If $\vec{a}$ and $\vec{b}$ are vectors in space given by $\vec{a}=\frac{\hat{i}-2 \hat{j}}{\sqrt{5}}$ and $\vec{b}=\frac{2 \hat{i}+\hat{j}+3 \hat{k}}{\sqrt{14}}$,then the value of $(2 \vec{a}+\vec{b}) \cdot[(\vec{a} \times \vec{b}) \times(\vec{a}-2 \vec{b})]$ is

If $\vec{x} \cdot \vec{y} = 0$ then,$(\vec{y} \times \vec{x}) \times \vec{x} = $ . . . . . . . where,$|\vec{x}| = 1$.

Let $\overrightarrow{a}=2 \hat{i}-3 \hat{j}+4 \hat{k}$,$\overrightarrow{b}=3 \hat{i}+4 \hat{j}-5 \hat{k}$,and a vector $\vec{c}$ be such that $\vec{a} \times(\vec{b}+\vec{c})+\vec{b} \times \vec{c}=\hat{i}+8 \hat{j}+13 \hat{k}$. If $\vec{a} \cdot \vec{c}=13$,then $(24-\vec{b} \cdot \vec{c})$ is equal to ...........

If $a=(1,2,3), b=(2,-1,1), c=(3,2,1)$ and $a \times(b \times c)=\alpha a+\beta b+\gamma c$,then

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non-zero vectors such that no two of them are collinear and $(\vec{a} \times \vec{b}) \times \vec{c} = \frac{1}{3}|\vec{b}| |\vec{c}| \vec{a}$. If $\theta$ is the angle between vectors $\vec{b}$ and $\vec{c}$,then a value of $\sin \theta$ is:

Let $\vec{a} = \hat{j} - \hat{k}$ and $\vec{c} = \hat{i} - \hat{j} - \hat{k}$. If $\vec{a} \times \vec{b} + \vec{c} = \vec{0}$ and $\vec{a} \cdot \vec{b} = 3$,find the vector $\vec{b}$.