$|OC| = \sqrt{2^2 + \left(\frac{8}{3}\right)^2 + 2^2} = \sqrt{4 + \frac{64}{9} + 4} = \sqrt{\frac{136}{9}} = \frac{2\sqrt{34}}{3}$

So, the correct answer is: $\frac{2}{3}\sqrt{34}$