Determine the equation of the plane in the form

1 answer:

8 0

A plane parallel to vectors <9, 6, 2> and <-8, -4, -5> is given by
$\left|\begin{array}{ccc}x&y&z\\9&6&2\\-8&-4&-5\end{array}\right| =0\\ (-30 + 8)x - (-45 + 16)y + (-36 + 48)z = 0\\-22x+29y+12z=0$

Since the plane passes through point (5, -5, 8), then
-22(5) + 29(-5) + 12(8) = -110 - 145 + 96 = -159

The required plane is -22x + 29y + 12z = -159
z = 11/6x - 29/12y - 53/4

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