Question

Let a unit vector $\hat{u} = x\hat{i} + y\hat{j} + z\hat{k}$ make angles $\frac{\pi}{2}, \frac{\pi}{3}$, and $\frac{2\pi}{3}$ with the vectors $\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{k}$, $\frac{1}{\sqrt{2}} \hat{j} + \frac{1}{\sqrt{2}} \hat{k}$, and $\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j}$ respectively. If $\vec{v} = \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} + \frac{1}{\sqrt{2}} \hat{k}$, then $|\hat{u} - \vec{v}|^2$ is equal to

• A. $\frac{11}{2}$
• B. $\frac{5}{2}$
• C. 9
• D. 7

Answer 1

Answer: (B)

$\frac{5}{2}$

Answer 2

Given that $\vec{u} = x\hat{i} + y\hat{j} + z\hat{k}$ is a unit vector, it satisfies: $x^2 + y^2 + z^2 = 1$
Step 1. Using the angle conditions:
- The angle between $\vec{u}$ and $\frac{\hat{i}}{\sqrt{2}} + \frac{\hat{j}}{\sqrt{2}}$ is $\frac{\pi}{2}$:
$\vec{u} \cdot \left( \frac{\hat{i}}{\sqrt{2}} + \frac{\hat{j}}{\sqrt{2}} \right) = 0 \implies \frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = 0$
$x + y = 0$ ---(1)
- The angle between $\vec{u}$ and $\frac{\hat{i}}{\sqrt{2}} + \hat{j} + \frac{\hat{k}}{\sqrt{2}}$ is $\frac{\pi}{3}$:
$\vec{u} \cdot \left( \frac{\hat{i}}{\sqrt{2}} + \hat{j} + \frac{\hat{k}}{\sqrt{2}} \right) = \frac{1}{2} \implies \frac{x}{\sqrt{2}} + y + \frac{z}{\sqrt{2}} = \frac{1}{2}$
$x + \sqrt{2}y + z = \frac{\sqrt{2}}{2}$
- The angle between $\vec{u}$ and $\frac{\hat{i}}{\sqrt{2}} + \frac{\hat{j}}{\sqrt{2}} + \hat{k}$ is $\frac{\pi}{2}$:
$\vec{u} \cdot \left( \frac{\hat{i}}{\sqrt{2}} + \frac{\hat{j}}{\sqrt{2}} + \hat{k} \right) = 0 \implies \frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} + z = 0$
$x + y + \sqrt{2}z = 0$
Step 2. Solving the system of equations:** From equations (1), (2), and (3):
- Substitute $z = -x$ in (2):
$x + \sqrt{2}y - x = \frac{\sqrt{2}}{2} \implies y = \frac{1}{\sqrt{2}}$
- Substitute $y = \frac{1}{\sqrt{2}}$ and $z = -x$ in (3):
$x + \frac{1}{\sqrt{2}} + \sqrt{2}(-x) = 0 \implies x = -\frac{1}{2\sqrt{2}}, \, z = \frac{1}{2\sqrt{2}}$

Step 3. Calculate $|\vec{u} - \vec{v}|^2$:
$|\vec{u} - \vec{v}|^2 = \left( x - \frac{1}{\sqrt{2}} \right)^2 + \left( y - \frac{1}{\sqrt{2}} \right)^2 + \left( z - \frac{1}{\sqrt{2}} \right)^2$
Substituting $x = -\frac{1}{2\sqrt{2}}, \, y = \frac{1}{\sqrt{2}}, \, z = \frac{1}{2\sqrt{2}}$:
$|\vec{u} - \vec{v}|^2 = \frac{5}{2}$
The Correct answer is :$\frac{5}{2}$.