# Vectors Q (1 Viewer)

#### dsakvyilsa

##### Active Member
Hey, just need some help understanding the worked solution for this question. Thanks! (first image is Q, second is worked solution)

#### dsakvyilsa

##### Active Member
the part that throws me off is the line where |b|^2 comes in, I can understand how they used it to get the solution but I don't understand where it actually came from.

#### Masaken

##### Clown™
the part that throws me off is the line where |b|^2 comes in, I can understand how they used it to get the solution but I don't understand where it actually came from.
b.b = |b|^2
they're expanding it out but subbing that identity immediately

#### 5uckerberg

##### Well-Known Member
I believe that in my memory this is where it comes from
$\bg_white a\cdot{b}=|a||b|\cos{\theta}$.
This comes from the angle for vectors.
There, what you do is replace $\bg_white a$ with $\bg_white b$ and then what you will have is
$\bg_white b\cdot{b}=|b||b|\cos{\theta}$.
The interesting part is that since $\bg_white b$ lies in the same direction of $\bg_white b$ then $\bg_white \theta=0$ which leads to $\bg_white b\cdot{b}=|b|^{2}$ beccause $\bg_white |b|\times{|b|}=|b|^{2}$. Note this is for two vectors going in the same direction.
For vectors going in the opposite direction
$\bg_white b\cdot{-b}=|b||b|\cos{\pi}$.

Vectors in this case are parallel and collinear.

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#### dsakvyilsa

##### Active Member
b.b = |b|^2
they're expanding it out but subbing that identity immediately
wait omg I didn't even notice that the brackets had two b's I thought they were 4 different terms for some reason damn