JEE Mathematics 2024 Question with Solution

Given: $u = x\hat{i} + y\hat{j} + z\hat{k}$ is a unit vector. It makes angles $\pi/2$, $\pi/3$ and $2\pi/3$ with $p_1 = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{k}$, $p_2 = \frac{1}{\sqrt{2}}\hat{j} + \frac{1}{\sqrt{2}}\hat{k}$ and $p_3 = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j}$ respectively. Also, $v = \frac{1}{\sqrt{2}}(\hat{i}+\hat{j}+\hat{k})$.

Find: $|u-v|^2$.

Use the dot product relation

$$\cos \theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|}$$

Since the given vectors $p_1, p_2, p_3$ are unit vectors, the dot products directly give:

$$u \cdot p_1 = \cos \frac{\pi}{2} = 0$$

So,

$$\frac{x}{\sqrt{2}} + \frac{z}{\sqrt{2}} = 0$$

which gives

$$x = -z$$

Also,

$$u \cdot p_2 = \cos \frac{\pi}{3} = \frac{1}{2}$$

Hence,

$$\frac{y}{\sqrt{2}} + \frac{z}{\sqrt{2}} = \frac{1}{2}$$

so

$$y + z = \frac{\sqrt{2}}{2}$$

Similarly,

$$u \cdot p_3 = \cos \frac{2\pi}{3} = -\frac{1}{2}$$

Thus,

$$\frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = -\frac{1}{2}$$

which gives

$$x + y = -\frac{\sqrt{2}}{2}$$

Now use $x=-z$ in the last two equations and solve for the components. Substituting these values into

$$|u-v|^2 = \left(x-\frac{1}{\sqrt{2}}\right)^2 + \left(y-\frac{1}{\sqrt{2}}\right)^2 + \left(z-\frac{1}{\sqrt{2}}\right)^2$$

we get

$$|u-v|^2 = \frac{5}{2}$$

Therefore, the correct option is B.

Given: From the extracted working, the coordinates obtained are used to evaluate $|u-v|^2$.

Find: The numerical value of $|u-v|^2$.

The solution states the final evaluated value as

$$|u-v|^2 = \frac{5}{2}$$

with the correct option marked as B.

The same page also shows the expression

$$|u-v|^2 = \left(x-\frac{1}{\sqrt{2}}\right)^2 + \left(y-\frac{1}{\sqrt{2}}\right)^2 + \left(z-\frac{1}{\sqrt{2}}\right)^2$$

After substituting the solved values of $x, y, z$ from the working, this simplifies to

$$\frac{5}{2}$$

Hence, $|u-v|^2 = \frac{5}{2}$ and the correct option is B.