JEE Mathematics 2024 Question with Solution

Given: $\vec{OA} = 2\hat{i} + 2\hat{j} + \hat{k}$ and $\vec{OB} = 2\hat{i} + 4\hat{j} + 4\hat{k}$.

Find: The length of $OC$ where C is the point at which the internal bisector of $\angle AOB$ meets line AB.

First, compute the magnitudes:

$$|\vec{OA}| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{9} = 3$$ $$|\vec{OB}| = \sqrt{2^2 + 4^2 + 4^2} = \sqrt{36} = 6$$

The internal angle bisector of $\angle AOB$ divides AB in the ratio

$$AC : CB = OA : OB = 3 : 6 = 1 : 2$$

Using the section formula,

$$\vec{OC} = \frac{2\vec{OA} + 1\vec{OB}}{2+1}$$ $$\vec{OC} = \frac{2(2\hat{i} + 2\hat{j} + \hat{k}) + (2\hat{i} + 4\hat{j} + 4\hat{k})}{3}$$ $$\vec{OC} = \frac{(4\hat{i} + 4\hat{j} + 2\hat{k}) + (2\hat{i} + 4\hat{j} + 4\hat{k})}{3}$$ $$\vec{OC} = \frac{6\hat{i} + 8\hat{j} + 6\hat{k}}{3} = 2\hat{i} + \frac{8}{3}\hat{j} + 2\hat{k}$$

Now find its magnitude:

$$|OC| = \sqrt{2^2 + \left(\frac{8}{3}\right)^2 + 2^2}$$ $$|OC| = \sqrt{4 + \frac{64}{9} + 4}$$ $$|OC| = \sqrt{\frac{72+64}{9}} = \sqrt{\frac{136}{9}} = \frac{\sqrt{136}}{3} = \frac{2\sqrt{34}}{3}$$

Therefore, the length of OC is $\frac{2\sqrt{34}}{3}$, so the correct option is B.

Given: $A = (2,2,1)$ and $B = (2,4,4)$ with O as the origin.

Find: The length $OC$.

Treat the given position vectors as coordinates of points A and B. From the magnitudes,

$$OA = 3, \qquad OB = 6$$

So the internal bisector of $\angle AOB$ divides AB in the ratio

$$OA : OB = 3:6 = 1:2$$

Hence C divides AB internally in the ratio $1:2$, and by section formula,

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

Therefore,

$$OC = \sqrt{2^2 + \left(\frac{8}{3}\right)^2 + 2^2} = \frac{2\sqrt{34}}{3}$$

Thus, the required length is $\frac{2\sqrt{34}}{3}$ and the correct option is B.