If $\overline{a}=\hat{i}+\hat{j}+\hat{k}$,$\overline{a} \cdot \overline{b}=1$ and $\overline{a} \times \overline{b}=\hat{j}-\hat{k}$,then $\overline{b}$ is

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.

Vectors $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are such that $(\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) = \vec{0}$. $P_1$ and $P_2$ are two planes determined by vectors $\vec{a}, \vec{b}$ and $\vec{c}, \vec{d}$ respectively. Then the angle between the planes $P_1$ and $P_2$ is

Find $|\vec{a} \times \vec{b}|,$ if $\vec{a}=2 \hat{i}+\hat{j}+3 \hat{k}$ and $\vec{b}=3 \hat{i}+5 \hat{j}-2 \hat{k}$.

If $\overrightarrow{a} = \hat{i} + \hat{j} + \hat{k}$,$\overrightarrow{b} = \hat{i} + 3\hat{j} + 5\hat{k}$ and $\overrightarrow{c} = 7\hat{i} + 9\hat{j} + 11\hat{k}$,then the area of the parallelogram having diagonals $\overrightarrow{a} + \overrightarrow{b}$ and $\overrightarrow{b} + \overrightarrow{c}$ is:

Let $\bar{a} = \alpha \hat{i} + 3 \hat{j} - \hat{k}$,$\bar{b} = 3 \hat{i} - \hat{j} + \beta \hat{k}$,and $\bar{c} = \hat{i} + 2 \hat{j} - 2 \hat{k}$ where $\alpha, \beta \in R$,be three vectors. If the projection of $\bar{a}$ on $\bar{c}$ is $\frac{10}{3}$ and $\bar{b} \times \bar{c} = -6 \hat{i} + 10 \hat{j} + 7 \hat{k}$,then the value of $(\alpha + \beta)$ is equal to