If $a=\hat{i}+\hat{j}+t \hat{k}$ and $b=\hat{i}+2 \hat{j}+3 \hat{k}$, then the values of $t$ for which $(a+b)$ and $(a-b)$ are perpendicular are:

In quadrilateral $ABCD$,$\overrightarrow{AB}=\vec{a}$,$\overrightarrow{BC}=\vec{b}$,$\overrightarrow{DA}=\vec{a}-\vec{b}$. $M$ is the midpoint of $BC$ and $X$ is a point on $DM$ such that $\overrightarrow{DX}=\frac{4}{5} \overrightarrow{DM}$. Then the points $A, X$ and $C$:

The projection of the vector $\vec{a} = \hat{i} - 2\hat{j} + \hat{k}$ on the vector $\vec{b} = 4\hat{i} - 4\hat{j} + 7\hat{k}$ is:

Let $a = \hat{i} + \hat{j} + \hat{k}$,$b = 2\hat{i} + 2\hat{j} + \hat{k}$,and $c = 5\hat{i} + \hat{j} - \hat{k}$ be three vectors. The area of the region formed by the set of points whose position vectors $\vec{r}$ satisfy the equations $\vec{r} \cdot \vec{a} = 5$ and $|\vec{r} - \vec{b}| + |\vec{r} - \vec{c}| = 4$ is closest to which integer?

Let $|\vec{a}|=2, |\vec{b}|=3$ and the angle between $\vec{a}$ and $\vec{b}$ be $\frac{\pi}{3}$. If a parallelogram is constructed with adjacent sides $2\vec{a}+3\vec{b}$ and $\vec{a}-\vec{b}$,then its shorter diagonal is of length

The cosine of the angle between any two diagonals of a cube is