1. vectors

Vectors $\vec{a}$ and $\vec{b}$ both have the same length, which is 1. The angle between them is $60^0$.

Find $t$ so that $2\vec{a} + \vec{b}$ and $t\vec{a} + 5\vec{b}$ are
perpendicular

I got that far:

$cos \phi = \frac{\vec{a} \cdot \vec{b}}{\mid \vec{a} \mid \cdot \mid \vec{b} \mid} = \frac{\vec{a} \cdot \vec{b}} {1 \cdot 1} \Rightarrow \phi = \frac{1}{2} \cdot 1 \cdot 1 = \vec{a} \cdot \vec{b}$

then i put $2\vec{a} = \frac{1}{\vec{b}}$ into

$(2\vec{a} + \vec{b}) \cdot (t\vec{a} + 5\vec{b}) = 0$ and got that:

$(\frac{1}{\vec{b}} + \vec{b}) \cdot (t \cdot \frac{1}{2\vec{b}} + 5\vec{b}) = 0$

and continued for a couple more lines hoping it would become easier but i got nothing useful..how do you solve this?

2. Originally Posted by metlx
Vectors $\vec{a}$ and $\vec{b}$ both have the same length, which is 1. The angle between them is $60^0$.

Find $t$ so that $2\vec{a} + \vec{b}$ and $t\vec{a} + 5\vec{b}$ are
perpendicular

I got that far:

$cos \phi = \frac{\vec{a} \cdot \vec{b}}{\mid \vec{a} \mid \cdot \mid \vec{b} \mid} = \frac{\vec{a} \cdot \vec{b}} {1 \cdot 1} \Rightarrow \phi = \frac{1}{2} \cdot 1 \cdot 1 = \vec{a} \cdot \vec{b}$

then i put $2\vec{a} = \frac{1}{\vec{b}}$ into

$(2\vec{a} + \vec{b}) \cdot (t\vec{a} + 5\vec{b}) = 0$ and got that:

$(\frac{1}{\vec{b}} + \vec{b}) \cdot (t \cdot \frac{1}{2\vec{b}} + 5\vec{b}) = 0$

and continued for a couple more lines hoping it would become easier but i got nothing useful..how do you solve this?
$\left( {2\overrightarrow a + \overrightarrow b } \right) \cdot \left( {t\overrightarrow a + 5\overrightarrow b } \right) = 2t\overrightarrow a \overrightarrow { \cdot a} + 2\overrightarrow a \cdot \overrightarrow b + t\overrightarrow a \cdot \overrightarrow b + 5\overrightarrow b \cdot \overrightarrow b$