If the angle between $\vec{a}$ and $\vec{b}$ is $\frac{2\pi}{3}$ and the projection of $\vec{a}$ in the direction of $\vec{b}$ is $-2$,then find $|\vec{a}|$.

If the vectors $3i + 2j + 8k$ and $2i + xj + k$ are perpendicular to each other,then $x = \dots$

If $\vec{a}, \vec{b}$ and $\vec{c}$ are unit vectors such that $\vec{a}$ is perpendicular to $\vec{b}$ and $\vec{c}$,and $|\vec{a} + \vec{b} + \vec{c}| = 1$,then the angle between $\vec{b}$ and $\vec{c}$ 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 $\vec{a} \neq \vec{0}, \vec{b} \neq \vec{0}.$

The projection of the vector $2i + j - 3k$ on the vector $i - 2j + k$ is:

Let $\vec{c}$ and $\vec{d}$ be vectors such that $|\vec{c}+\vec{d}|=\sqrt{29}$ and $\vec{c}\times(2\hat{i}+3\hat{j}+4\hat{k})=(2\hat{i}+3\hat{j}+4\hat{k})\times\vec{d}$. If $\lambda_1, \lambda_2$ $(\lambda_1 > \lambda_2)$ are the possible values of $(\vec{c}+\vec{d}) \cdot (-7\hat{i}+2\hat{j}+3\hat{k})$, then the equation $K^{2}x^{2}+(K^{2}-5K+\lambda_{1})xy+(3K+\frac{\lambda_{2}}{2})y^{2}-8x+12y+\lambda_{2}=0$ represents a circle, for $K$ equal to: