If the vectors $\vec{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}$,$\vec{b}=-\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{c}=3 \hat{i}+\hat{j}$ are such that $(\vec{a}+\lambda \vec{b})$ is perpendicular to $\vec{c}$,then the value of $\lambda$ is

Two forces $\vec{F_1} = 2\hat{i} - 5\hat{j} + 6\hat{k}$ and $\vec{F_2} = -\hat{i} + 2\hat{j} - \hat{k}$ act on a particle. The particle is displaced from point $P(4\hat{i} - 3\hat{j} + 2\hat{k})$ to point $Q(6\hat{i} + \hat{j} + 3\hat{k})$. The work done by the forces is ............. units.

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three unit vectors such that $|\vec{a}-\vec{b}|^{2}+|\vec{a}-\vec{c}|^{2}=8$. Then $|\vec{a}+2\vec{b}|^{2}+|\vec{a}+2\vec{c}|^{2}$ is equal to

If the angle $\theta$ between the vectors $\overrightarrow{a}=2 x^2 \hat{i}+4 x \hat{j}+\hat{k}$ and $\overrightarrow{b}=7 \hat{i}-2 \hat{j}+x \hat{k}$ is such that $90^{\circ} < \theta < 180^{\circ}$,then $x$ lies in the interval

The area of the rectangle having vertices $P, Q, R, S$ with position vectors $-\hat{i}+\hat{j}+\hat{k}, \hat{i}+\hat{j}+\hat{k}, \hat{i}-\hat{j}+\hat{k}, -\hat{i}-\hat{j}+\hat{k}$ respectively is

Prove that $(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=|\vec{a}|^{2}+|\vec{b}|^{2},$ if and only if $\vec{a}$ and $\vec{b}$ are perpendicular,given $\vec{a} \neq \vec{0}, \vec{b} \neq \vec{0}.$