Scalar projection of the line segment joining the points $A(-2,0,3)$ and $B(1,4,2)$ on the line whose direction ratios are $6,-2,3$ is

If $\vec{a} = 2\hat{i} - \hat{j} - 2\hat{k}$ and $\vec{b} = 6\hat{i} + 2\hat{j} - 3\hat{k}$ are two vectors,and we consider a vector $\vec{c} = \vec{a} + t\vec{b}$,find the value of $t$ such that the magnitude $|\vec{c}|$ is minimum.

Find the angle between the vectors $2 \hat{i}-\hat{j}+\hat{k}$ and $3 \hat{i}+4 \hat{j}-\hat{k}$.

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 $\vec{a} = -\hat{i} - \hat{k}$,$\vec{b} = -\hat{i} + \hat{j}$,and $\vec{c} = \hat{i} + 2\hat{j} + 3\hat{k}$ be three given vectors. If $\vec{r}$ is a vector such that $\vec{r} \times \vec{b} = \vec{c} \times \vec{b}$ and $\vec{r} \cdot \vec{a} = 0$,then the value of $\vec{r} \cdot \vec{b}$ is

In a plane, $\vec{a}$ and $\vec{b}$ are the position vectors of two points $A$ and $B$ respectively. $A$ point $P$ with position vector $\vec{r}$ moves on that plane in such a way that $|\vec{r}-\vec{a}| - |\vec{r}-\vec{b}| = c$ (where $c$ is a real constant). The locus of $P$ is a conic section whose eccentricity is: