27 exam-style questions on Equations of a plane — topic 3.17 of Geometry and Trigonometry in the IB Mathematics: Analysis and Approaches syllabus — with answers and explanations. Samples below; the full set is free with an account.
A plane passes through the points (1, 0, 0), (0, 2, 0) and (0, 0, 3). Determine its Cartesian equation.
Answer: 6x + 3y + 2z = 6
With intercepts 1, 2 and 3 the plane is x/1 + y/2 + z/3 = 1; multiplying by 6 gives 6x + 3y + 2z = 6, which all three points satisfy.
A plane contains the two direction vectors u = (1, 1, 0) and v = (0, 1, 1). Which vector is normal to the plane?
Answer: (1, -1, 1)
The normal is u × v = (1·1 - 0·1, 0·0 - 1·1, 1·1 - 1·0) = (1, -1, 1); remember the middle component of a cross product carries a sign change.
Calculate the perpendicular distance from the origin to the plane 2x - y + 2z = 9.
Answer: 3
Distance = |d| / |n| = 9 / √(4 + 1 + 4) = 9/3 = 3; forgetting to divide by |n| gives the distractor 9.
A plane has equation r · (1, 2, -2) = 4. Which of the following points lies on the plane?
Answer: (2, 1, 0)
Substituting (2, 1, 0) gives 2 + 2 - 0 = 4, so it satisfies r · n = 4; the other points give 1, -4 and 9 respectively.
Determine the Cartesian equation of the plane r = (1, 2, 0) + λ(1, 0, 1) + μ(0, 1, 1).
Answer: x + y - z = 3
The normal is (1, 0, 1) × (0, 1, 1) = (-1, -1, 1), giving -x - y + z = -3, i.e. x + y - z = 3; x + y + z = 3 and x - y + z = -1 also pass through (1, 2, 0), but their normals are not perpendicular to both direction vectors.
Which of the following is a correct vector (parametric) form of the plane x + 2y - z = 4?
Answer: r = (4, 0, 0) + λ(2, -1, 0) + μ(1, 0, 1)
The point (4, 0, 0) satisfies the equation and both (2, -1, 0) and (1, 0, 1) are perpendicular to the normal (1, 2, -1); using the normal itself as a direction vector, or a point not on the plane, is wrong.
That was 6 of 27 Equations of a plane questions.
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