If $\vec{a} = -4 \hat{i} + 2 \hat{j} + 4 \hat{k}$ and $\vec{b} = \sqrt{2} \hat{i} - \sqrt{2} \hat{j}$ are two vectors,then the angle between the vectors $2 \vec{a}$ and $\frac{\vec{b}}{2}$ is (in $^{\circ}$)

If $a, b, c$ are coplanar vectors,then

Let $\vec{a}$ and $\vec{b}$ be vectors of the same magnitude such that $\frac{|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}|}{|\vec{a}+\vec{b}|-|\vec{a}-\vec{b}|}=\sqrt{2}+1$. Then $\frac{|\vec{a}+\vec{b}|^2}{|\vec{a}|^2}$ is:

If $\alpha=\hat{i}-3 \hat{j}$ and $\beta=\hat{i}+2 \hat{j}-\hat{k}$,express $\beta$ in the form $\beta=\beta_1+\beta_2$,where $\beta_1$ is parallel to $\alpha$ and $\beta_2$ is perpendicular to $\alpha$. Then $\beta_1$ is given by:

The magnitude of vectors $\vec{a}$ and $\vec{b}$ are $1$ and $2$ respectively,and $\vec{a} \cdot \vec{b} = 1$. Then,the angle between the two vectors $\vec{a}$ and $\vec{b}$ is . . . . . . .

If $\vec{a} = \hat{i} + \hat{j}$ and $\vec{b} = \hat{i} - \hat{k}$, then the point of intersection of the lines $\vec{r} \times \vec{a} = \vec{b} \times \vec{a}$ and $\vec{r} \times \vec{b} = \vec{a} \times \vec{b}$ is