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Lines and planes in three dimensions questions
A line is a point plus a direction to slide along. A plane is a point plus a normal to stay perpendicular to. Two short equations describe both objects.
7 original questions · 26 marks · the lines and planes in three dimensions notes · Further vectors
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Find a vector equation of the line through the points P(2, 1, 0) and Q(4, 2, 3).
Worked answer
Direction PQ = (2, 1, 3), so r = (2, 1, 0) + λ(2, 1, 3). B1 for the direction, B1 for a complete vector equation. Any point of the line and any non-zero multiple of the direction give an equally correct equation.Determine whether the point (8, 4, 9) lies on the line r = (2, 1, 0) + λ(2, 1, 3).
Worked answer
From x: 2 + 2λ = 8, so λ = 3. Then y = 1 + 3 = 4 and z = 0 + 9 = 9, both matching. All three coordinates agree at the same λ, so the point is on the line. M1 for finding λ from one coordinate, M1 for testing the other two, A1 for the conclusion.A plane has equation r·(2, −1, 2) = 9. Write down its cartesian equation, and determine whether the point (3, 1, 2) lies in it.
Worked answer
Cartesian: 2x − y + 2z = 9. At the point: 6 − 1 + 4 = 9, so yes, it lies in the plane. B1 for the cartesian equation, M1 for substituting the point, A1 for the conclusion. The scalar product form and the cartesian form are the same statement in different notation.Find a cartesian equation of the plane through A(1, 1, 1), B(2, 3, 1) and C(3, 1, 4).
Worked answer
In-plane directions: AB = (1, 2, 0) and AC = (2, 0, 3). A normal (a, b, c) needs a + 2b = 0 and 2a + 3c = 0; choosing b = 1 gives a = −2, c = 4/3, and scaling by 3 gives n = (−6, 3, 4). Then d = n·A = 1, so −6x + 3y + 4z = 1. Checking B and C: −12 + 9 + 4 = 1 and −18 + 3 + 16 = 1. M1 for two in-plane directions, M1 for the perpendicularity equations, A1 for the normal, M1 for finding the constant, A1 for the cartesian equation.Convert the line r = (2, 1, 0) + λ(2, 1, 3) to cartesian form, and explain what the equal fractions represent.
Worked answer
Solve each coordinate for λ: (x − 2)/2 = (y − 1)/1 = z/3. Each fraction equals λ, so setting them equal states that one value of the parameter produces all three coordinates at once, which is exactly membership of the line. M1 for solving each coordinate for λ, A1 for the cartesian form, B1 for the explanation.A line has direction vector (0, 2, 5). Explain the difficulty in writing its cartesian equation in the standard form, and give the correct way to write it.
Worked answer
The x-component of the direction is zero, so (x − a)/0 is meaningless. What it is trying to say is that x never changes. Write x = a separately and set the remaining fractions equal, giving x = a, (y − b)/2 = (z − c)/5. B1 for the zero denominator being meaningless, B1 for x being constant, B1 for the correct form. A zero denominator signals a frozen coordinate rather than a formula to force.Show that the lines r = (1, 2, 3) + λ(1, 0, −1) and r = (2, 0, 1) + μ(0, 1, 1) are skew, and find the shortest distance between them.
Worked answer
The directions are not multiples of each other, so the lines are not parallel. Equating components gives 1 + λ = 2, 2 = μ and 3 − λ = 1 + μ. The first two give λ = 1 and μ = 2, but then the third reads 2 = 3, so the lines never meet and are skew. A vector perpendicular to both is the cross product (1, 0, −1) × (0, 1, 1) = (1, −1, 1), of length √3. Take the vector between the two anchor points, (2, 0, 1) − (1, 2, 3) = (1, −2, −2), and project it onto that common perpendicular. The shortest distance is |(1, −2, −2)·(1, −1, 1)|/√3 = |1 + 2 − 2|/√3 = 1/√3 = √3/3, about 0.577. B1 for the directions not being parallel, M1 for equating components, A1 for the contradiction showing the lines are skew, M1 for the cross product, A1 for the common perpendicular, M1 for projecting the vector between the anchors, A1 for the shortest distance. Any pair of points, one on each line, gives the same projection, so the choice of anchors does not matter.
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