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Angle Between Two Vectors

Greetings,

I would like to ask whether the cosine of the angle between two vectors can be derived from the knowledge of directional cosines. If that is the case, how can I prove the statement? I am attaching a scanned page from the book V*ector and Tensor Analysis* by Hay, which illustrates the concept beginning with the phrase, "By a formula from of analytic geometry..." and writes the formula, around which I have put a red box, below this statement. I have put a red rectangular box to emphasize my point.

Thanks

Re: Angle Between Two Vectors

The theorem in the red box is pretty much a restatement of (7.1). Not particularly helpful.

To find $\displaystyle \cos \theta$, draw vectors $\displaystyle \vec{a}$ and $\displaystyle \vec{b}$, along with their difference, $\displaystyle \vec{a} - \vec{b}$, forming a triangle. Note that the dot product obeys the distributive property, that is

$\displaystyle (\vec{a} - \vec{b})^2 = (\vec{a})^2 + (\vec{b})^2 - 2(\vec{a} \cdot \vec{b}) = ||\vec{a}||^2 + ||\vec{b}||^2 - 2(\vec{a} \cdot \vec{b})$. However by the law of cosines,

$\displaystyle (\vec{a} - \vec{b})^2 = ||\vec{a}||^2 + ||\vec{b}||^2 - 2||a|| \hspace{1 mm}||b|| \cos \theta$

Set these two expressions equal, cancel stuff, and you will obtain $\displaystyle \vec{a} \cdot \vec{b} = ||\vec{a}||\hspace{1 mm}||\vec{b}|| \cos \theta$, or $\displaystyle \cos \theta = \frac{\vec{a} \cdot \vec{b}}{||\vec{a}||\hspace{1 mm}||\vec{b}||}$

Re: Angle Between Two Vectors

Thanks, you are right, but I think you emphasize that no such formula exists in analytical geometry if I am not mistaken. However, I still wonder why the author wrote, "By a formula of analytical geometry..." do you have an explanation?