Let $$\overrightarrow{\mathrm{a}}=\mathrm{a}_1 \hat{i}+\mathrm{a}_2 \hat{j}+\mathrm{a}_3 \hat{k}$$ and $$\overrightarrow{\mathrm{b}}=\mathrm{b}_1 \hat{i}+\mathrm{b}_2 \hat{j}+\mathrm{b}_3 \hat{k}$$ be two vectors such that $$|\overrightarrow{\mathrm{a}}|=1, \vec{a} \cdot \vec{b}=2$$ and $$|\vec{b}|=4$$. If $$\vec{c}=2(\vec{a} \times \vec{b})-3 \vec{b}$$, then the angle between $$\vec{b}$$ and $$\vec{c}$$ is equal to:

Let a unit vector $$\hat{u}=x \hat{i}+y \hat{j}+z \hat{k}$$ make angles $$\frac{\pi}{2}, \frac{\pi}{3}$$ and $$\frac{2 \pi}{3}$$ with the vectors $$\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{k}, \frac{1}{\sqrt{2}} \hat{j}+\frac{1}{\sqrt{2}} \hat{k}$$ and $$\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}$$ respectively. If $$\vec{v}=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}+\frac{1}{\sqrt{2}} \hat{k}$$ then $$|\hat{u}-\vec{v}|^2$$ is equal to

Let $$\overrightarrow{O A}=\vec{a}, \overrightarrow{O B}=12 \vec{a}+4 \vec{b} \text { and } \overrightarrow{O C}=\vec{b}$$, where O is the origin. If S is the parallelogram with adjacent sides OA and OC, then $$\mathrm{{{area\,of\,the\,quadrilateral\,OA\,BC} \over {area\,of\,S}}}$$ is equal to _________.

Let $$\vec{a}, \vec{b}$$ and $$\vec{c}$$ be three non-zero vectors such that $$\vec{b}$$ and $$\vec{c}$$ are non-collinear. If $$\vec{a}+5 \vec{b}$$ is collinear with $$\vec{c}, \vec{b}+6 \vec{c}$$ is collinear with $$\vec{a}$$ and $$\vec{a}+\alpha \vec{b}+\beta \vec{c}=\overrightarrow{0}$$, then $$\alpha+\beta$$ is equal to