I want to check my proof of dot product (without using cosine law). We want to prove that $\vec a\cdot \vec b=|\vec a||\vec b|\cos \theta$, where theta is the angle the two vectors form. If we name $\vec a'$ the orthogonal projection of $\vec a$ on $\vec b$, then $\vec a-\vec a'$ is orthogonal to $\vec b$ (that is by the definition of projection, right??), so $\vec b\cdot (\vec a-\vec a')=0$ and thus $\vec b\cdot \vec a$=$\vec b\cdot \vec a'$. We know that $\vec a'=\displaystyle\frac{\vec b|\vec a|*\cos\theta}{|\vec b|}$. So $\vec a\cdot \vec b=|\vec a||\vec b|\cos \theta$.
Asked
Active
Viewed 46 times
0
-
How do you know that $\vec a'=\displaystyle\frac{\vec b|\vec a|*\cos\theta}{|\vec b|}$? How is that identity derived? – md2perpe Nov 15 '21 at 20:10
-
Its a known fact. It has the direction of b but its modulo is that of a times cosine of theta. My main concern is about a-a' being orthogonal to b. – user10000024 Nov 15 '21 at 20:13
-
Yes, those two are orthogonal---drawing the picture might convince you. – Jakob Streipel Nov 15 '21 at 20:15
-
Yes, I am convinced. I do not know if that would be proved "formally" – user10000024 Nov 15 '21 at 20:38
-
One way to prove it formally is to compute $\vec{b} \cdot (\vec{a} - \vec{a}')$, wherein you substitute whatever your preferred definition of $\vec{a}'$ is and then expand. – Jakob Streipel Nov 15 '21 at 21:06