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anonymous

  • 4 years ago

Multivariable Calculus: Vectors and the Geometry of Space Find parametric equations for the line through (5,1,0) that is perpendicular to the plane 2x-y+z=1. Also, in what points does this line intersect the coordinate planes. Please explain the steps. (:

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  1. anonymous
    • 4 years ago
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    Yes, for a parametric equation of a line the equation of line is actually a combination of 3 equations x = 1 + ar y = 1 + br z = 1 + cr Now, this line passes through the point (5,1,0) hence putting x = 5, y = 1 and z = 0 5 = 1 + ar => a = 4 / r 1 = 1 + br => br = 0 => b = 0 (as r not equal to zero from other two equations) 0 = 1 + cr => c = -1 / r Now for a line to e perpendicular to a plane, the line vector ((a,b,c) in this case) and the normal vector of the line (2, -1, 1) are parallel. Which means their dot product has a value 1 (since cos0 = 1) hence, having the dot product = 0 we have a.2 + b. (-1) + c.1 = 1 Replacing a by 4 / r and c by -1 / r and b = 0. We have 8 / r - 1 / r = 1 => r = 7 Hence, a = 4 / 7 and c = -1 / 7 Hence, the required equation of the line is x = 1 + 4r / 7 y = 1 z = 1 - r / 7 For crossing XY planes, Z = 0 hence r = 7, x = 5, y = 1 point (5,1,0) [our point] For Crossing XZ planes, Y = 0, this condition is never met, the line does not cross XZ plane For crossing YZ plane, X = 0 hence 1 + 4r / 7 = 0, r = - 7 / 4, hence x = 0, y = 1, z = 1.25 point is (0,1,1.25)

  2. anonymous
    • 4 years ago
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    Thank you so much! I just have one more question. How did you figure out that x=1+ar, y=1+br, and z=1+zr. I'm wondering where you got the 1 on all of them. Thanks. (:

  3. anonymous
    • 4 years ago
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    The 1 is unitary displacement in each direction, since the locus of the point moves linearly in all 3 directions hence, that is handled by adding 1. We could use any constant in place of 1 here, but the values of r will change in that case, but the final answer will be in the multiples of the a,b and c found out. Please let me know if you need more explanation.

  4. anonymous
    • 4 years ago
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    There are two three corrections, which are typo in the original solutions - "Now for a line to e perpendicular to a plane, the line vector ((a,b,c) in this case) and the normal vector of the line (Should read Plane) (2, -1, 1) are parallel." "hence, having the dot product = 0 (should read 1) we have"

  5. anonymous
    • 4 years ago
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    Oh ok I see. So in place of the 1 i could use any number? And that would change the r, but in the end, the answer would still be correct?

  6. anonymous
    • 4 years ago
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    Hey I am sorry, let me check it a little more. I will come back to you.

  7. anonymous
    • 4 years ago
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    However, we can use 1, that is the general convention

  8. anonymous
    • 4 years ago
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    Ok thank you so much! (:

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