The vectors $\vec{AB} = 3\hat{i} - 2\hat{j} + 2\hat{k}$ and $\vec{BC} = \hat{i} - 2\hat{k}$ are the adjacent sides of a parallelogram. The angle between its diagonals is

Let $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$. If $\vec{b}$ is a vector such that $\vec{a} \cdot \vec{b} = |\vec{b}|^2$ and $|\vec{a} - \vec{b}| = \sqrt{7}$,then find $|\vec{b}|$.

If the angle between the vectors $\vec{a} = 2\lambda^2 \hat{i} + 4\lambda \hat{j} + \hat{k}$ and $\vec{b} = 7\hat{i} - 2\hat{j} + \lambda \hat{k}$ is obtuse,then the values of $\lambda$ lie in:

$a, b, c$ are three vectors such that $a + b + c = 0$,$|a| = 1, |b| = 2, |c| = 3$. Then $a \cdot b + b \cdot c + c \cdot a$ is equal to:

$\vec{a}, \vec{b}, \text{ and } \vec{c}$ are three vectors such that $|\vec{a}|=3, |\vec{b}|=5, |\vec{c}|=7$. If $\vec{a}, \vec{b}, \vec{c}$ are perpendicular to the vectors $\vec{b}+\vec{c}, \vec{c}+\vec{a}, \vec{a}+\vec{b}$ respectively,then $\sqrt{|\vec{a}+\vec{b}+\vec{c}|^2-2} = $

Let $x = \hat{i} + \hat{j}$ and $y = 3\hat{i} - 2\hat{k}$. Then, the vector $r$ of magnitude $\sqrt{21}$ satisfying $r \times x = y \times x$ and $r \times y = x \times y$ is