If $x$ and $y$ are two unit vectors and the angle between them is $\phi$,then $\frac{1}{2} |x - y| = $

Let $a=\hat{i}+2 \hat{j}-2 \hat{k}$ and $b=2 \hat{i}-\hat{j}-2 \hat{k}$. If the orthogonal projection vector of $a$ on $b$ is $x$ and the orthogonal projection vector of $b$ on $a$ is $y$,then $|x-y|=$

If $A(3,2,3)$,$B(1,4,6)$,and $C(7,4,5)$ are the three vertices of a parallelogram $ABCD$,then the angle between its diagonal through $D$ and the side $DC$ is

Let $a = i + 2j + k$,$b = i - j + k$,$c = i + j - k$. $A$ vector in the plane of $a$ and $b$ has projection $\frac{1}{\sqrt{3}}$ on $c$. Then,one such vector is

The value of $b$ such that the scalar product of the vector $(i + j + k)$ with the unit vector parallel to the sum of the vectors $(2i + 4j - 5k)$ and $(bi + 2j + 3k)$ is $1$,is:

Let the position vectors of the vertices $A, B$ and $C$ of a tetrahedron $ABCD$ be $\hat{i}+2\hat{j}+\hat{k}$,$\hat{i}+3\hat{j}-2\hat{k}$ and $2\hat{i}+\hat{j}-\hat{k}$ respectively. The altitude from the vertex $D$ to the opposite face $ABC$ meets the median line segment through $A$ of the triangle $ABC$ at the point $E$. If the length of $AD$ is $\frac{\sqrt{110}}{3}$ and the volume of the tetrahedron is $\frac{\sqrt{805}}{6\sqrt{2}}$,then the position vector of $E$ is