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Question

So I have done part of this question. Using the projection vector equation, I found that the projection vector of (4,2,-1) onto V was: (1,-1,-1)

And a vector orthogonal to V is: (3,3,0)=(4,2,-1)-(1,-1,-1) But, just not sure how to find the point in V which is closest to (4,2,-1). Any help????

J-Dorman
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  • You are on the right track! Note that the cartesian equation for V is $x+y=0$ then use the given hint. – user Sep 17 '18 at 08:40
  • @gimusi does that still contain (0,0,2)?? thanks for your help – J-Dorman Sep 17 '18 at 08:42
  • Yes of course indeed $0+0=0$ satisfy the equation. Since the plane is spanned by (0,0,2) all points (0,0,t) belongs to the plane, that is z coordinate is free, that is the plane is parallel to x-y plain. we can see that in many ways, for example also using cross product. – user Sep 17 '18 at 08:46
  • In the future please take the time to enter important parts of your question as text instead of pasting in pictures of it. Images are neither searchable nor accessible to screen-reading devices. You can find a quick reference on using MathJax to format mathematical expressions here. – amd Sep 18 '18 at 01:20
  • The orthogonal projection is the nearest point. – amd Sep 18 '18 at 01:21

1 Answers1

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HINT

Note that $(3,3,0)$ is orthogonal to the plane

$$(4,2,-1)+t(3,3,0)$$

for some $t$ belongs to it.

user
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