If $\bar{a} = \bar{i} + \sqrt{11} \bar{j} - 2 \bar{k}$ and $\bar{b} = \bar{i} + \sqrt{11} \bar{j} - 10 \bar{k}$ are two vectors, then the component of $\bar{b}$ perpendicular to $\bar{a}$ is:

If $a, b$ and $c$ are three vectors such that $|a + b + c| = 1$, $c = \lambda(a \times b)$, and $|a| = \frac{1}{\sqrt{3}}$, $|b| = \frac{1}{\sqrt{2}}$, $|c| = \frac{1}{\sqrt{6}}$, then the angle between $a$ and $b$ is

If the position vectors of $A, B, C, D$ are $\bar{i}+2\bar{j}+2\bar{k}, 2\bar{i}-\bar{j}, \bar{i}+\bar{j}+3\bar{k}$ and $4\bar{j}+5\bar{k}$ respectively,then the quadrilateral $ABCD$ is a

In a triangle $PQR$,let $\vec{a}=\vec{QR}, \vec{b}=\vec{RP}$ and $\vec{c}=\vec{PQ}$. If $|\vec{a}|=3, |\vec{b}|=4$ and $\frac{\vec{a} \cdot(\vec{c}-\vec{b})}{\vec{c} \cdot(\vec{a}-\vec{b})}=\frac{|\vec{a}|}{|\vec{a}|+|\vec{b}|}$,then the value of $|\vec{a} \times \vec{b}|^2$ is.

Two adjacent sides of a parallelogram $ABCD$ are given by $\vec{AB} = 2\hat{i} + 10\hat{j} + 11\hat{k}$ and $\vec{AD} = -\hat{i} + 2\hat{j} + 2\hat{k}$. The side $\vec{AD}$ is rotated by an acute angle $\alpha$ in the plane of the parallelogram so that $\vec{AD}$ becomes $\vec{AD'}$. If $\vec{AD'}$ makes a right angle with the side $\vec{AB}$, then the cosine of the angle $\alpha$ is...

Let $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$,$\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$ and $\vec{c} = \hat{i} + \hat{j} - 2\hat{k}$ be three vectors. $A$ vector of the type $\vec{b} + \lambda \vec{c}$ for some scalar $\lambda$,whose projection on $\vec{a}$ is of magnitude $\sqrt{\frac{2}{3}}$ is