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anonymous
 5 years ago
Find the parametric equations for the tangent line to the curve of intersection of the surfaces x+y+z=4 and xy+2z=1 at the point (1,2,1)?
anonymous
 5 years ago
Find the parametric equations for the tangent line to the curve of intersection of the surfaces x+y+z=4 and xy+2z=1 at the point (1,2,1)?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it looks like those two are planes, and the intersection of two planes is a line itself

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how would i find the tangent line of the intersection though, would the tangent be the same as the line of intersection?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that's why i don't understand the question. i would think so, but i've never heard it phrased that way

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0apparently they are supposed to intersect in a curve, and i need to find the tangent line to that curve

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Would i just find the gradient of the equation of the curve of intersection and use that to find the tangent?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Find the dell, fsub(x,y,z) of each surface (partial fractions, ending with 2 vectors. In put (1,2,1) in each vector. Cross the two vectors: the new vector and the point (1,2,1) gives you your parametric eq.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What is fsub? and it will give me the parametric equations for the tangent?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think it is called the dell for example the first surface x+Y+z4 becomes <1, 1, 1> Partial fraction of x, then y, then z.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I dont understand what you mean by that though. How does the first surface become <1,1,1>?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I dont understand what you mean by that though. How does the first surface become <1,1,1>?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Some people use different notation. First find the partial fraction in relation to x, they all (x,y,z) happen to be 1. Some people right it i+j+k, which is the same as <1,1,1>

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh ok, so the first equation will be <1,1,1> and the second would be <1,1,2>?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That's right, now you input (1,2,1), but there is no variable x, y, or z so it reains the same. Now you cross the vectors

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cross each vector with (1,2,1), then cross those resulting vectors, then cross the final vector with (1,2,1)? that would give me the equation of the tangent?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Or would the final vector just be used for the parametric equations?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No. Sometimes you get an answer like <2x, y, 3z>. In a case like that you input the given points (1,2,1) but in your case this is not necessary. So just cross <1,1,1,> and <1, 1,2>

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Or would the final vector just be used for the parametric equations?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Go ahead cross it, you would get a vector.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the resulting vector will be for the parametric equations of the tangent?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The vector is tangent to the curve of intersection. Let's say the vector is <a,b,c> Use this vector ad the given point (1,2,1) to get the parametric eq. The parametric equations would be x=1+at etc

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I got <3,1,2> as the resulting vector after crossing <1,1,1> and <1,1,2>.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, use this and the given point (1,2,1) to get your parametric equations: x=1+3t etc

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so (1,2,1) will give me the x0,y0, and z0. While the <3,1,2> gives me the a,b,and c? Thank you very much for your help

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x=1+3t, y+2t etc. Good luck
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