Question

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)?

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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.

Answer 7

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

Answer 8

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.

Answer 9

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

Answer 10

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

Answer 11

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>

Answer 12

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

Answer 13

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

Answer 14

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?

Answer 15

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

Answer 16

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>

Answer 17

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

Answer 18

Go ahead cross it, you would get a vector.

Answer 19

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

Answer 20

The 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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

y=2-t