Question

Let a unit vector which makes an angle of $60^\circ$ with $2\hat{i} + 2\hat{j} - \hat{k}$ and an angle of $45^\circ$ with $\hat{i} - \hat{k}$ be $\vec{C}$. Then $\vec{C} + \left( -\frac{1}{2} \hat{i} + \frac{1}{3\sqrt{2}} \hat{j} - \frac{\sqrt{2}}{3} \hat{k} \right)$ is:

• A. $\frac{\sqrt{2}}{3} \hat{i} + \frac{\sqrt{2}}{3} \hat{j} + \left( \frac{1}{2} + \frac{2\sqrt{2}}{3} \right) \hat{k}$
• B. $\frac{\sqrt{2}}{3} \hat{i} + \frac{1}{3\sqrt{2}} \hat{j} - \frac{1}{2} \hat{k}$
• C. $\left( \frac{1}{\sqrt{3}} + \frac{1}{2} \right) \hat{i} + \left( \frac{1}{\sqrt{3}} - \frac{1}{3\sqrt{2}} \right) \hat{j} + \left( \frac{1}{\sqrt{3}} + \frac{\sqrt{2}}{3} \right) \hat{k}$
• D. $\frac{\sqrt{2}}{3} \hat{i} - \frac{1}{2} \hat{k}$

Answer 1

Answer: (D)

$\frac{\sqrt{2}}{3} \hat{i} - \frac{1}{2} \hat{k}$

Answer 2

Express $\vec{C}$ as a Unit Vector:
Let $\vec{C} = C_1\vec{i} + C_2\vec{j} + C_3\vec{k}$ such that:
$C_1^2 + C_2^2 + C_3^2 = 1.$

Using the Angle Condition with $2\vec{i} + 2\vec{j} - \vec{k}$:
The dot product $\vec{C} \cdot (2\vec{i} + 2\vec{j} - \vec{k})$ is given by:
$\vec{C} \cdot (2\vec{i} + 2\vec{j} - \vec{k}) = |\vec{C}||2\vec{i} + 2\vec{j} - \vec{k}|\cos 60^\circ.$

Since $\vec{C}$ is a unit vector: $2C_1 + 2C_2 - C_3 = \frac{3}{2}.$

Using the Angle Condition with $\vec{i} - \vec{k}$:
The dot product $\vec{C} \cdot (\vec{i} - \vec{k})$ is given by:
$\vec{C} \cdot (\vec{i} - \vec{k}) = |\vec{C}||\vec{i} - \vec{k}|\cos 45^\circ.$

Simplifying gives: $C_1 - C_3 = 1.$

Solving the Equations:
From $C_1 - C_3 = 1$ and $2C_1 + 2C_2 - C_3 = \frac{3}{2}$, we find: $C_1 = \frac{\sqrt{2}}{3}, \quad C_2 = -\frac{1}{3\sqrt{2}}, \quad C_3 = \frac{\sqrt{2}}{3} - \frac{1}{2}.$

Vector Addition:
Adding $\vec{C}$ to $\left(\frac{1}{2}\vec{i} + \frac{1}{3\sqrt{2}}\vec{j} - \frac{\sqrt{2}}{3}\vec{k}\right)$:

$\vec{C} + \left(\frac{1}{2}\vec{i} + \frac{1}{3\sqrt{2}}\vec{j} - \frac{\sqrt{2}}{3}\vec{k}\right) = \frac{\sqrt{2}}{3}\vec{i} - \frac{1}{2}\vec{k}.$