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Find the plane through the point P = (2, 1, -1) which is perpendicular to the planes 2x + y - z = 3 and x + 2y + z = 2.

So far, what I'm doing is taking the two direction vectors from the 2 given planes (2, 1, -1) and (1, 2, 1) and taking the cross product between them to get (3, 3, 3). I then used this directional derivative to get an equation of a plane through the point (2, 1, -1) like so: 3(x - 2) + 3(y - 1) + 3(z + 1) = 0. However, my answer is x + y + z = 3 and the correct answer is x - y + z = 0. What am I doing wrong?

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You made a mistake in the cross product. It should be (3,-3,3).

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  • $\begingroup$ Does it matter which way do the cross product of the vectors in? Because when I do (2, 1, -1) x (1, 2, 1), I get (3, 3, 3), but if I do (1, 2, 1) x (2, 1, -1), I get (3, -3, 3). $\endgroup$ – Ellie Queens Dec 15 '17 at 2:13
  • $\begingroup$ @EllieQueens Reversing the arguments to the cross product changes the sign of the result—that is, the signs of all of the coordinates. However, you’re getting the wrong answer both ways. $\endgroup$ – amd Dec 15 '17 at 7:16

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