If $a, b, c, d$ are the position vectors of the points $A, B, C$ and $D$ respectively,referred to the same origin $O$,such that no three of these points are collinear and $a + c = b + d$,then the quadrilateral $ABCD$ is a

If $A(3,4,5), B(4,6,3), C(-1,2,4)$ and $D(1,0,5)$ are such that the angle between the lines $DC$ and $AB$ is $\theta$,then $\cos \theta$ 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 $|\vec{r}|$.

Let $\vec{a}, \vec{b}, \vec{c}$ be the position vectors of the vertices of a triangle $ABC$. Through the vertices,lines are drawn parallel to the sides to form the triangle $A'B'C'$. Then the centroid of $\Delta A'B'C'$ is

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

The projection of $\overline{AB}$ on $\overline{CD}$,where $A \equiv (2, -3, 0)$,$B \equiv (1, -4, -2)$,$C \equiv (4, 6, 8)$,and $D \equiv (7, 0, 10)$ is