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belladona

  • 3 years ago

find equation of plane through the origin and the points <4,-2,5> and <6,1,1>

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  1. calculusfunctions
    • 3 years ago
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    Which one. Vector, parametric, symmetric, or scalar equation of the plane?

  2. belladona
    • 3 years ago
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    simple equation of plane

  3. experimentX
    • 3 years ago
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    ax+by+cz+d=0 --------------- for passing through origin, out x=y=z=0 and get d=0 ---------------------- then put a = 6-4, b=1-(-2) ...

  4. belladona
    • 3 years ago
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    thanx

  5. belladona
    • 3 years ago
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    find r(t) if r'(t) = 4t i + 3tj +t k and r(1) = i+j

  6. experimentX
    • 3 years ago
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    integrate it ... use the initial value to get rid of constant.

  7. estudier
    • 3 years ago
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    "then put a = 6-4, b=1-(-2) ... " How does this work?

  8. estudier
    • 3 years ago
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    If I cross the vectors to get a normal, I get (-7,26,16) so plane is -7(x-4) + 26(y+2) + 16(z-5) = 0 which satisfies all three points given.

  9. experimentX
    • 3 years ago
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    a(x_1) + b(y_1)+ c(z_1) = 0 a(x_2) + b(y_2)+ c(z_2) = 0 ----------------------- a b c ------------ = ----------- = ---------- y1z2 - z1y2 z1x2 - x1z2 x1y2 - y1x2 |dw:1351160338396:dw|

  10. experimentX
    • 3 years ago
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    |dw:1351160579673:dw|

  11. belladona
    • 3 years ago
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    ||a||=3 and ||b||=2. a) find ||a x b||

  12. experimentX
    • 3 years ago
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    what is angle between them?

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