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
- \(\frac{2}{3}\sqrt31\)
- \(\frac{2}{3}\sqrt34\)
- \(\frac{3}{4}\sqrt34\)
- \(\frac{3}{2}\sqrt31\)
The Correct Option is B
Solution and Explanation
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: \(\frac{2}{3}\sqrt{34}\)
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