# Question #b35f7

Jul 17, 2017

The angle is $= \frac{2}{3} \pi$

#### Explanation:

$\vec{a}$ and $\left(\vec{a} + \vec{b}\right)$ are perpendicular

Therefore, the dot product is $= 0$

$\vec{a} . \left(\vec{a} + \vec{b}\right) = \vec{a} . \vec{a} + \vec{a} . \vec{b} = 0$

$\vec{a} . \vec{a} = | | \vec{a} | {|}^{2}$

But,

$| | \vec{a} | | = \frac{1}{2} | | \vec{b} | |$

The angle between $\vec{a}$ and $\vec{b}$ is given by the dot product

$\vec{a} . \vec{b} = | | \vec{a} | | \cdot | | \vec{b} | | \cos \theta$

$\cos \theta = \frac{\vec{a} . \vec{b}}{| | \vec{a} | | \cdot | | \vec{b} | |}$

$= \frac{- | | \vec{a} | {|}^{2}}{2 | | \vec{a} | {|}^{2}}$

Therefore,

$\theta = \frac{2}{3} \pi$