If $|a \times b|^2 + |a \cdot b|^2 = 144$ and $|a| = 4$,then $|b|$ is equal to

Let $\vec a, \vec b, \vec c$ be three vectors such that $\vec a \perp (\vec b + \vec c)$,$\vec b \perp (\vec c + \vec a)$,and $\vec c \perp (\vec a + \vec b)$. If $|\vec a| = 1, |\vec b| = 2, |\vec c| = 3$,then $|\vec a + \vec b + \vec c| = \dots$

$\vec{a}, \vec{b}, \vec{c}$ are three unit vectors such that $|\vec{a}+\vec{b}+\vec{c}|=1$ and $\vec{a}$ is perpendicular to $\vec{b}$. If $\vec{c}$ makes angles $\alpha, \beta$ with $\vec{a}, \vec{b}$ respectively, then $\cos \alpha+\cos \beta=$

If $a=\hat{i}+2 \hat{j}+3 \hat{k}$,$b=2 \hat{i}+3 \hat{j}+2 \hat{k}$ and $c$ is a vector perpendicular to $b$,then $\left\{\frac{a \cdot(b \times c)}{|b \times c|^2}\right\}(b \times c)+\left\{\frac{a \cdot b}{|b|^2}\right\} b+\left\{\frac{a \cdot c}{|c|^2}\right\} c$ is equal to:

In a parallelogram $OACB$,$\overrightarrow{OA} = \vec{a}$,$\overrightarrow{OB} = \vec{b}$,and the foot of the perpendicular drawn from point $B$ to $AC$ is $M$. If $\vec{a} \cdot \vec{b} = 1$ and $|\vec{a}| = |\vec{b}| = 2$,then $|\overrightarrow{BM}|$ is:

The value of $b$ such that the scalar product of the vector $(i + j + k)$ with the unit vector parallel to the sum of the vectors $(2i + 4j - 5k)$ and $(bi + 2j + 3k)$ is $1$,is: