If two vectors $\vec{a}$ and $\vec{b}$ are such that $|\vec{a}| = 2$, $|\vec{b}| = 3$ and $\vec{a} \cdot \vec{b} = 4$, then $|\vec{a} - \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 $|\vec{x}-\vec{y}|$.

If $a, b, c$ are three mutually perpendicular vectors such that the magnitudes of $b$ and $c$ are $1/2$ times and $\sqrt{3}/2$ times that of $a$, respectively, then the angle between the vectors $a+b+c$ and $b$ is

The angle between the diagonals of the parallelogram whose adjacent sides are $2 \hat{i}+4 \hat{j}-5 \hat{k}$ and $\hat{i}+2 \hat{j}+3 \hat{k}$ is

If $\theta$ is the angle between the vectors $\vec{a} = 2\hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = 6\hat{i} - 3\hat{j} + 2\hat{k}$,then:

If $A, B, C, D$ are the points $(2, 3, -1), (3, 5, -3), (1, 2, 3), (3, 5, 7)$ respectively,then the angle between $AB$ and $CD$ is