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# Need help: angle between two vectors formula PROOF

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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

#1
+394
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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
+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.

hairyberry Mar 6, 2024