If $\vec{a}$ and $\vec{b}$ are unit vectors such that the vector $\vec{a} + 3\vec{b}$ is perpendicular to $7\vec{a} - 5\vec{b}$,then find the angle between $\vec{a}$ and $\vec{b}$.

Let $\vec{a}=\hat{i}+2 \hat{j}-2 \hat{k}$ and $\vec{b}=2 \hat{i}-\hat{j}-2 \hat{k}$ be two vectors. If the orthogonal projection vector of $\vec{a}$ on $\vec{b}$ is $\vec{x}$ and the orthogonal projection vector of $\vec{b}$ on $\vec{a}$ is $\vec{y}$, then find $|\vec{x}-\vec{y}|$.

Let $\overrightarrow{a} = \hat{i} + 2\hat{j} + 3\hat{k}$,$\overrightarrow{b} = \hat{i} - \hat{j} + 2\hat{k}$,and $\overrightarrow{c} = 5\hat{i} - 3\hat{j} + 3\hat{k}$ be three vectors. If $\overrightarrow{r}$ is a vector such that $\overrightarrow{r} \times \overrightarrow{b} = \overrightarrow{c} \times \overrightarrow{b}$ and $\overrightarrow{r} \cdot \overrightarrow{a} = 0$,then $25|\overrightarrow{r}|^2$ is equal to:

The angle between unit vectors $\bar{a}$ and $\bar{b}$ in $\mathbb{R}^3$ is $\theta$. Then,the value of $\left|\frac{\bar{a} \cdot \bar{a}}{\bar{a} \cdot \bar{b}} \cdot \frac{\bar{b} \cdot \bar{a}}{\bar{b} \cdot \bar{b}}\right| + |\bar{a} \times \bar{b}|^2$ is:

The angle between two diagonals of a cube is:

If $|\vec{a}|=4, |\vec{b}|=5, |\vec{a}-\vec{b}|=3$ and $\theta$ is the angle between the vectors $\vec{a}$ and $\vec{b}$, then $\cot^2 \theta=$