Let $\vec{a}=2 \hat{i}+5 \hat{j}-\hat{k}$,$\vec{b}=2 \hat{i}-2 \hat{j}+2 \hat{k}$ and $\vec{c}$ be three vectors such that $(\vec{c}+\hat{i}) \times (\vec{a}+\vec{b}+\hat{i}) = \vec{a} \times (\vec{c}+\hat{i})$ and $\vec{a} \cdot \vec{c} = -29$. Then $\vec{c} \cdot (-2 \hat{i}+\hat{j}+\hat{k})$ is equal to:

$|a \times b|^2 + (a \cdot b)^2 = ?$

If $b$ and $c$ are any two non-collinear unit vectors and $a$ is any vector,then $(a \cdot b)b + (a \cdot c)c + \frac{a \cdot (b \times c)}{|b \times c|} (b \times c) = $

The scalars $l$ and $m$ such that $la + mb = c,$ where $a, b$ and $c$ are given vectors,are equal to

Three vectors $\vec a, \vec b, \vec c$ are inclined at an acute angle with each other such that $|\vec a| = 2, |\vec b| = 3, |\vec c| = 9$ and the lengths of the projections of $\vec a$ on $\vec b$,$\vec b$ on $\vec c$,and $\vec c$ on $\vec a$ respectively are in geometric progression. If the angle between $\vec a$ and $\vec b$ is $\frac{5\pi}{12}$ and the angle between $\vec c$ and $\vec a$ is $\frac{\pi}{12}$,then the angle between $\vec b$ and $\vec c$ is:

If vector $a$ has magnitude $5$ and points north-east,and vector $b$ has magnitude $5$ and points north-west,then $|a - b| = $