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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 x-y+2z=1 at the point (1,2,1)?

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  1. anonymous
    • 5 years ago
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    it looks like those two are planes, and the intersection of two planes is a line itself

  2. anonymous
    • 5 years ago
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    how would i find the tangent line of the intersection though, would the tangent be the same as the line of intersection?

  3. anonymous
    • 5 years ago
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    that's why i don't understand the question. i would think so, but i've never heard it phrased that way

  4. anonymous
    • 5 years ago
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    apparently they are supposed to intersect in a curve, and i need to find the tangent line to that curve

  5. anonymous
    • 5 years ago
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    Would i just find the gradient of the equation of the curve of intersection and use that to find the tangent?

  6. anonymous
    • 5 years ago
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    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.

  7. anonymous
    • 5 years ago
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    What is fsub? and it will give me the parametric equations for the tangent?

  8. anonymous
    • 5 years ago
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    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.

  9. anonymous
    • 5 years ago
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    I dont understand what you mean by that though. How does the first surface become <1,1,1>?

  10. anonymous
    • 5 years ago
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    I dont understand what you mean by that though. How does the first surface become <1,1,1>?

  11. anonymous
    • 5 years ago
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    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>

  12. anonymous
    • 5 years ago
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    oh ok, so the first equation will be <1,1,1> and the second would be <1,-1,2>?

  13. anonymous
    • 5 years ago
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    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

  14. anonymous
    • 5 years ago
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    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?

  15. anonymous
    • 5 years ago
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    Or would the final vector just be used for the parametric equations?

  16. anonymous
    • 5 years ago
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    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>

  17. anonymous
    • 5 years ago
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    Or would the final vector just be used for the parametric equations?

  18. anonymous
    • 5 years ago
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    Go ahead cross it, you would get a vector.

  19. anonymous
    • 5 years ago
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    so the resulting vector will be for the parametric equations of the tangent?

  20. anonymous
    • 5 years ago
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    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

  21. anonymous
    • 5 years ago
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    I got <3,-1,-2> as the resulting vector after crossing <1,1,1> and <1,-1,2>.

  22. anonymous
    • 5 years ago
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    Ok, use this and the given point (1,2,1) to get your parametric equations: x=1+3t etc

  23. anonymous
    • 5 years ago
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    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

  24. anonymous
    • 5 years ago
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    x=1+3t, y+2-t etc. Good luck

  25. anonymous
    • 5 years ago
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    y=2-t

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