Let $A = \hat{i} + 2 \hat{j}$. If $B$ is a vector in the $XY$ plane such that $(A + B) \cdot B = 15$ and $A \cdot B = 6$,then $|B|$ is

If $a$,$b$,and $c$ are unit vectors,then $|a - b|^2 + |b - c|^2 + |c - a|^2$ does not exceed

$a, b$ and $c$ are three vectors with magnitudes $|a| = 4, |b| = 4, |c| = 2$ such that $a$ is perpendicular to $(b + c)$,$b$ is perpendicular to $(c + a)$,and $c$ is perpendicular to $(a + b)$. It follows that $|a + b + c|$ is equal to

If $\vec{a}, \vec{b}$ and $\vec{c}$ are unit vectors such that $\vec{a}$ is perpendicular to $\vec{b}$ and $\vec{c}$,and $|\vec{a} + \vec{b} + \vec{c}| = 1$,then the angle between $\vec{b}$ and $\vec{c}$ is:

Let $u$ and $v$ be two non-zero vectors in $\mathbb{R}^3$. Then $|u \times v|^2 + |u \cdot v|^2$ is equal to

If $a+b+c=0$ and $|a|=5, |b|=3$ and $|c|=7$,then the angle between $a$ and $b$ is