If $|a|=2$ and $|b|=3$ and the angle between $a$ and $b$ is $120^{\circ}$,then the length of the vector $\left|\frac{a}{2}-\frac{b}{3}\right|$ is

Let $a, b$ and $c$ be unit vectors such that $a$ is perpendicular to the plane containing $b$ and $c$ and the angle between $b$ and $c$ is $\frac{\pi}{3}$. Then,$|a+b+c|=$

Let $|\vec{a}| = |\vec{b}| = 1$ and $|\vec{a} + \vec{b}| = \sqrt{3}$. If $\vec{c}$ is a vector satisfying the condition $\vec{c} - \vec{a} - 2\vec{b} = 3(\vec{a} \times \vec{b})$,then $\vec{c} \cdot \vec{b} = \dots$ (in $/2$)

If $\overrightarrow{F_1} = i - j + k,$ $\overrightarrow{F_2} = -i + 2j - k,$ $\overrightarrow{F_3} = j - k,$ $\vec{A} = 4i - 3j - 2k$ and $\vec{B} = 6i + j - 3k,$ then the scalar product of $(\overrightarrow{F_1} + \overrightarrow{F_2} + \overrightarrow{F_3})$ and $\overrightarrow{AB}$ will be:

If $a, b, c$ are non-coplanar vectors,then for what value of $m$ are the three points with position vectors $-2b + 3c$,$2a + mb - 4c$,and $-7b + 10c$ collinear?

If $a$,$b$,and $c$ are unit vectors,then $|a - b|^2 + |b - c|^2 + |c - a|^2$ does not exceed