The vector projection of $\overline{AB}$ on $\overline{CD}$,where $A \equiv(2,-3,0), B \equiv(1,-4,-2), C \equiv(4,6,8)$ and $D \equiv(7,0,10)$,is

If $\vec{a}$ and $\vec{b}$ are two vectors such that $|\vec{a}|=3, |\vec{b}|=4, |\vec{a}+\vec{b}|=\sqrt{37}, |\vec{a}-\vec{b}|=k$ and the angle between $\vec{a}$ and $\vec{b}$ is $\theta$, then find the value of $\frac{4}{13}(k \sin \theta)^2$.

Let $\overrightarrow{a} = \hat{i} + 2\hat{j} - 3\hat{k}$ and $\overrightarrow{b} = 2\hat{i} - 3\hat{j} + 5\hat{k}$. If $\overrightarrow{r} \times \overrightarrow{a} = \overrightarrow{b} \times \overrightarrow{r}$,$\overrightarrow{r} \cdot (\alpha\hat{i} + 2\hat{j} + \hat{k}) = 3$ and $\overrightarrow{r} \cdot (2\hat{i} + 5\hat{j} - \alpha\hat{k}) = -1$,where $\alpha \in R$,then the value of $\alpha + |\overrightarrow{r}|^{2}$ is equal to:

The angle between two vectors $\vec{a} = \hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + \hat{j} - \hat{k}$ is . . . . . . .

Let $\vec{a}=\hat{i}-2\hat{j}+3\hat{k}$, $\vec{b}=2\hat{i}+\hat{j}-\hat{k}$, $\vec{c}=\lambda\hat{i}+\hat{j}+\hat{k}$ and $\vec{v}=\vec{a}\times\vec{b}$. If $\vec{v} \cdot \vec{c}=11$ and the length of the projection of $\vec{b}$ on $\vec{c}$ is $p$, then $9p^{2}$ is equal to:

If $4 \hat{i}+7 \hat{j}+8 \hat{k}$,$2 \hat{i}+3 \hat{j}+4 \hat{k}$,and $2 \hat{i}+5 \hat{j}+7 \hat{k}$ are respectively the position vectors of the vertices $A, B, C$ of $\triangle ABC$,then the position vector of the point where the bisector of angle $A$ meets $BC$ is