# Scalar Product

H4NS
Hello!

I am preparing for an exam, I didn't really had much time for, and it would be nice of you if you could help me!

## Homework Statement

Draw a figure, so that the following is true: (AC - AB) * AB = 0

2. The attempt at a solution
Since I had to miss some classes, I don't really have an idea on how to start.
I don't know if AC*AB=AB² is teh right way to go

Gold Member
Since the dot product is zero, the vectors are at 90° angle with each other. (considering non-zero vectors). Let us draw a vector AB. Now draw AC vector such that AC's projection on AB is equal and opposite to AB. Meaning, if you consider any angle θ between AB and AC, then |AC|cosθ=|AB|. Then the resultant of AC-AB will be |AC|sinθ, perpendicular to AB, thus their dot product is zero.

H4NS
Since the dot product is zero, the vectors are at 90° angle with each other. (considering non-zero vectors). Let us draw a vector AB. Now draw AC vector such that AC's projection on AB is equal and opposite to AB. Meaning, if you consider any angle θ between AB and AC, then |AC|cosθ=|AB|. Then the resultant of AC-AB will be |AC|sinθ, perpendicular to AB, thus their dot product is zero.
Thanks a lot for your input, but could you please expand this a little more, I don't quite understand

Staff Emeritus
Homework Helper
Gold Member
Hello!

I am preparing for an exam, I didn't really had much time for, and it would be nice of you if you could help me!

## Homework Statement

Draw a figure, so that the following is true: (AC - AB) * AB = 0

2. The attempt at a solution
Since I had to miss some classes, I don't really have an idea on how to start.
I don't know if AC*AB=AB² is teh right way to go

I assume that you need the scalar product, $\displaystyle (\vec{\text{AC}}-\vec{\text{AB}})\cdot\vec{\text{AB}}\ .$

That's the scalar product of two vectors, $\displaystyle \vec{\text{AC}}-\vec{\text{AB}}$ and $\displaystyle \vec{\text{AB}}\ .$

For this product to be zero, either the two vectors must be perpendicular to each other, as AGNuke pointed out, or else the magnitude of one of them must be zero.

H4NS
I assume that you need the scalar product, $\displaystyle (\vec{\text{AC}}-\vec{\text{AB}})\cdot\vec{\text{AB}}\ .$

That's the scalar product of two vectors, $\displaystyle \vec{\text{AC}}-\vec{\text{AB}}$ and $\displaystyle \vec{\text{AB}}\ .$

For this product to be zero, either the two vectors must be perpendicular to each other, as AGNuke pointed out, or else the magnitude of one of them must be zero.

OK, I assume the figure might be a 90° triangle, with AB and AC being the cathetes.

Let's suppose A: (1|1), B: (2|1) and C: (1|3), my question is, how do I get a vector out of this data?

OK, I assume the figure might be a 90° triangle, with AB and AC being the cathetes.

Let's suppose A: (1|1), B: (2|1) and C: (1|3), my question is, how do I get a vector out of this data?

In my experience, usually the line segments joining sides are the vectors.

Gold Member
A vector is the length of the side, with a directional arrow. Let's see, You may understand North-South-East-West Direction and displacement vector. Now I will explain how to get your AC-AB vector.

Suppose you rose in the morning for a morning walk. You know the sun is in the east, so you walked 10 steps towards the east. That is your vector AB, 10 steps in East.

Now if you were to walk 10 steps in west instead of east, you inverted your vector, your new vector is -AB, 10 steps in west OR -10 steps in East.

Now you walk 10√2 steps (Mathematically?) in North-East direction, where you find yourself? You actually find yourself 10 steps North of your initial position. That was you AC vector, 10√2 steps in North-East.

Now adding the two vectors, AC and -AB, you get AC-AB = 10 steps in North, Perpendicular to AB vector. So their Dot Product is zero.

Also, you must know that AC vector can be broken into a sum of perpendicular vector. One vector defines its motion in East and other in North, and by chance, North and East directions are perpendicular. Since North-East direction makes 45° angle with East, AC=10√2×cos45° East + 10√2×sin45° North, which is equal to AC=10 steps East + 10 steps North.

If you want to make your life easier, try learning vector notation, unit vectors, projection.