If $\theta$ is the angle between the vectors $\vec{a}$ and $\vec{b}$,then $\frac{|\vec{a} \times \vec{b}|}{|\vec{a} \cdot \vec{b}|}$ is equal to:

Let $\vec{a} = 6 \hat{i} - 3 \hat{j} - 6 \hat{k}$ and $\vec{d} = \hat{i} + \hat{j} + \hat{k}$. Suppose that $\vec{a} = \vec{b} + \vec{c}$,where $\vec{b}$ is parallel to $\vec{d}$ and $\vec{c}$ is perpendicular to $\vec{d}$. Then $\vec{c}$ is

If the line joining points $A$ and $B$ having position vectors $6 \vec{a}-4 \vec{b}+4 \vec{c}$ and $-4 \vec{c}$ respectively,and the line joining the points $C$ and $D$ having position vectors $-\vec{a}-2 \vec{b}-3 \vec{c}$ and $\vec{a}+2 \vec{b}-5 \vec{c}$ intersect,then their point of intersection is

$A$ tetrahedron has vertices at $O(0, 0, 0)$,$A(1, 2, 1)$,$B(2, 1, 3)$,and $C(-1, 1, 2)$. The angle between the faces $OAB$ and $ABC$ is:

If $\vec{a} = \hat{i} + 3\hat{j} - 2\hat{k}$ and $\vec{b} = 4\hat{i} - 2\hat{j} + 4\hat{k}$,then $(2\vec{a} + \vec{b}) \cdot (\vec{a} - 2\vec{b}) = \dots$

If magnitudes of vectors $\vec{a}, \vec{b}, \vec{c}$ are $3, 4,$ and $5$ respectively,and $\vec{a}$ is perpendicular to $\vec{b} + \vec{c}$,$\vec{b}$ is perpendicular to $\vec{c} + \vec{a}$,and $\vec{c}$ is perpendicular to $\vec{a} + \vec{b}$,then find the value of $|\vec{a} + \vec{b} + \vec{c}|$.