1. Gr 12. Vector problem!

Given a and b unit vectors.
a) if the angle between them is 60deg, calculate (6a + b) dot (a - 2b)
b) if |a + b| = root 3, determine (2a - 5b) dot (b + 3a)
a and b are unit vectors

2. Originally Posted by narutoblaze
Given a and b unit vectors.
a) if the angle between them is 60deg, calculate (6a + b) dot (a - 2b)
b) if |a + b| = root 3, determine (2a - 5b) dot (b + 3a)
a and b are unit vectors

Dont forget that the dot product is a LINEAR.

$\displaystyle (\vec a + \vec b) \cdot \vec c =\vec a \cdot \vec c +\vec b \cdot \vec c$ for all vectors in you space and

for any real number $\displaystyle \alpha,\beta \in \mathbb{R}$
$\displaystyle (\alpha \vec a) \cdot (\beta \vec b)=\alpha \beta (\vec a \cdot \vec b)$

Also the defintion $\displaystyle \vec a \cdot \vec b= ||a||||b||\cos(\theta)$

using these two properties we get

$\displaystyle (2\vec a-5 \vec b) \cdot (\vec b+\vec 3a) =(2\vec a)\cdot (\vec b+\vec 3a)+(-5\vec b)\cdot (\vec b+\vec 3a)=$

$\displaystyle (2\vec a)\cdot (\vec b)+(2\vec a)\cdot (3\vec a)+(-5\vec b)\cdot (\vec b)+(-5\vec b)\cdot (\vec 3a)=$

$\displaystyle 2(\vec a \cdot \vec b)+6(\vec a \cdot \vec a)-5(\vec b \cdot \vec b)-15 (\vec b \cdot \vec a)$

Since a and be are both unit vectors we get

$\displaystyle 2\cos(60)+6\cos(0)-5\cos(0)-15\cos(60)$
Just simplify from here