A vector perpendicular to a plane containing a triangle ABC having vertices as $A(1,1,0)$, $B(2,1,1)$ and $C(0,3,2)$, is:

If $\vec{a}$ and $\vec{b}$ are two perpendicular vectors such that $|\vec{a}| = 3$, $|\vec{b}| = 4$ and $\theta$ is the angle between $\vec{a}$ and $(\vec{a} - \vec{b})$, then $\cos\theta$ is equal to:

If $|\vec{a}| = 3|\vec{b}|$, $|\vec{b}| = 2$ and angle between $\vec{a}$ and $\vec{b}$ is $60^\circ$, then $|\vec{a} - \vec{b}|$ is equal to:

The position vectors of the points A and B are $\vec{a}=2\hat{i}-3\hat{j}+2\hat{k}$ and $\vec{b}=2\hat{i}+3\hat{j}+\hat{k}$ respectively, then area of triangle OAB is:

Vector projection of $7\hat i+\hat j+4\hat k$ on the vector $2\hat i+6\hat j-3\hat k$ is

If $\vec{a}.\vec{b} = \vec{a}.\vec{c}$ and $\vec{a}\times\vec{b} = \vec{a}\times\vec{c}$, $\vec{a}\neq\vec{0}$, then

If $\theta$ is the acute angle between two unit vectors $\vec{a}$ and $\vec{b}$, then $\cos\frac{\theta}{2} =$