Find the equation of the plane passing through (2, 1, -1) with normal vector n = <3, -1, 4>.
Answer and explanation
Correct answer: C. 3x - y + 4z - 1 = 0
- Step 1: Standard plane equation: A(x - x0) + B(y - y0) + C(z - z0) = 0.
- Step 2: Substitute n = <3, -1, 4> and point (2, 1, -1):
- Step 3: 3(x - 2) - 1(y - 1) + 4(z - (-1)) = 0.
- Step 4: Expand: 3x - 6 - y + 1 + 4z + 4 = 0.
- Step 5: Combine constants: 3x - y + 4z - 1 = 0.
Answer
3x - y + 4z - 1 = 0.
Why not the others: '3x - y + 4z + 1 = 0' makes a sign error when combining constants (-6 + 1 + 4 = -1); '2x + y - z - 6 = 0' uses the point coordinates as normal coefficients; '3x + y - 4z - 9 = 0' inverts signs of normal components.
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