I don't understand the step on the right that starts: "We next apply Eq 3.20..." (that equation is the same as the e-28 on the left side).
How are the angles determined to be $0^\circ$ and $90^\circ$?
How can I find angle between vector?
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physics
2 Answers
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We know that in $2$-space, the vectors $\vec i = [1,0] $ and $\vec j =[0,1] $, thus we get, $$\vec i \cdot \vec i = 1 ; \vec i \cdot \vec j =0$$ $$\vec j \cdot \vec i =0 ; \vec j \cdot \vec j =1$$
Also see here regarding why the angle between like vectors is $0 ^\circ $ and between unlike vectors is $90^\circ $. Hope it helps.
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The unit vectors $i,j,k$ form a basis for $\mathbb{R^3}$, along the lines of the $x,y,z$ coordinate system.
As each axis is perpendicular to each other, the angle between any different two is $90^\circ$, otherwise it is $0^\circ$.
Then $\cos 0^\circ=1$ and $\cos 90^\circ=0$.