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Calculus Grade 12

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Find the equation of the plane that passes through the point (3,7,-1) and is perpendicular to the line of intersection of the planes x-y-2z+3=0 and 3x-2y+z+5=0

  • Calculus Grade 12 -

    The line of intersection of the planes is in the direction of the cross-product of the two normals:

    {1,-1,-2}x{3,-2,1} = {-5,-7,1}

    Now we have a point and a normal vector.
    The equation of the plane is thus

    -5(x-3) - 7(y-7) + 1(z-1) = 0
    -5x + 15 - 7y + 49 + z - 1 = 0
    5x+7y-z = 63

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