The Theoretical Minimum book question

AI Thread Summary
The discussion revolves around understanding the dot product of orthogonal vectors in the context of the book "The Theoretical Minimum." It explains that the dot product is zero when vectors are orthogonal because the angle θ is π/2, making cos(θ) equal to zero. The concept of vector projection is clarified by comparing it to the shadow of an object, where an orthogonal vector has no shadow on the ground. Participants engage in visualizing vectors i, j, and k to enhance comprehension of these concepts. Overall, the discussion successfully connects mathematical principles with graphical interpretations.
OrigamiCaptain
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I'm reading the book The Theoretical Minimum. I wonder if two one of the problems in the book can be thought of graphically. The mathematical solution didn't even occur to me, although it make perfect sense and probably should have been obvious.

1. Can you explain why the dot product of two vectors that are orthogonal is 0?

2. A dot B= abs(a)abs(b)cosθ

I know now that if it is orthogonal than θ=∏/2 and cos∏/2=0

Is there a way of visualizing i, j and k to get the same answer?3. For a second I thought that connecting x, y z together end to end would create a situation that would eliminate distance so the magnitude would equal zero, but after thinking about it a little more before posting this I don't believe that would work. I actually feel like there is something to this line of think though, which is why I'm asking this question.

Thank you for your time and consideration
 
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The dot product of two vectors can be thought of as the projection of one vector on the other- that's where the |b|cos(θ) comes from. If two vectors are orthogonal then the projections are of 0 length.
 
HallsofIvy said:
The dot product of two vectors can be thought of as the projection of one vector on the other- that's where the |b|cos(θ) comes from. If two vectors are orthogonal then the projections are of 0 length.

I think I get it, but could you be a little more specific what you mean by projection?

Thank you for the help.
 
The projection is like the shadow of an object on the ground (with the sun directly overhead).
If you are orthogonal (perpendicular) to the plane of the ground (and very, very skinny) then your shadow should be 0.
 
Got it! Thanks!
 
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