Finding the Orthogonal Projection of a Vector onto Another Vector

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The discussion focuses on proving that the vector B minus its projection onto vector A results in the orthogonal projection of B onto A. It is noted that the condition of the dot product being non-zero implies that the angle between the vectors is not zero. A suggested method for proof involves using vector geometry visualization or demonstrating that the dot product of the difference vector and vector A equals zero. The formula for the projection of B onto A is provided, leading to a simplification that confirms the orthogonality. The conversation emphasizes understanding the relationship between vector projections and their geometric implications.
linuspauling
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Given:

\vec A \cdot \vec B = non zero
and
\theta does not equal 0

I can't seem to prove that Vector B minus the Projection of B onto A makes the orthogonal projection of B onto A.

Can you help?
 
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I am not sure I understand your question properly. The first statement in fact implies the second statement in your question.
You can see the proof by simple visualization, using laws of vector geometry of course.
Alternatively, prove that the dot product of "Vector B - Projection of B onto A and vector A" is 0. Just put that statement in algebra and simplify.
 
The projection of B onto A is
\frac{\vec{A}\cdot\vec{B}}{|\vec{A}|^2}\vec{A}
so
(\vec{B}- proj_{\vec{B}}(\vec{B}))\cdot \vec{A}= \vec{A}\cdot\vect{B}- \frac{\vec{A}\cdot\vec{B}}{|\vec{A}|^2}\vec{A}\cdot\vec{A}
 
thank you halls of ivy.

thanks hall of ivy
 

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