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Math Help - Discrete Math Exam Review - Planes

  1. #1
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    Discrete Math Exam Review - Planes

    Determine a plan detailing on how you would find the value(s) of k, if any, that would make the planes kx + 2ky + 5z - 6 = 0 and -x + (k - 1)y - 7z - 9 = 0 perpendicular. Implement your plan and solve for k.

    I tried use the dot product to solve this problem, but it just got messy and got a headache ...

    Is there an easier way to solve for it and if so, please show in full steps?

    Answer:

    k = 5
    k = - \frac{7}{2}
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  2. #2
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    Recall that a plane given by the equation ax + by + cz + d = 0 has the vector (a,b,c) perpendicular to it.

    Now looking at your two planes, we see that the normals for each plane respectively are (k,2k,5) and (-1, k-1, -7). If the planes are perpendicular to each other, then so should their normals. So, take the dot product of these two vectors and set it to 0:

    (k,2k,5) \cdot (-1, k-1, -7) = 0
    -k + 2k(k-1) - 35 = 0

    etc. etc. and solve the quadratic.
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