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Consider the points $D = (1, −1, 5)$ and $E = (4, 3, −2)$ Find the plane $S_1$ which contains $D$, $E$ and the origin, in parametric form. Now find the equational form for $S_1$.

Murtuza Vadharia
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zen
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  • my question is how do i approach this... – zen May 05 '16 at 11:24
  • i know that if we take the cross product of the first two vectors we get -13i+22j+7k which is normal to the plane. so let -13i+22j+7k = n. also DE=(3,4,-7). since it passes through the origin i beleive we can use n⋅(3,4,-7)=0 however i think the answer should be in the form ax+by+cz=0 so i think i have done something wrong in the working out. is anyone able to help me instead of downvoting for no reason ? :) – zen May 05 '16 at 11:29
  • There is some useful explanation in the first answer to this question http://math.stackexchange.com/questions/152467/parametric-form-of-a-plane – almagest May 05 '16 at 11:29
  • Great! You found the normal $(-13,22,7)$. So since the plane goes through the origin the equation is just $(x,y,z)\cdot(-13,22,7)=0$ or $-13x+22y+7z=0$. But you still want the parametric form. – almagest May 05 '16 at 11:33
  • oh !! thanks so much that makes much more sense now :) – zen May 05 '16 at 11:35
  • is the parametric form {(x, y, z) ∈ R^3| (x, 13x-7z, z) | x,z ∈ R} ? – zen May 05 '16 at 11:46
  • and { (x, y, z) ∈ R^3 | -13x + 22y + 7z = 0 } would be the equational form ? – zen May 05 '16 at 11:48
  • The parametric form expresses each of $x,y,z$ in terms of two parameters, say $s,t$. You know that the origin and $D$ are in the plane, so any point of the form $(s,-s,5s)$ must be in the plane. Similarly the origin and $E$ are in it, so any point of the form $(4t,3t,-2t)$ is in the plane. Hence also any point $(x,y,z)$ with $x=s+4t,y=-s+3t,z=5s-2t$. Yes, you have got the equational form. – almagest May 05 '16 at 11:53
  • oh ok i see so the parametric could be written as { (s+4t, −s+3t, 5s-2t) | s, t ∈ R } ? – zen May 05 '16 at 12:04
  • Yes, that looks fine to me. – almagest May 05 '16 at 12:05
  • thank you, you have been very helpful :) – zen May 05 '16 at 12:19

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