Let $(x, y) \in (R \times R)$ and $\vec{a} = x \hat{i} + 2 \hat{j} - \hat{k}$, $\vec{b} = 6 \hat{i} - y \hat{j} + 2 \hat{k}$ be two vectors. If $|\vec{a} \times \vec{b}|^2 + |\vec{a} \cdot \vec{b}|^2 = f(x) g(y)$, then $f(x) + g(y) - 46 = 0$ represents:

In a parallelogram $ABCD$,the position vectors of vertices $A$ and $C$ are $3\hat{i} + 3\hat{j} + 5\hat{k}$ and $\hat{i} - 5\hat{j} - 5\hat{k}$ respectively. If $M$ is the midpoint of the diagonal $DB$,find the projection of $\overline{OM}$ on $\overline{OC}$,where $O$ is the origin.

Let the arc $AC$ of a circle subtend a right angle at the centre $O$. If the point $B$ on the arc $AC$ divides the arc $AC$ such that $\frac{\text{length of arc } AB}{\text{length of arc } BC} = \frac{1}{5}$,and $\overrightarrow{OC} = \alpha \overrightarrow{OA} + \beta \overrightarrow{OB}$,then $\alpha + \sqrt{2}(\sqrt{3}-1) \beta$ is equal to

In a $\triangle ABC$,$|CB|=a$,$|CA|=b$,$|AB|=c$ and $CD$ is the median through the vertex $C$. Then,$CA \cdot CD=$

The position vectors of the vertices of a $\Delta ABC$ are $4\hat{i}-2\hat{j}$,$\hat{i}+4\hat{j}-3\hat{k}$,and $-\hat{i}+5\hat{j}+\hat{k}$ respectively. Then,$\angle ABC$ is equal to:

Let $\vec{a}=2 \hat{i}-\hat{j}+\hat{k}$ be the position vector of a point $A$. Let $\vec{b}=\hat{i}+2 \hat{j}-\hat{k}$ and $\vec{c}=\hat{i}+\hat{j}-2 \hat{k}$ be two vectors and $\vec{r}$ be a vector passing through the point $A$ with position vector $\vec{a}$ and parallel to the vector $\vec{b}$. If the projection of $\vec{r}$ on $\vec{c}$ is $\frac{9}{\sqrt{6}}$, then find $|\vec{r}|$.