Cheung D asked in 科學及數學數學 · 1 decade ago

Interection

Determine where the line L : ‘x =(-1, 8,9) + t(3, -9,-4)

and the plane

E : -2(x1) + 6(x2) + 3(x3) + 19 = 0 intersect and calculate the angle between

them.

1 Answer

Rating
  • 1 decade ago
    Favorite Answer

    I use x,y, z instead for ease of reading

    L is

    x = -1 + 3t; y = 8 - 9t; z = 9 - 4t

    The plane is -2x + 6y + 3z + 19 = 0

    Sub the line into the plane,

    -2(-1 + 3t) + 6(8 - 9t) + 3(9 - 4t) + 19 = 0

    2 - 6t + 48 - 54t + 27 - 12t + 19 = 0

    96 = 72t

    t = 4/3 => intersection point is (x,y,z) = (3, -4, 11/3)

    The normal vector of the plane is (-2, 6, 3)

    And the directional vector of the line is (3, -9, -4)

    Let x be angle between normal vector and directional vector

    (-2,6,3).(3,-9,-4) = |(-2,6,3)||(3,-9,-4)| cos x

    cos x = -0.999

    x = 177.49 degrees or 2.51 degrees depending on which side of the normal is considered

    The angle bewteen the line and the plane = 90 - 2.51 = 87.49 degrees

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