If $\overline{a}$ and $\overline{b}$ are two unit vectors such that $\overline{a}+2\overline{b}$ and $5\overline{a}-4\overline{b}$ are perpendicular to each other,then the angle between $\overline{a}$ and $\overline{b}$ is

If $a = 4i + 6j$ and $b = 3j + 4k$,what is the vector component of $a$ along the direction of $b$?

The figure formed by the four points $i + j - k$,$2i + 3j$,$3i + 5j - 2k$,and $k - j$ is:

If two vectors $\vec{a}$ and $\vec{b}$ which are perpendicular to each other are such that $|\vec{a}|=8$ and $|\vec{b}|=3$, then $|\vec{a}-2\vec{b}|=$

Let $\vec{a}=\hat{i}+2\hat{j}+\hat{k}$ and $\vec{b}=2\hat{i}+\hat{j}-\hat{k}$. Let $\hat{c}$ be a unit vector in the plane of the vectors $\vec{a}$ and $\vec{b}$ and be perpendicular to $\vec{a}$. Then such a vector $\hat{c}$ is :

Let $\overrightarrow{a}=\hat{i}+\alpha \hat{j}+\beta \hat{k}$,where $\alpha, \beta \in R$. Let a vector $\overrightarrow{b}$ be such that the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{4}$ and $|\vec{b}|^2=6$. If $\vec{a} \cdot \vec{b}=3 \sqrt{2}$,then the value of $(\alpha^2+\beta^2)|\vec{a} \times \vec{b}|^2$ is equal to