For non-zero vectors $\bar{a}, \bar{b}, \bar{c}$,if $\bar{a} \times \bar{b} = \bar{c}$ and $\bar{b} \times \bar{c} = \bar{a}$,then:

Let $\vec{a}=2 \hat{i}+\alpha \hat{j}+\hat{k}$,$\vec{b}=-\hat{i}+\hat{k}$,and $\vec{c}=\beta \hat{j}-\hat{k}$,where $\alpha$ and $\beta$ are integers and $\alpha \beta=-6$. Let the values of the ordered pair $(\alpha, \beta)$ for which the area of the parallelogram with diagonals $\vec{a}+\vec{b}$ and $\vec{b}+\vec{c}$ is $\frac{\sqrt{21}}{2}$,be $(\alpha_1, \beta_1)$ and $(\alpha_2, \beta_2)$. Then $\alpha_1^2+\beta_1^2-\alpha_2 \beta_2$ is equal to

$A$ vector $\vec{a}$ is parallel to the line of intersection of the plane determined by the vectors $\hat{i}$ and $\hat{i}+\hat{j}$,and the plane determined by the vectors $\hat{i}-\hat{j}$ and $\hat{i}+\hat{k}$. The obtuse angle between $\vec{a}$ and the vector $\vec{b}=\hat{i}-2\hat{j}+2\hat{k}$ is

Let $\overrightarrow{a}=2 \hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{b}=((\overrightarrow{a} \times(\hat{i}+\hat{j})) \times \hat{i}) \times \hat{i}$. Then the square of the projection of $\vec{a}$ on $\vec{b}$ is:

The area of the parallelogram whose diagonals are $\frac{3}{2}i + \frac{1}{2}j - k$ and $2i - 6j + 8k$ is

Find the vector perpendicular to the vectors $4 i-j+3 k$ and $-2 i+j-2 k$ whose magnitude is $9$.