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

- anonymous

- jamiebookeater

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- anonymous

it looks like those two are planes, and the intersection of two planes is a line itself

- anonymous

how would i find the tangent line of the intersection though, would the tangent be the same as the line of intersection?

- anonymous

that's why i don't understand the question. i would think so, but i've never heard it phrased that way

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- anonymous

apparently they are supposed to intersect in a curve, and i need to find the tangent line to that curve

- anonymous

Would i just find the gradient of the equation of the curve of intersection and use that to find the tangent?

- anonymous

Find 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

What is fsub? and it will give me the parametric equations for the tangent?

- anonymous

I think it is called the dell for example the first surface x+Y+z-4 becomes <1, 1, 1> Partial fraction of x, then y, then z.

- anonymous

I dont understand what you mean by that though. How does the first surface become <1,1,1>?

- anonymous

I dont understand what you mean by that though. How does the first surface become <1,1,1>?

- anonymous

Some 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

oh ok, so the first equation will be <1,1,1> and the second would be <1,-1,2>?

- anonymous

That'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

cross 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

Or would the final vector just be used for the parametric equations?

- anonymous

No. 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

Or would the final vector just be used for the parametric equations?

- anonymous

Go ahead cross it, you would get a vector.

- anonymous

so the resulting vector will be for the parametric equations of the tangent?

- anonymous

The vector is tangent to the curve of intersection. Let's say the vector is 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

I got <3,-1,-2> as the resulting vector after crossing <1,1,1> and <1,-1,2>.

- anonymous

Ok, use this and the given point (1,2,1) to get your parametric equations: x=1+3t etc

- anonymous

so (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

x=1+3t, y+2-t etc. Good luck

- anonymous

y=2-t

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