Questions on lines and planes in three dimensions

Question 1

Find a vector equation of the line through the points P(2, 1, 0) and Q(4, 2, 3).

Accepted 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.

Question 2

Determine whether the point (8, 4, 9) lies on the line r = (2, 1, 0) + λ(2, 1, 3).

Accepted 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.

Question 3

Write down a normal vector to the plane 3x − 2y + z = 7, and express the plane in the form r·n = d.

Accepted Answer

n = (3, −2, 1), read directly from the coefficients. B1. r·(3, −2, 1) = 7. B1.

Question 4

State how many parameters the vector equation of a line needs, and how many a plane needs, and explain why.

Accepted Answer

A line needs one parameter and a plane needs two. B1. A line is one-dimensional, so a single direction vector generates every point of it from one anchor; a plane is two-dimensional, and one direction alone would only ever generate a line inside it, so a second, non-parallel direction is needed to reach every point. B1.

Question 5

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.

Accepted 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.

Question 6

Find a cartesian equation of the plane through A(1, 1, 1), B(2, 3, 1) and C(3, 1, 4).

Accepted 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.

Question 7

Convert the line r = (2, 1, 0) + λ(2, 1, 3) to cartesian form, and explain what the equal fractions represent.

Accepted 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.

Question 8

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.

Accepted 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.

Question 9

A plane has vector equation r = (1, 0, 2) + λ(1, 1, 0) + μ(0, 1, 1). Find a cartesian equation of the plane.

Accepted Answer

The two directions are (1, 1, 0) and (0, 1, 1). A normal (a, b, c) must satisfy a + b = 0 and b + c = 0; choosing b = −1 gives a = 1 and c = 1, so n = (1, −1, 1). M1 for the two perpendicularity equations, A1 for the normal. d = n·(1, 0, 2) = 1 − 0 + 2 = 3, so the plane is x − y + z = 3. M1 for d, A1 for the cartesian equation.

Question 10

Two lines are given by r = (1, 2, 0) + λ(2, −1, 3) and r = (3, 1, 3) + μ(−4, 2, −6). Explain what the two direction vectors show about the lines, and determine whether the lines are the same line or two distinct parallel lines.

Accepted Answer

(−4, 2, −6) = −2(2, −1, 3), so the two directions are parallel: the lines are parallel or identical, never skew or crossing at one point. B1 for the scalar-multiple relationship and its meaning. To decide which, test whether a point of the first line lies on the second. Substituting the anchor (1, 2, 0) into the second line's parametrised form: 3 − 4μ = 1 gives μ = 1/2; 1 + 2μ = 2 gives μ = 1/2; 3 − 6μ = 0 gives μ = 1/2. M1 for setting up the three component equations, A1 for the same μ from all three. Because a single value of μ satisfies all three coordinates, (1, 2, 0) lies on the second line too, so the two equations describe the same line, not merely two parallel lines. M1 for checking every component agrees, A1 for the conclusion.

Question 11

Two vectors lying in a plane are b = (1, −1, 2) and c = (2, 1, 0). Given that the plane passes through (0, 3, 1), find its cartesian equation.

Accepted Answer

A normal (p, q, r) must satisfy n·b = 0 and n·c = 0: p − q + 2r = 0 and 2p + q = 0. M1 for both perpendicularity equations. From the second, q = −2p. Substituting into the first: p − (−2p) + 2r = 0, so 3p + 2r = 0 and r = −3p/2. Choosing p = 2 gives q = −4 and r = −3, so n = (2, −4, −3). M1 for solving with a free choice, A1 for the normal. d = n·(0, 3, 1) = 0 − 12 − 3 = −15, so the plane is 2x − 4y − 3z = −15. M1 for d, A1 for the cartesian equation. Checking with b: moving from the given point along b changes the left side by n·b = 0, and the same holds for c, confirming both directions really do lie in this plane.

Question 12

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.

Accepted 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.

Question 13

A plane has cartesian equation 2x + y − z = 4. Write the plane in the form r·n = d, and find a two-parameter vector equation r = a + λb + μc for it, showing that b and c are not parallel.

Accepted Answer

Reading off the coefficients, r·(2, 1, −1) = 4. B1. To find a point on the plane, try x = 2, y = 0, z = 0: 2(2) + 0 − 0 = 4, so a = (2, 0, 0) is a valid anchor. M1 for a point on the plane. Two directions must satisfy n·(direction) = 0. Trying b = (1, −2, 0): 2(1) + 1(−2) − 0 = 0. Trying c = (0, 1, 1): 2(0) + 1(1) − 1(1) = 0. M1 for finding each direction, A1 for both. Since b and c are not scalar multiples of each other they span the whole plane rather than a single line inside it. B1 for the non-parallel check. So r = (2, 0, 0) + λ(1, −2, 0) + μ(0, 1, 1). A1 for the vector equation. Substituting back, x = 2 + λ, y = −2λ + μ, z = μ, gives 2x + y − z = 2(2 + λ) + (−2λ + μ) − μ = 4 for every λ and μ, confirming the two forms describe the same plane.

Question 14

Show that the points A(1, 2, 3), B(2, 4, 5) and C(4, 8, 9) are collinear, and explain why they do not determine a unique plane.

Accepted Answer

AB = (1, 2, 2) and AC = (3, 6, 6). M1 for both vectors. Since AC = 3AB, the two vectors point along the same line through A, so A, B and C are collinear. A1 for the scalar-multiple relationship and the conclusion. Finding a plane through three points needs two directions that are not parallel, since a single direction only ever sweeps out a line; here every pair of the three points gives a direction along the same line, so no second, independent direction is available. B1. Infinitely many planes contain a given line, so infinitely many planes pass through A, B and C. B1 for the geometric consequence. The usual method of solving n·AB = 0 and n·AC = 0 collapses to one equation, not two independent ones, leaving n underdetermined. B1 for what fails in the calculation.

Question 15

The lines r = (1, 0, 0) + λ(1, 1, 0) and r = (0, k, 1) + μ(2, 1, 1) are given for a constant k. Find the value of k for which the lines meet, and explain why they are skew for every other value of k.

Accepted Answer

Equating components: 1 + λ = 2μ, λ = k + μ, and 0 = 1 + μ. M1 for the three equations. The third does not involve k, and gives μ = −1 whatever k is. A1. Substituting into the first, 1 + λ = −2, so λ = −3, again independent of k. A1. The second equation is the only one carrying k: −3 = k + (−1), so k = −2. M1 for using the second equation, A1 for k = −2. For every other value of k, λ = −3 and μ = −1 satisfy the first and third equations but not the second, so the three equations have no common solution. The directions (1, 1, 0) and (2, 1, 1) are not scalar multiples of each other for any k, since 2/1 ≠ 1/1, so the lines are never parallel; not meeting and not parallel makes them skew for every k ≠ −2. B1.