When using the dot product to find vector lengths should the answer always be the same as when you apply cosine law? I was doing some linear algebra review and decided to solve some questions about vector lengths using both dot product and cosine law.
Lets say you have a vector $\vec u = 2\vec a + 3\vec b$ and $\vec v = 3\vec a -\vec b$ where $\|\vec a\| = 3$, $\|\vec b\| = 4$, and $\vec a \cdot \vec b = 5$. Find $\|\vec u\|$ and the angle between $\vec u$ and $\vec v$.
We know that the dot product of a vector and its self is $\vec x\cdot \vec x = \|\vec x\|^2$ because the angle between a vector and its self is 0 and cosine of 0 is 1 so the dot product just the magnitude (length) squared.
Using this method I computed the following: $$\vec u\cdot \vec u = (2\vec a + 3\vec b)\cdot(2\vec a + 3\vec b)$$ $$\|\vec u\|^2 = 4(\vec a\cdot \vec a) + 6(\vec a\cdot \vec b) + 6(\vec b\cdot \vec a) + 9(\vec b\cdot \vec b)$$ $$\|\vec u\|^2 = 4\|\vec a\|^2 + 12(\vec a\cdot \vec b) + 9\|\vec b\|^2$$ $$\|\vec u\|^2 = 4(3)^2 + 12(5) + 9(4)^2$$ $$\|\vec u\| = \sqrt{240} = 15.5$$
I used the same strategy for solving the magnitude of $\|\vec v\|$. This looks like: $$\vec v\cdot \vec v = (3\vec a - \vec b)\cdot(3\vec a - \vec b)$$ $$\|\vec v\|^2 = 9\|\vec a\|^2 -3(\vec a\cdot \vec b) -3(\vec b\cdot \vec a) + \|\vec b\|^2$$ $$\|\vec v\|^2 = 9\|\vec a\|^2 -6(\vec a\cdot \vec b) + \|\vec b\|^2$$ $$\|\vec v\|^2 = 9(3)^2 -6(5) + (4)^2$$ $$\|\vec v\|^2 = \sqrt{67}=8.12$$
Then using $\vec u\cdot \vec v$ I can solve for the angle. $$\vec u\cdot \vec v = \|\vec u\| \|\vec v\| cos(\theta)$$ $$(2\vec a + 3\vec b)\cdot (3\vec a -\vec b) = \|\vec u\| \|\vec v\| cos(\theta)$$ $$6\|\vec a\|^2 +7(\vec a\cdot \vec b) - 3\|\vec b\|^2 = \sqrt{240}\sqrt{67}cos(\theta)$$ $$cos^{-1}(\frac{6(3)^2 +7(5) - 3(4)^2}{\sqrt{240}\sqrt{67}}) = \theta$$ $$71.13^{\circ}= \theta$$
If I use cosine law to calculate the same vector lengths as I did above then I get different lengths. For $\|\vec u\|$ I calculated the following using cosine law: $$\|\vec u\|^2 =(2\|\vec a\|)^2 + (3\|\vec b\|)^2 - 2(2\|\vec a\|)(3\|\vec b\|)cos(180-\gamma)$$ Gamma is the angle between vectors $\vec a$ and $\vec b$ $$\vec a\cdot \vec b = \|\vec a\| \|\vec b\| cos(\gamma)$$ $$cos^{-1}(\frac{5}{(3)(4)})=\gamma$$ $$65.37^\circ = \gamma$$ Back to solving for $\|\vec u\|$ $$\|\vec u\|^2 =(2)^2(3)^2 + (3)^2(4)^2 - 2(2(3))(3(4))cos(114.62^\circ)$$ $$\|\vec u\| =\sqrt{240}=15.5$$
Using cosine law for the vector magnitude of $\|\vec v\|^2$ I get: $$\|\vec v\|^2 = ((3\|\vec a\|)^2) + ((-\|\vec b\|)^2 -2(3\|\vec a\|)(-\|\vec b\|)cos(\gamma)$$ $$\|\vec v\|^2 = (3)^2(3)^2 + (-1)^2(4)^2 -2(3(3))(-1(4))cos(65.37^\circ)$$ $$\|\vec v\| = \sqrt{127.00}$$ $$\|\vec v\| = 11.23$$
And then using dot product to solve for the angle between $\vec u$ and $\vec v$ would be the same method as I did earlier but with a different magnitude for $\|\vec v\|$. $$\vec u\cdot \vec v = \|\vec u\| \|\vec v\| cos(\theta)$$ $$(2\vec a + 3\vec b)\cdot(3\vec a -\vec b)= (15.5)(11.23)cos(\theta)$$ $$6\|\vec a\|^2 +7(\vec a\cdot \vec b) - 3\|\vec b\|^2 = (15.5)(11.23)cos(\theta)$$ $$cos^{-1}(\frac{6(3)^2 +7(5) - 3(4)^2}{(15.5)(11.23)}) = \theta$$ $$76.37^{\circ}= \theta$$
I am trying to figure out why the cosine law method and dot product give me different magnitudes for $\|\vec v\|$. This results in a different angle between the two vectors.
Sorry if this has been posted before. I tried to look for something like this but couldn't find anything. Any help is greatly appreciated!
Your error is here:
$$\|\vec v\|^2 \stackrel?= (3\|\vec a\|)^2 + (-\|\vec b\|)^2 -2(3\|\vec a\|)(-\|\vec b\|)cos(\gamma).$$
You don't have a triangle with a side of length $-\|\vec b\|$ making an angle $\gamma$ with another side of length $3\|\vec a\|$. You have a triangle with a side of length $\|\vec b\|$ making an angle $\gamma$ with another side of length $3\|\vec a\|$.
Yes, the side whose length you specified as "$-\|\vec b\|$" is part of a vector diagram produced by a vector in the direction opposite from $\vec b.$ Nevertheless, once you have identified that part of the diagram as the side of a triangle and apply the cosine rule to it, the length of the side is positive because the cosine rule assumes the lengths of all three sides of your triangle are positive numbers. The fact that you got this side from $-\vec b$ instead of $\vec b$ was accounted for when you identified the angle between sides as $\gamma$ instead of $180^\circ - \gamma.$
Because of this sign error, you added $30$ when you should subtract $30.$ That is how you conclude that $\|\vec v\|^2 = 127$ although the correct answer is $\|\vec v\|^2 = 67.$