If $a = i - j$ and $b = i + k$,then a unit vector coplanar with $a$ and $b$ and perpendicular to $a$ is

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.

Let $\sqrt{3} \hat{i} + \hat{j}$,$\hat{i} + \sqrt{3} \hat{j}$ and $\beta \hat{i} + (1 + \beta) \hat{j}$ respectively be the position vectors of the points $A, B$ and $C$ with respect to the origin $O$. If the distance of $C$ from the bisector of the acute angle between $OA$ and $OB$ is $\frac{3}{\sqrt{2}}$,then the sum of all possible values of $\beta$ is:

In $\triangle OAB$, $O(0, 0, 0)$, $A(6, 2, -3)$ and $B(4, 0, 3)$ are the vertices. Let $\vec{a}$ and $\vec{b}$ be position vectors of points $A$ and $B$ respectively. If $OM$ is the projection of $\vec{a}$ on $\vec{b}$, then the length $l(AM)$ is equal to...

Let $\vec{a}$ and $\vec{b}$ be two vectors. Let $|\vec{a}|=1, |\vec{b}|=4$ and $\vec{a} \cdot \vec{b}=2$. If $\vec{c}=(2 \vec{a} \times \vec{b})-3 \vec{b}$,then the value of $\vec{b} \cdot \vec{c}$ is

If the vectors $\hat{i}+3 \hat{j}+4 \hat{k}$ and $\lambda \hat{i}-4 \hat{j}+\hat{k}$ are orthogonal to each other,then $\lambda$ is equal to