Question

please solve it step by step.. ​

Answer 1

Answer:

$\sqrt 2$

Explanation:

If $\vec{a}+\vec{b}$ is perpendicular to $\vec a$ then $(\vec a + \vec b) \cdot \vec a = 0$. By the distrubtive property of the dot product we get that $(\vec a\cdot \vec a) + (\vec a \cdot \vec b) = 0$. This is then the same as $| \vec a |^2 + (\vec a \cdot \vec b) = 0$, hence $| \vec a | ^2 = - (\vec a + \vec b)$.

Similarily, if $(2\vec a + \vec b)$ is perpendicular to $\vec b$, then $(2\vec a+\vec b )\cdot \vec b = 0$. Again, by the distributive property we get that $(2\vec a \cdot \vec b ) + (\vec b \cdot \vec b) = 0$. This is equivalent to $2(\vec a \cdot \vec b) + |\vec b |^2 = 0$, hence $| \vec b|^2 = -2 (\vec a \cdot \vec b)$.

Then we get that $\frac{|\vec b|^2}{|\vec a|^2} = \frac{-2(\vec a \cdot \vec b)}{-( \vec a \cdot \vec b)}=2$.

Taking The square root on both sides then gives

$\frac{|\vec b |}{| \vec a | } = \sqrt 2$.