Step 1: The unit vector \( \mathbf{v} \) in the XY-plane making an angle of \( 45^\circ \) with \( \hat{i} + \hat{j} \) is given by:
\[ \mathbf{v} = \hat{i} \cos(45^\circ) + \hat{j} \sin(45^\circ) \]
Step 2: The angle between \( \mathbf{v} \) and the vector \( 3\hat{i} - 4\hat{j} \) is \( 60^\circ \). Using the dot product formula:
\[ \mathbf{v} \cdot (3\hat{i} - 4\hat{j}) = |\mathbf{v}| \cdot |3\hat{i} - 4\hat{j}| \cdot \cos(60^\circ) \]
Step 3: Solving this system of equations gives the components of the unit vector \( \mathbf{v} \).
Let $ \vec{a} = \hat{i} + 2\hat{j} + \hat{k} $, $ \vec{b} = 3\hat{i} - 3\hat{j} + 3\hat{k} $, $ \vec{c} = 2\hat{i} - \hat{j} + 2\hat{k} $ and $ \vec{d} $ be a vector such that $ \vec{b} \times \vec{d} = \vec{c} \times \vec{d} $ and $ \vec{a} \cdot \vec{d} = 4 $. Then $ |\vec{a} \times \vec{d}|^2 $ is equal to _______
Let $L_1: \frac{x-1}{1} = \frac{y-2}{-1} = \frac{z-1}{2}$ and $L_2: \frac{x+1}{-1} = \frac{y-2}{2} = \frac{z}{1}$ be two lines. Let $L_3$ be a line passing through the point $(\alpha, \beta, \gamma)$ and be perpendicular to both $L_1$ and $L_2$. If $L_3$ intersects $L_1$, then $\left| 5\alpha - 11\beta - 8\gamma \right|$ equals:

A quantity \( X \) is given by: \[ X = \frac{\epsilon_0 L \Delta V}{\Delta t} \] where:
- \( \epsilon_0 \) is the permittivity of free space,
- \( L \) is the length,
- \( \Delta V \) is the potential difference,
- \( \Delta t \) is the time interval.
The dimension of \( X \) is the same as that of: