The area inside the parabola $y^2=4 \mathrm{a} x$, between the lines $x=\mathrm{a}$ and $x=4 \mathrm{a}$ is equal to

$\int_2^4 \frac{\log x^2}{\log x^2+\log \left(36-12 x+x^2\right)} \mathrm{d} x$ is equal to

Let $\bar{a}, \bar{b}$, and $\bar{c}$ be unit vectors. Suppose that $\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}=\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}=0$ and if the angle between $\overline{\mathrm{b}}$ and $\overline{\mathrm{c}}$ is $\frac{\pi}{6}$, then $\overline{\mathrm{a}}$ is

If $\overline{\mathrm{a}}$ and $\overline{\mathrm{b}}$ are unit vectors such that $|\overline{\mathrm{a}}+\overline{\mathrm{b}}|=\sqrt{3}$, then the angle between $\bar{a}$ and $\bar{b}$ is