anonymous
  • anonymous
Show that the vector (orthogonal b onto a) = (b - (projection of b onto a)) is orthogonal to a. It is called an orthogonal projection of b. from chapter Vectors and the Geometry of Space. Thanks.
OCW Scholar - Multivariable Calculus
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
Let & be the angle between B und A. We have to show that B - |B|cos & A/|A| is normal to A. Using the dot product we must have (B - |B|cos & A/|A| )·A =0. Since the dot product is distributive with respect to addition, we can write it as: B·A - |B|cos & A/|A|·A = |B| |A|cos & - (|B|cos & A/|A|)|A|^2 = 0, since (|B|cos & A/|A|)|A|^2 = |B| |A|cos &.
anonymous
  • anonymous
15 minutes ago I posted a reply. I do not understand why the proof is not right. However, I will try to clear some doubts. A/|A| is the unit vector in the direction of A. A·A = |A| |A| cos 0 = |A|^2, then (|B|cos&/|A|)A·A = |B||A| cos&.

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