If the position vectors of the vertices $A, B$ and $C$ of $\triangle ABC$ are $\hat{i}+2\hat{j}-5\hat{k}$,$-2\hat{i}+2\hat{j}+\hat{k}$ and $2\hat{i}+\hat{j}-\hat{k}$ respectively,then $\angle B=$

If $A(3,2,3)$, $B(1,4,6)$, and $C(7,4,5)$ are the three vertices of a parallelogram $ABCD$, then the angle between its diagonal through $D$ and the side $DC$ is

Let $\vec{a}=\alpha \hat{i}+\hat{j}+\beta \hat{k}$ and $\vec{b}=3 \hat{i}-5 \hat{j}+4 \hat{k}$ be two vectors,such that $\vec{a} \times \vec{b}=-\hat{i}+9 \hat{j}+12 \hat{k}$. Then the projection of $\vec{b}-2 \vec{a}$ on $\vec{b}+\vec{a}$ is equal to.

If $\bar{a}=\hat{i}+2 \hat{j}-3 \hat{k}$,$\bar{b}=3 \hat{i}-\hat{j}+2 \hat{k}$,$\bar{c}=\hat{i}+3 \hat{j}+\hat{k}$ and $\bar{a}+\lambda \bar{b}$ is perpendicular to $\bar{c}$,then $\lambda=$

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:

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 $\frac{4}{13}(k \sin \theta)^2$.