Let $\vec{u}$ be a vector coplanar with the vectors $\vec{a} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\vec{b} = \hat{j} + \hat{k}$. If $\vec{u}$ is perpendicular to $\vec{a}$ and $\vec{u} \cdot \vec{b} = 24$,then $|\vec{u}|^2 = \dots$

If $\vec{b}=2 \hat{i}-\hat{j}-\hat{k}$,$\vec{a}=3 \hat{i}+4 \hat{j}-5 \hat{k}$ and $\vec{b} \times(\vec{a} \times \vec{b})=\frac{\vec{a}-k \vec{b}}{l}$,then $\frac{k}{l|\vec{b}|}$ is

Let $\bar{a}, \bar{b}, \bar{c}$ be three vectors such that $|\bar{a}|=1, |\bar{c}|=1, |\bar{b}|=4$,and $|\bar{b} \times \bar{c}|=\sqrt{15}$. If $\lambda \bar{a}=\bar{b}-2 \bar{c}$,then the value of $\lambda$ is

$A$ vector $a$ makes equal acute angles with the coordinate axes. Then the projection of vector $b = 5\hat{i} + 7\hat{j} - \hat{k}$ on $a$ is

Two adjacent sides of a triangle are represented by the vectors $\vec{a} = 2\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{b} = 2\sqrt{3}\hat{i} - 2\sqrt{3}\hat{j} + \sqrt{3}\hat{k}$. Then the least angle of the triangle and the perimeter of the triangle are respectively:

Let $|\vec{a}| = |\vec{b}| = 1$ and $|\vec{a} + \vec{b}| = \sqrt{3}$. If $\vec{c}$ is a vector satisfying the condition $\vec{c} - \vec{a} - 2\vec{b} = 3(\vec{a} \times \vec{b})$,then $\vec{c} \cdot \vec{b} = \dots$ (in $/2$)