If $\vec{a}$ and $\vec{b}$ are two unit vectors inclined at an angle $\theta$, then the value of $|\vec{a}-\vec{b}|$ is

If $\vec{a}, \vec{b}$ and $\vec{c}$ are three vectors such that $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$, where $\vec{a}$ and $\vec{b}$ are unit vectors and $|\vec{c}|=2$, then the angle between the vectors $\vec{b}$ and $\vec{c}$ is :

If $(\vec{a}-\vec{b}) \cdot(\vec{a}+\vec{b})=27$ and $|\vec{a}|=2|\vec{b}|$, then $|\vec{b}|$ is :

The vectors from origin to 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 the area (in Sq. unit) of triangle OAB is:

let $\vec{a}$ be a non-zero vector of magnitude '$a$' and $\lambda$ is a non-zero scalar, then $\lambda\vec{a}$ is a unit vector if

If $\vec{a}$ is a unit vector and $(\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = 8$ then $|\vec{x}|$ is

Let $\vec{a},\vec{b},\vec{c}$ are three vectors such that $|\vec{a}|=3, |\vec{b}|=2, |\vec{c}|=1$. If the projection of $\vec{b}$ along $\vec{c}$ is equal to the projection of $\vec{a}$ along $\vec{c}$ and $\vec{b} \perp \vec{a}$ , then the value of $|2\vec{a}-2\vec{b}+3\vec{c}|$ is equal to