equation of the plane help?

Find an equation of the plane.

The plane that passes through the line of intersection of the planes 

x − z = 1 and y + 3z = 3

 and is perpendicular to the plane 

x + y − 4z = 4

2 Answers

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  • Pope
    Lv 7
    1 month ago

    Any plane (save one) including that line of intersection may be written in this form:

    (x - z - 1) + k(y + 3z - 3) = 0

    x + ky + (3k - 1)z + (-3k - 1) = 0

    The dot product of normal vectors of the perpendicular planes must be zero.

    <1, k, 3k - 1> · <1, 1, -4> = 0

    1 + k - 12k + 4 = 0

    k = 5/11

    x + ky + (3k - 1)z + (-3k - 1) = 0

    x + (5/11)y + [3(5/11) - 1]z + [-3(5/11) - 1] = 0

    11x + 5y + 4z - 26 = 0

  • 1 month ago

    Let me first find a couple of points on that line...

    Let x = 1, z = 0, then y = 3.  So A(1,3,0) is on the line.

    Let y = 0, z = 1, then x = 2.  So B(2,0,1) is on the line.

    The given plane x + y - 4z = 4

    has as its normal vector <1,1,-4>.

    If we start at (1,3,0) and go in the direction of the normal to the given plane, we find C(2,4,-4).

    Note that vector AB = <1,-3,1>

    and vector BC = <0,4,-5>.

    The cross-product of AB and BC will be normal to the plane we are SEEKING.  That cross-product is

    <11,5,4>.

    The equation of the desired plane is

    11x + 5y + 4z = D,

    where D must be 11*1 + 5*3 + 4*0 = 26.

    The equation is 11x + 5y + 4z = 26.

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