Let $P, Q, R$ and $S$ be the points on the plane with position vectors $-2 \hat{i}-\hat{j}, 4 \hat{i}, 3 \hat{i}+3 \hat{j}$ and $-3 \hat{i}+2 \hat{j}$ respectively. The quadrilateral $PQRS$ must be a

If $a$ is a unit vector, then $|a \times \hat{i}|^2+|a \times \hat{j}|^2+|a \times \hat{k}|^2=$

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:

Let $\overrightarrow{x}$ be a vector in the plane containing vectors $\overrightarrow{a} = 2\hat{i} - \hat{j} + \hat{k}$ and $\overrightarrow{b} = \hat{i} + 2\hat{j} - \hat{k}$. If the vector $\overrightarrow{x}$ is perpendicular to $(3\hat{i} + 2\hat{j} - \hat{k})$ and its projection on $\overrightarrow{a}$ is $\frac{17\sqrt{6}}{2}$,then the value of $|\overrightarrow{x}|^{2}$ is equal to ...... .

If $\vec{a} = 2\hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = 6\hat{i} - 3\hat{j} + 2\hat{k}$,then find $\vec{a} \cdot \vec{b}$.

If $\vec{a} = 2\hat{i} - \hat{j} - 2\hat{k}$ and $\vec{b} = 6\hat{i} + 2\hat{j} - 3\hat{k}$ are two vectors,and we consider a vector $\vec{c} = \vec{a} + t\vec{b}$,find the value of $t$ such that the magnitude $|\vec{c}|$ is minimum.