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#1 |
FFR Veteran
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Age: 33
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![]() Gosh, it might be a brain fart but I can't figure out how to do this.
Code:
Find a vector equation and parametric equations in t for the line through the point and perpendicular to the given plane. (P0 corresponds to t = 0.) P0 = (4, 0, 7) x + 2y + z = 6 v = ?i + ?j + ?k x = ? y = ? z = ? |
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#2 |
tool
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![]() Have you tried taking the derivative with respect to each of the values and plugging in the intial conditions?
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#3 |
FFR Veteran
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Age: 33
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![]() I don't think that is what he is looking for since this assignment is in the chapter before derivatives of vectors. Dot products or cross products are probably involved.
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#4 |
Registered User
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Location: Hamilton Ontario
Age: 36
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![]() I haven't done this for a while, so I hope what I say is not complete gibberish lol
For the equation of a plane, aren't the coefficients simply the normal (perpendicular) to the plane? If so then we have P0 = 4,0,7 and normal 1,2,2. Leading us to get the general equation r = (4,0,7) + t(1,2,2) so x = 4 + t y = 2t z = 7 + 2t
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#5 | |
smoke wheat hail satin
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Age: 36
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![]() Quote:
As a result the parametrics change to: x = 4 + t y = 2t z = 7 + t So, v is going to then be a substitution of the parametrics back into i, j, k form? I don't remember either lol... EDIT: Yeah, I think v = (4 + t)i + (2t)j + (7 + t)k Last edited by foilman8805; 09-29-2009 at 02:10 PM.. |
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#6 |
FFR Veteran
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Age: 33
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![]() Yeah, your right thank you =).
Answer was v= (4+t)i + (2t)j + (7+t)k If I have any more issues I'll edit it in here. (math can be a bitch sometimes) |
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#7 |
smoke wheat hail satin
![]() ![]() Join Date: Sep 2006
Location: LA baby
Age: 36
Posts: 5,704
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![]() If you have more issues, bump the thread, otherwise we probably won't come back to look for it.
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#8 |
Registered User
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Location: Hamilton Ontario
Age: 36
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![]() lol brain fart on my part haha :P
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