Vector $\vec{a} + 3\vec{b}$ is perpendicular to $7\vec{a} - 5\vec{b}$ and $\vec{a} - 5\vec{b}$ is perpendicular to $7\vec{a} + 3\vec{b}$. The angle between non-zero vectors $\vec{a}$ and $\vec{b}$ is:

The vector $\overline{a}=\alpha \hat{i}+2 \hat{j}+\beta \hat{k}$ lies in the plane of the vectors $\bar{b}=\hat{i}+\hat{j}$ and $\bar{c}=\hat{j}+\hat{k}$ and bisects the angle between $\bar{b}$ and $\bar{c}$. Then which one of the following gives possible values of $\alpha$ and $\beta$?

Let $\vec{a}=6 \hat{i}+9 \hat{j}+12 \hat{k}$,$\vec{b}=\alpha \hat{i}+11 \hat{j}-2 \hat{k}$ and $\vec{c}$ be vectors such that $\vec{a} \times \vec{c}=\vec{a} \times \vec{b}$. If $\vec{a} \cdot \vec{c}=-12$ and $\vec{c} \cdot (\hat{i}-2 \hat{j}+\hat{k})=5$,then $\vec{c} \cdot (\hat{i}+\hat{j}+\hat{k})$ is equal to $.............$.

The points $A(5, -1, 1)$,$B(7, -4, 7)$,$C(1, -6, 10)$,and $D(-1, -3, 4)$ are vertices of a:

Suppose $\overrightarrow{a}=\lambda \hat{i}-7 \hat{j}+3 \hat{k}$ and $\overrightarrow{b}=\lambda \hat{i}+\hat{j}+2 \lambda \hat{k}$. If the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is greater than $90^{\circ}$, then $\lambda$ satisfies the inequality:

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