For any two non-zero vectors $\vec{a}$ and $\vec{b}$,$(a \vec{b} + b \vec{a}) \cdot (a \vec{b} - b \vec{a})$ is equal to:

If either vector $\vec{a}=\vec{0}$ or $\vec{b}=\vec{0},$ then $\vec{a} \cdot \vec{b}=0 .$ But the converse need not be true. Justify your answer with an example.

If the angle between the vectors $\vec{a} = 2\lambda^2 \hat{i} + 4\lambda \hat{j} + \hat{k}$ and $\vec{b} = 7\hat{i} - 2\hat{j} + \lambda \hat{k}$ is obtuse,then the values of $\lambda$ lie in:

Let $a$,$b$,and $c$ be $3$ non-zero vectors such that no $2$ of these are collinear. If vector $a + 2b$ is collinear with $c$ and $b + 3c$ is collinear with $a$,then $a + 2b + 6c$ equals:

The position vectors of the vertices $A, B$ and $C$ of a triangle are $2 \hat{i}-3 \hat{j}+3 \hat{k}$,$2 \hat{i}+2 \hat{j}+3 \hat{k}$ and $-\hat{i}+\hat{j}+3 \hat{k}$ respectively. Let $l$ denote the length of the angle bisector $AD$ of $\angle BAC$,where $D$ is on the line segment $BC$. Then $2 l^2$ equals:

$A$ force of magnitude $6$ acts along the vector $(9, 6, -2)$ and passes through a point $A(4, -1, -7)$. The moment of the force about the point $O(1, -3, 2)$ is