Let $\vec{a}$ and $\vec{b}$ be two vectors of length $\sqrt{2}$ such that $|\vec{a} + \vec{b}| = \sqrt{5}$. If $\vec{c} = \vec{a} + 2\vec{b} + 2(\vec{a} \times \vec{b})$,then $|\vec{c}|$ is

The diagonals of a parallelogram are $\vec{d_1} = \hat{j} + \hat{k}$ and $\vec{d_2} = \hat{i} + \hat{j}$. The area of the parallelogram is . . . . . . sq. units.

If either $\vec{a}=\overrightarrow{0}$ or $\vec{b}=\overrightarrow{0},$ then $\vec{a} \times \vec{b}=\overrightarrow{0} .$ Is the converse true? Justify your answer with an example.

Let $\vec{a}=2 \hat{i}-3 \hat{j}-5 \hat{k}$ and $\vec{b}=3 \hat{i}+2 \hat{j}-5 \hat{k}$ be two vectors and $\vec{r}$ be a vector in the plane of $\vec{a}$ and $\vec{b}$. If $\vec{r}$ is orthogonal to the vector $5 \hat{i}-2 \hat{j}+3 \hat{k}$ and the magnitude of $\vec{r}$ is $\sqrt{94}$,then $|\vec{r} \cdot \vec{b}|=$

Let $L_1: \frac{x+1}{3}=\frac{y+2}{2}=\frac{z+1}{1}$ and $L_2: \frac{x-2}{2}=\frac{y+2}{1}=\frac{z-3}{3}$ be the given lines. Then the unit vector perpendicular to $L_1$ and $L_2$ is

If the points $P(1, -1, 2)$,$Q(2, 0, -1)$,and $R(0, 2, 1)$ are coplanar,find the unit vector perpendicular to the plane containing these points.