If $|\vec{a}|=4$ and $|\vec{b}|=5$,then the values of $k$ for which $\vec{a}+k \vec{b}$ is perpendicular to $\vec{a}-k \vec{b}$ are

Given the following simultaneous equations for vectors $x$ and $y$:
$(i) x + y = a$
$(ii) x \times y = b$
$(iii) x \cdot a = 1$
Then $x = ?, y = ?$

The vector $a + b$ bisects the angle between the vectors $a$ and $b$,if

Let $ABCD$ be a parallelogram where $\overrightarrow{AB} = \overrightarrow{a}$,$\overrightarrow{AD} = \overrightarrow{b}$,$|\overrightarrow{a}| = |\overrightarrow{b}| = 2$ and $|\overrightarrow{a} \times \overrightarrow{b}| + \overrightarrow{a} \cdot \overrightarrow{b} = \sqrt{2} |\overrightarrow{a}| |\overrightarrow{b}|$,where $\overrightarrow{a} \cdot \overrightarrow{b} > 0$. Then the area of this parallelogram is (in square units):

Let a unit vector $\hat{u}=x \hat{i}+y \hat{j}+z \hat{k}$ make angles $\frac{\pi}{2}, \frac{\pi}{3}$ and $\frac{2 \pi}{3}$ with the vectors $\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{k}, \frac{1}{\sqrt{2}} \hat{j}+\frac{1}{\sqrt{2}} \hat{k}$ and $\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}$ respectively. If $\overrightarrow{v}=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}+\frac{1}{\sqrt{2}} \hat{k}$,then $|\hat{u}-\overrightarrow{v}|^2$ is equal to

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