If vector $\vec{a} = 2\hat{i} - 3\hat{j} + 6\hat{k}$ and vector $\vec{b} = -2\hat{i} + 2\hat{j} - \hat{k},$ then $\frac{\text{Projection of vector } \vec{a} \text{ on vector } \vec{b}}{\text{Projection of vector } \vec{b} \text{ on vector } \vec{a}} = $

Two adjacent sides of a parallelogram $ABCD$ are given by $\overrightarrow{AB} = 2\hat{i} + 10\hat{j} + 11\hat{k}$ and $\overrightarrow{AD} = -\hat{i} + 2\hat{j} + 2\hat{k}$. The side $AD$ is rotated by an acute angle $\alpha$ in the plane of the parallelogram so that $AD$ becomes $AD'$. If $AD'$ makes a right angle with the side $AB$,then the cosine of the angle $\alpha$ is given by

If $a, b, c$ are three vectors such that $a = b + c$ and the angle between $b$ and $c$ is $\pi / 2$,then:

$p=2 \hat{i}-3 \hat{j}+\hat{k}, q=\hat{i}+\hat{j}-\hat{k}$. If the vectors $a$ and $b$ are the orthogonal projections of $p$ on $q$ and $q$ on $p$ respectively,then $\frac{a \times b}{a \cdot b}=$

If $\overrightarrow{a}=\hat{i}-\hat{j}-\hat{k}$ and $\overrightarrow{b}=\lambda \hat{i}-3 \hat{j}+\hat{k}$ and the orthogonal projection of $\overrightarrow{b}$ on $\overrightarrow{a}$ is $\frac{4}{3}(\hat{i}-\hat{j}-\hat{k})$,then $\lambda$ is equal to

If $e$ is a unit vector perpendicular to the plane determined by the points $2 \hat{i}+\hat{j}+\hat{k}$,$\hat{i}-\hat{j}+\hat{k}$ and $-\hat{i}+\hat{j}-\hat{k}$. If $a=2 \hat{i}-3 \hat{j}+6 \hat{k}$,then the projection vector of $a$ on $e$ is