1. ## Vector problem

Greetings,
I am struggling with a problem and I would appreciate if someone would explain how should I solve the following problem. "Angle between vector A and vector B is 120 degrees. |a|=3; |b|=4. What is the result of (3a-2b)*(a+2b)?"

Kostas.

2. ## Re: Vector problem

Originally Posted by Kostass
Greetings,
I am struggling with a problem and I would appreciate if someone would explain how should I solve the following problem. "Angle between vector A and vector B is 120 degrees. |a|=3; |b|=4. What is the result of (3a-2b)*(a+2b)?"
$\vec{A} \cdot \vec{B} = \left\| \vec{A} \right\|\left\| \vec{B} \right\|\cos \left( {\dfrac{{2\pi }}{3}} \right)$

3. ## Re: Vector problem

Hello, Kostass!

$\text{Angle between }\vec{a} \text{ and }\vec{b}\text{ is }120^o. \;\;|a|=3;\;|b|=4.$

$\text{Find: }\:(3a-2b)\cdot(a+2b)$

Preliminary calculations:

$a^2 \:=\:|a|^2 \:=\: 3^2 \quad\Rightarrow\quad a^2 \:=\: 9$

$b^2 \:=\:|b|^2 \:=\:4^2 \quad\Rightarrow\quad b^2 \:=\:16$

$\cos\theta \:=\:\frac{a\cdot b}{|a||b|} \;\;\;\Rightarrow\;\;\; \cos 120^o \:=\:\frac{a\cdot b}{3\cdot 4} \;\;\;\Rightarrow\;\;\; a\cdot b \:=\:12(\text{-}\tfrac{1}{2}) \;\;\;\Rightarrow\;\;\; a\cdot b \:=\:\text{-}6$

$(3a-2b)\cdot(a + 2b) \;=\; 3a^2 + 4ab - 4b^2$

. . . . . . . . . . . . . . $=\;3(9) + 4(\text{-}6) - 4(16)$

. . . . . . . . . . . . . . $=\;27 - 24 - 64$

. . . . . . . . . . . . . . $=\;-61$