Vectors Q (1 Viewer)

dsakvyilsa · Jan 25, 2023

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

dsakvyilsa · Jan 25, 2023

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 · Jan 25, 2023

dsakvyilsa said:
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 · Jan 25, 2023

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

Vectors in this case are parallel and collinear.

dsakvyilsa · Jan 25, 2023

Masaken said:
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

Vectors Q (1 Viewer)

dsakvyilsa

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dsakvyilsa

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Masaken

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5uckerberg

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dsakvyilsa

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