# Thread: Vector question - any hints welcome

1. ## Vector question - any hints welcome

Ok, if I know that $\displaystyle 3 \vec c = \vec a + 2 \vec b$ and $\displaystyle \vec d = - \vec a + 2 \vec b$ and that $\displaystyle | \vec a | = 2 | \vec b |$. (o is origin)

How do I prove that $\displaystyle |<cod| = \frac{ \pi }{2}$ ?

2. Originally Posted by FreeT
Ok, if I know that $\displaystyle 3 \vec c = \vec a + 2 \vec b$ and $\displaystyle \vec d = - \vec a + 2 \vec b$ and that $\displaystyle | \vec a | = 2 | \vec b |$. (o is origin)

How do I prove that $\displaystyle |<cod| = \frac{ \pi }{2}$ ?

3. Thanks, I did $\displaystyle ( \frac {\vec a}{3} + \frac {2 \vec b}{3}).( - \vec a + 2 \vec b)$ and said that $\displaystyle \vec a= x_1 i + y_1 j$ and that $\displaystyle \vec b= x_2 i + y_2 j$
It worked down to $\displaystyle 1/3 (|x_1|^2 - |y_1|^2) + 4/3 (|x_2|^2 + |y_2|^2)$ and now I need to prove this equal to zero.
$\displaystyle \left( {\overrightarrow a + 2\overrightarrow b } \right) \cdot \left( { - \overrightarrow a + 2\overrightarrow b } \right) = - \overrightarrow a \cdot \overrightarrow a + 4\overrightarrow b \cdot \overrightarrow b = - \left\| {\overrightarrow a } \right\|^2 + 4\left\| {\overrightarrow b } \right\|^2 = ?$