If $\vec{a}, \vec{b}$ and $\vec{c}$ are unit vectors satisfying $|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2=9$,then $|2 \vec{a}+5 \vec{b}+5 \vec{c}|$ is

In a triangle $ABC$,if $|\overrightarrow{BC}|=3$,$|\overrightarrow{AC}|=5$,and $|\overrightarrow{BA}|=7$,then the projection of the vector $\overrightarrow{BA}$ on $\overrightarrow{BC}$ is equal to:

The vector $\vec{a} = (\alpha, 2, \beta)$ lies in the plane of the vectors $\vec{b} = (1, 1, 0)$ and $\vec{c} = (0, 1, 1)$ and bisects the angle between $\vec{b}$ and $\vec{c}$. Then which one of the following gives the possible values of $\alpha$ and $\beta$?

Let $\vec{a} = 2\hat{i} + \lambda_{1}\hat{j} + 3\hat{k}$,$\vec{b} = 4\hat{i} + (3 - \lambda_{2})\hat{j} + 6\hat{k}$,and $\vec{c} = 3\hat{i} + 6\hat{j} + (\lambda_{3} - 1)\hat{k}$ be three vectors such that $\vec{b} = 2\vec{a}$ and $\vec{a}$ is perpendicular to $\vec{c}$. Then a possible value of $(\lambda_{1}, \lambda_{2}, \lambda_{3})$ is

The orthogonal projection of vector $a$ on vector $b$ is given by:

If $a$ and $b$ are unit vectors and $\alpha$ is the angle between them,then $a+b$ is a unit vector when $\cos \alpha=$