If $\vec{a}$ and $\vec{b}$ are two vectors such that $|\vec{a}|=2$,$|\vec{b}|=3$ and $\vec{a}+t \vec{b}$ and $\vec{a}-t \vec{b}$ are perpendicular,where $t$ is a positive scalar,then

Consider a $\triangle ABC$ where $A(1,3,2)$,$B(-2,8,0)$,and $C(3,6,7)$. If the angle bisector of $\angle BAC$ meets the line $BC$ at $D$,then the length of the projection of the vector $\overrightarrow{AD}$ on the vector $\overrightarrow{AC}$ is:

Let $\vec{a} = 2\hat{i} + \hat{k}$, $\vec{b} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{c} = 4\hat{i} - 3\hat{j} + 7\hat{k}$. 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 $\vec{r} \cdot \vec{c} = $

Let $u = -2 \hat{i} + 2 \hat{j} + \hat{k}$ and $v = \hat{i} - 2 \hat{j} + 2 \hat{k}$. Then the component of $v$ on $u$ is

The vector $2\hat{i} + a\hat{j} + \hat{k}$ is perpendicular to the vector $2\hat{i} - \hat{j} - \hat{k},$ if $a = $

If $|a| = 3$,$|b| = 4$ and the angle between $a$ and $b$ is $120^\circ$,then $|4a + 3b| = $