#Cube, linear algebra

21 messages · Page 1 of 1 (latest)

balmy charm
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A cube has a corner in the point O=(-5,7,-2). The corners of the cube which is closest to O is:
A=(-3,10,4), B=(-2,λ,0), and C.

Find the parameter λ and decide the two possible coordinates for C.

My first thought was that the vector OA and OB are perpendicular to each other and that they both are perpendicular to OC, aka "normal vector"
So the dot product between OA and OB has to be 0.

So after i calculated OA and OB, I applied the cross product on them which would give me the normal vector.

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Have I correctly found the normal vector? I know I haven't completed the exercise yet

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So C should be OC+O right? so (6,2,-3)+(-5,7,-2)=(1,9,-5) which is one of the possible locations for C?

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and the other one can be found by subtracting O-OC?

mystic mica
# balmy charm

Is the z-coordinate of O 2 or -2? You wrote different things in the comment and the picture.

balmy charm
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ah my bad, it's -2

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Thanks for noticing, I changed it now

mystic mica
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Ah, ok.

mystic mica
# balmy charm

λ is correct.
As for the point C: we have two choices depending on the direction of the perpendicular vector.
Choice 1: C = O - OA⨯OB.
Choice 2: C = O - OB⨯OA.

balmy charm
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Something like this then? 🙂

mystic mica
# balmy charm

Ah! Sorry, forgot about one thing: the length of the cross product is too large: the square of the side length.
So, the correct formulas would be O - (OA⨯OB)/|OA| and O - (OB⨯OA)/|OA|.

balmy charm
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Just curious, why is C=O-(OA⨯OB)/|OA|? shouldn't C=O+OC just like OC=O-C and OA=A-O etc

mystic mica
balmy charm
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Btw, the coordinates for C2 are the same that I got when I tried to complete it without your help, so it feels good that I was on the right track anyway 🙂

mystic mica
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Ah, I see the problem. OA⨯OB isn't correct.

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Also, you don't really need to separately calculate OB⨯OA, since OB⨯OA = -OA⨯OB.

balmy charm
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I think I see it too, C1 should be (1,5,1) right?