For a unit vector $\vec{a}$,if $(\vec{x}-\vec{a}) \cdot (\vec{x}+\vec{a}) = 15$,then find $|\vec{x}|$.

The point of intersection of $r \times a = b \times a$ and $r \times b = a \times b$,where $a = i + j$ and $b = 2i - k$ is

The unit vector in $ZOX$ plane, making angles $45^{\circ}$ and $60^{\circ}$ respectively with $\vec{\alpha}=2 \hat{i}+2 \hat{j}-\hat{k}$ and $\vec{\beta}=\hat{j}-\hat{k}$ is

If $\overrightarrow{a} = \hat{i} + 2\hat{k}$,$\overrightarrow{b} = \hat{i} + \hat{j} + \hat{k}$,and $\overrightarrow{c} = 7\hat{i} - 3\hat{j} + 4\hat{k}$,such that $\overrightarrow{r} \times \overrightarrow{b} + \overrightarrow{b} \times \overrightarrow{c} = \overrightarrow{0}$ and $\overrightarrow{r} \cdot \overrightarrow{a} = 0$,then $\overrightarrow{r} \cdot \overrightarrow{c}$ is equal to:

Let the vectors $\vec{a} = -\hat{i} + \hat{j} + 3\hat{k}$ and $\vec{b} = \hat{i} + 3\hat{j} + \hat{k}$. For some $\lambda, \mu \in \mathbb{R}$, let $\vec{c} = \lambda \vec{a} + \mu \vec{b}$. If $\vec{c} \cdot (3\hat{i} - 6\hat{j} + 2\hat{k}) = 10$ and $\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = -2$, then $|\vec{c}|^2$ is equal to:

Let $\hat{a}$ and $\hat{b}$ be two unit vectors. If the vectors $\vec{c} = \hat{a} + 2\hat{b}$ and $\vec{d} = 5\hat{a} - 4\hat{b}$ are perpendicular to each other,then the angle between $\hat{a}$ and $\hat{b}$ is: