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:

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

If $a \perp b$ and $(a+b) \perp (a+mb)$,then $m$ is equal to

If $\overrightarrow{a}=-\hat{i}+\hat{j}+2 \hat{k}$, $\overrightarrow{b}=2 \hat{i}-\hat{j}-\hat{k}$ and $\overrightarrow{c}=-2 \hat{i}+\hat{j}+3 \hat{k}$, then the angle between $2 \overrightarrow{a}-\overrightarrow{c}$ and $\overrightarrow{a}+\overrightarrow{b}$ is

If $\vec{a}$ is a nonzero vector such that its projections on the vectors $2 \hat{i}-\hat{j}+2 \hat{k}$,$\hat{i}+2 \hat{j}-2 \hat{k}$,and $\hat{k}$ are equal,then a unit vector along $\vec{a}$ is:

An arc $PQ$ of a circle subtends a right angle at its centre $O$. The midpoint of the arc $PQ$ is $R$. If $\vec{OP}=\vec{u}$,$\vec{OR}=\vec{v}$ and $\vec{OQ}=\alpha \vec{u}+\beta \vec{v}$,then $\alpha, \beta^2$ are the roots of the equation