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If a and b are both vectors, then cos(x) = (a*b)/(|a|*|b|). 

I don't understand why this cos(x) equals the right side of the equation. Can somebody explain, or better yet, could somebody prove the formula?

 Mar 6, 2024

Best Answer 

 #1
avatar+394 
+3

This formula 

\(\cos(x)=\frac{a\cdot b}{||a||\,||b||}\). Just to clarify, The \(\cdot\) between the a and b is not a multiplication sign, it is a dot product.

This formula is very useful for calculating the angle between 2 vectors.

To prove: draw a vector, connecting vector  \(\vec{a}\) and \(\vec{b}\), remembering vector subtraction, this vector is \(\vec{a-b}\).

We see a cosine, so we hope to use the law of cosines to help us relate side lengths. However, we don't know the side lengths of these vectors. However, we have a very important symbol called the norm, \(||~||\), and this the distance to the endpoint to the origin. Remember, vectors aren't defined with position, so norm gives an effective way to represent length. Here is a graph:

Set \(\theta = \angle AOB\)

Therefore by the law of cosines, we have:

\(\cos(\theta) = \frac{{||a||}^{2}+{||b||}^{2}-{||a-b||}^{2}}{2||a||\,||b||}\).

To relate norm to dot product, we have another very important formula:

\(v\cdot v ={||v||}^{2}\). (dot product again). PS, notice the resemblance to the formula \(z*\overline{z}={|z|}^{2}\), for complex number z.

Using this formula, 

\(\cos(\theta) = \frac{a\cdot a+b\cdot b-(a-b)\cdot(a-b)}{2||a||\,||b||}\).

Dot products have a special property, because they are commutative, and distributive.

Therefore, \(\cos(\theta) = \frac{a\cdot a+b\cdot b-(a\cdot a- 2a\cdot b + b\cdot b)}{2||a||\,||b||}\)

Simplifying:

\(\cos(\theta) = \frac{2a \cdot b}{2||a|| \, ||b||}=\frac{a\cdot b}{||a||\,||b||}\).

This gives our desired formula.

 Mar 6, 2024
 #1
avatar+394 
+3
Best Answer

This formula 

\(\cos(x)=\frac{a\cdot b}{||a||\,||b||}\). Just to clarify, The \(\cdot\) between the a and b is not a multiplication sign, it is a dot product.

This formula is very useful for calculating the angle between 2 vectors.

To prove: draw a vector, connecting vector  \(\vec{a}\) and \(\vec{b}\), remembering vector subtraction, this vector is \(\vec{a-b}\).

We see a cosine, so we hope to use the law of cosines to help us relate side lengths. However, we don't know the side lengths of these vectors. However, we have a very important symbol called the norm, \(||~||\), and this the distance to the endpoint to the origin. Remember, vectors aren't defined with position, so norm gives an effective way to represent length. Here is a graph:

Set \(\theta = \angle AOB\)

Therefore by the law of cosines, we have:

\(\cos(\theta) = \frac{{||a||}^{2}+{||b||}^{2}-{||a-b||}^{2}}{2||a||\,||b||}\).

To relate norm to dot product, we have another very important formula:

\(v\cdot v ={||v||}^{2}\). (dot product again). PS, notice the resemblance to the formula \(z*\overline{z}={|z|}^{2}\), for complex number z.

Using this formula, 

\(\cos(\theta) = \frac{a\cdot a+b\cdot b-(a-b)\cdot(a-b)}{2||a||\,||b||}\).

Dot products have a special property, because they are commutative, and distributive.

Therefore, \(\cos(\theta) = \frac{a\cdot a+b\cdot b-(a\cdot a- 2a\cdot b + b\cdot b)}{2||a||\,||b||}\)

Simplifying:

\(\cos(\theta) = \frac{2a \cdot b}{2||a|| \, ||b||}=\frac{a\cdot b}{||a||\,||b||}\).

This gives our desired formula.

hairyberry Mar 6, 2024

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