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mazehaq

  • one year ago

Find a Cartesian equation of the the plane that contains the point (3,-7,2) and the line <x,y,z> = <3,5,-4> + t<-2,1,3>

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  1. IrishBoy123
    • one year ago
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    from the equation of the line, you can generate two more points on the plane , say let t= 0, t= 1 then solve ax + by + cz = d with your three points: or calculate the crossproduct aka normal for the plane, depending upon what you are currently learning.

  2. mazehaq
    • one year ago
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    how would you go about doing it without doing cross product

  3. IrishBoy123
    • one year ago
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    solve ax + by + cz = d

  4. mazehaq
    • one year ago
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    Yeah I think that's what I don't get... so I let t=1, I got another point (1,6,-1) My parametric equations are: x=3-2t, y=5+t, z=-4+3t

  5. mazehaq
    • one year ago
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    What do I do next?

  6. IrishBoy123
    • one year ago
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    make 3 points, then solve for the plane equation you have 4 unknowns - a,b,c,d - but if you plough through with the solution - you will see that you can still get something

  7. mazehaq
    • one year ago
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    I'm still very confused, I'm so sorry. I'm attempting at the cross product right now. Let me see if I get somewhere. I suppose I'm not understanding the concept too well. Thank you for your help though!

  8. IrishBoy123
    • one year ago
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    ok if the plane is ax + by + cz = d it is also a/d x + b/d y + c/d z = 1 you do the cross product, i'll try the other way, see where we get!

  9. IrishBoy123
    • one year ago
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    i've done it the other way you are solving \[\left[\begin{matrix}3 & -7 & 2\\ 3 & 5 & -4\\ 1 & 6 & -1\end{matrix}\right]\left(\begin{matrix}\frac{a}{d} \\ \frac{b}{d} \\ \frac{c}{d}\end{matrix}\right) = \left(\begin{matrix}1 \\ 1 \\ 1\end{matrix}\right)\] thats the given point and 2 generated off the line at t= 0 and t= 1 i get 7x + 2y + 4z = 15 do we agree?

  10. amistre64
    • one year ago
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    cross product using 2 noncolinear vectors in the plane, one vector is the direction vector of the line (-2,1,3) the other is a vector from a point on the line, and the stated point ... the anchor point is convenient: (3,-7,2) -(-2,1,3) --------- (5,-8,-1) seems useful cross the vectors -(5z-24x+2y) x y z x y -2 1 3 -2 1 5 -8 -1 5 -8 +(-x+15y+16z) 23x, 13y,11z should be the cross product for our normal not all vectors (x^,y^,z^) that perp to the normal (a,b,c) form the inner product (a,b,c).(x^,y^,z^) = 0 ax^ + by^ + cz^ = 0

  11. amistre64
    • one year ago
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    bu is early and i mighta messed up the cross of course

  12. amistre64
    • one year ago
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    i see my error, i used the vector as the anchor point ...

  13. amistre64
    • one year ago
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    (3,-7,2) -(3,5,-4) --------- (0,-12,6) ... or (0,2,-1) as the other vector .. maybe

  14. amistre64
    • one year ago
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    -(-4z-1x+0y) x y z x y 0 2 -1 0 2 -2 1 3 -2 1 +(6x+2y+0z) 7x, 2y, 4z ... now we agree :)

  15. IrishBoy123
    • one year ago
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    cool!

  16. mazehaq
    • one year ago
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    Thanks so much y'all!

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