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The two planes 3x - y + z + 1 = 0 and 5x + y + 3z = 0 meet along the line PQ. Find the equation of the plane that passes through the point (2, 1, 4) and is perpendicular to PQ.
- x + y - 2z = 5
- x + y + 2z = -5
- x + y + 2z = 5
- x + y - 2z = -5
Correct answer: x + y - 2z = -5
Solution
The line PQ has direction n1 x n2 = (-4,-4,8), proportional to (1,1,-2). The required plane has this as its normal and passes through (2,1,4): 1(x-2)+1(y-1)-2(z-4)=0 -> x+y-2z = -5.
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