If $\mathbf{a}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}, \mathbf{x}=\left(\frac{\mathbf{a b}}{|\mathbf{b}|^2}\right) \mathbf{b}, \mathbf{y}=\left(\frac{\mathbf{a b}}{|\mathbf{a}|^2}\right) \mathbf{a}$ and $\theta$ is angle between $\mathbf{a}$ and $\mathbf{b}$, then $x^2+y^2=$

Three non-coplanar vectors $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ are the coterminous edges of a parallelopiped. If $\mathbf{a}$ and $\mathbf{b}$ determine the base of the parallelopiped, then its height is

Let $A B C$ be a triangle and $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ be the position vectors of $A, B$ and $C$ respectively. If $D$ divides $B C$ in the ratio $2: 3$ internally and $E$ divides $C A$ in the ratio $2: 1$ internally, then the position vector of the point $P$ which divides $D E$ in the ratio $3: 5$ internally is

If $2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}, \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ are the position vectors of the vertices $A, B, C$ of a triangle respectively, then a unit vector along the median drawn through the vertex $A$ is