If $a \cdot b = a \cdot c$,$a \times b = a \times c$ and $a \neq 0$,then

Three non-coplanar vectors $\bar{a}, \bar{b}, \bar{c}$ are the coterminous edges of a parallelepiped. If $\bar{a}$ and $\bar{b}$ determine the base of the parallelepiped, then its height is:

Suppose $\overrightarrow{a}=\lambda \hat{i}-7 \hat{j}+3 \hat{k}$ and $\overrightarrow{b}=\lambda \hat{i}+\hat{j}+2 \lambda \hat{k}$. If the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is greater than $90^{\circ}$, then $\lambda$ satisfies the inequality:

If $\vec{a}=\hat{i}+(\tan \theta) \hat{j}+\left(\frac{3}{\sqrt{\sin \frac{\theta}{2}}}\right) \hat{k}$ and $\vec{b}=\tan \theta(\hat{j}-\hat{i})-\left(2 \sqrt{\sin \frac{\theta}{2}}\right) \hat{k}$ are orthogonal vectors and $\vec{c}=(\sin 2 \theta) \hat{i}-2 \hat{j}+2 \hat{k}$ makes an obtuse angle with $X$-axis, then $\theta=$

If $\vec{a} = \hat{i} - 2\hat{j} + 2\hat{k}$ and $\vec{b} = 2\hat{i} - 3\hat{j} + \hat{k}$, then the component of $\vec{b}$ perpendicular to $\vec{a}$ is

If $a, b, c$ are lengths of the sides $BC, CA, AB$ respectively of $\triangle ABC$ and $H$ is any point in the plane of $\triangle ABC$ such that $a \vec{AH} + b \vec{BH} + c \vec{CH} = \vec{0}$,then $H$ is the