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Lines and planes in space

Algebra · level 8, lesson 8 of 8

A question to start

On a flat page, two straight lines either meet at one point, are parallel, or are the same line. Is that also true in space?

Look around a room. Take the line where the floor meets the front wall, and the line where the ceiling meets a side wall. They are not parallel (one runs left to right, the other front to back), and yet they never meet: one is down at the floor, the other up at the ceiling. In three dimensions there is a fourth possibility. Such lines are called skew. To handle space properly, we need equations for lines and planes that work in three dimensions, and the slope \(m\) from \(y = mx + b\) will not do the job.

The idea

Points and vectors in space

A point in space has three coordinates \((x, y, z)\). A vector such as \(\vec{d} = (2, -3, 5)\) is a step: 2 in the \(x\) direction, \(-3\) in \(y\), 5 in \(z\). You met vectors and the dot product in geometry: \(\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3 = \lvert\vec{a}\rvert\lvert\vec{b}\rvert\cos\theta\), and two vectors are perpendicular exactly when their dot product is 0.

The equation of a line

A slope is one number, enough to describe a direction on a flat page. In space a direction needs more, so we use a direction vector \(\vec{d}\). A line is "start at a point \(\vec{r}_0\), then take any number \(t\) of steps \(\vec{d}\)":

\[\vec{r} = \vec{r}_0 + t\,\vec{d}, \qquad t \text{ any real number}.\]

This is the vector equation of the line. The number \(t\) is called a parameter: each value of \(t\) gives one point (\(t = 0\) gives the start, \(t = 1\) one step along, \(t = -2\) two steps backwards). Writing it coordinate by coordinate gives the parametric equations. For the line through \((1, 2, -1)\) with direction \((2, -3, 5)\):

\[x = 1 + 2t, \qquad y = 2 - 3t, \qquad z = -1 + 5t.\]

To test whether a point is on the line, find the \(t\) that works for one coordinate and check it in the others.

How two lines can sit in space

Directions Do they meet? Name
parallel (one is a multiple of the other) yes, everywhere the same line
parallel no parallel lines
not parallel yes, at one point intersecting lines
not parallel no skew lines

To decide, set the two parametric equations equal (with different parameters, \(t\) and \(s\)) and solve.

The cross product

To describe a plane we need a vector that sticks straight out of it. The cross product of \(\vec{a} = (a_1, a_2, a_3)\) and \(\vec{b} = (b_1, b_2, b_3)\) is

\[\vec{a} \times \vec{b} = (a_2b_3 - a_3b_2,\; a_3b_1 - a_1b_3,\; a_1b_2 - a_2b_1).\]

A way to remember it: each entry uses the other two positions, in the cyclic order 1, 2, 3, 1, 2. The first entry uses positions 2 and 3, the second uses 3 and 1, the third uses 1 and 2; each time it is "first one's \(a\) times second one's \(b\), minus the other way round".

a b a × b area = |a × b|

It is perpendicular to both. Dot it with \(\vec{a}\):

\[a_1(a_2b_3 - a_3b_2) + a_2(a_3b_1 - a_1b_3) + a_3(a_1b_2 - a_2b_1).\]

Expand: \(a_1a_2b_3 - a_1a_3b_2 + a_2a_3b_1 - a_1a_2b_3 + a_1a_3b_2 - a_2a_3b_1\). Every term cancels with another, so the sum is 0. The same happens with \(\vec{b}\). So \(\vec{a} \times \vec{b}\) is perpendicular to both \(\vec{a}\) and \(\vec{b}\).

Its length is the area of the parallelogram. Expanding both sides (a page of patient algebra) shows

\[\lvert\vec{a} \times \vec{b}\rvert^2 = \lvert\vec{a}\rvert^2\lvert\vec{b}\rvert^2 - (\vec{a} \cdot \vec{b})^2 = \lvert\vec{a}\rvert^2\lvert\vec{b}\rvert^2(1 - \cos^2\theta) = \lvert\vec{a}\rvert^2\lvert\vec{b}\rvert^2\sin^2\theta.\]

So \(\lvert\vec{a} \times \vec{b}\rvert = \lvert\vec{a}\rvert\lvert\vec{b}\rvert\sin\theta\). The parallelogram with sides \(\vec{a}\) and \(\vec{b}\) has base \(\lvert\vec{a}\rvert\) and height \(\lvert\vec{b}\rvert\sin\theta\), so this is exactly its area. A triangle with those two sides has half that area.

Its direction follows the right-hand rule: point the fingers of your right hand along \(\vec{a}\), curl them towards \(\vec{b}\), and your thumb points along \(\vec{a} \times \vec{b}\). So order matters: \(\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})\). For example \((1, 0, 0) \times (0, 1, 0) = (0, 0, 1)\): \(x\) across to \(y\) gives \(z\) upward.

The equation of a plane

A plane is fixed by one point \(P_0 = (x_0, y_0, z_0)\) in it and a normal vector \(\vec{n} = (a, b, c)\) perpendicular to it. A point \(P = (x, y, z)\) is in the plane exactly when the step \(\overrightarrow{P_0P}\) lies flat in the plane, that is, is perpendicular to \(\vec{n}\):

\[\vec{n} \cdot \overrightarrow{P_0P} = 0 \quad\Longleftrightarrow\quad a(x - x_0) + b(y - y_0) + c(z - z_0) = 0.\]

Multiplying out gives the scalar equation of the plane:

\[ax + by + cz = d, \qquad \text{where } d = ax_0 + by_0 + cz_0.\]

The coefficients of \(x\), \(y\), \(z\) are the normal vector. Given three points of a plane, two steps between them lie in the plane, and their cross product is a normal.

Where a line meets a plane

Put the line's parametric equations into the plane's equation. You get one equation in \(t\):

  • one solution: the line crosses the plane at one point;
  • no solution (like \(0 = 5\)): the line is parallel to the plane and never meets it;
  • every \(t\) works (like \(0 = 0\)): the line lies in the plane.

Three equations, three planes

Each equation \(ax + by + cz = d\) is a plane, so solving three equations in three unknowns means finding where three planes meet. The possibilities:

  • one point: like two walls and the floor at a corner of a room;
  • a line: like three pages of an open book meeting along the spine (infinitely many solutions);
  • no point: two planes parallel, or three planes meeting in pairs along three parallel lines, like the three sides of a triangular tent (a prism);
  • the whole plane: all three equations describe the same plane.

Elimination, from level 6, works exactly as before: combine equations to remove one variable at a time.

Worked examples

1. A line through two points. Find the line through \(A(1, 2, -1)\) and \(B(3, -1, 4)\). Is \((5, -4, 9)\) on it?

  • Direction \(\overrightarrow{AB} = (3 - 1, -1 - 2, 4 - (-1)) = (2, -3, 5)\).
  • \(\vec{r} = (1, 2, -1) + t(2, -3, 5)\), or \(x = 1 + 2t\), \(y = 2 - 3t\), \(z = -1 + 5t\).
  • \(x = 5\) needs \(t = 2\). Then \(y = 2 - 6 = -4\) ✓ and \(z = -1 + 10 = 9\) ✓. Yes, it is on the line.

2. A cross product. Find \((1, 2, 3) \times (4, 5, 6)\) and the area of the parallelogram they make.

  • \((2 \cdot 6 - 3 \cdot 5,\; 3 \cdot 4 - 1 \cdot 6,\; 1 \cdot 5 - 2 \cdot 4) = (-3, 6, -3)\).
  • Check: \((1, 2, 3) \cdot (-3, 6, -3) = -3 + 12 - 9 = 0\) ✓; \((4, 5, 6) \cdot (-3, 6, -3) = -12 + 30 - 18 = 0\) ✓.
  • Area \(= \sqrt{9 + 36 + 9} = \sqrt{54} = 3\sqrt6 \approx 7.35\).

3. A plane through three points. \(P(1, 0, 2)\), \(Q(2, 1, 0)\), \(R(0, 3, 1)\).

  • \(\overrightarrow{PQ} = (1, 1, -2)\) and \(\overrightarrow{PR} = (-1, 3, -1)\).
  • Normal \(\overrightarrow{PQ} \times \overrightarrow{PR} = (1(-1) - (-2)3,\; (-2)(-1) - 1(-1),\; 1 \cdot 3 - 1(-1)) = (5, 3, 4)\).
  • \(5x + 3y + 4z = d\), and \(P\) gives \(d = 5 + 0 + 8 = 13\). The plane is \(5x + 3y + 4z = 13\).
  • Check \(Q\): \(10 + 3 + 0 = 13\) ✓. \(R\): \(0 + 9 + 4 = 13\) ✓.

4. A line meets the plane. Where does the line of example 1 meet the plane of example 3?

  • \(5(1 + 2t) + 3(2 - 3t) + 4(-1 + 5t) = 13\), so \(5 + 10t + 6 - 9t - 4 + 20t = 13\), so \(7 + 21t = 13\), so \(t = \tfrac27\).
  • Point: \(\left(1 + \tfrac47,\; 2 - \tfrac67,\; -1 + \tfrac{10}{7}\right) = \left(\tfrac{11}{7}, \tfrac87, \tfrac37\right)\).
  • Check: \(\tfrac{55}{7} + \tfrac{24}{7} + \tfrac{12}{7} = \tfrac{91}{7} = 13\). ✓

5. Three planes meeting at a point. Solve \(x + y + z = 6\), \(2x - y + z = 3\), \(x + 2y - z = 2\).

  • Add the first and third: \(2x + 3y = 8\). Add the second and third: \(3x + y = 5\).
  • From \(3x + y = 5\): \(y = 5 - 3x\). Then \(2x + 3(5 - 3x) = 8\), so \(-7x = -7\), so \(x = 1\), \(y = 2\).
  • From the first equation, \(z = 6 - 1 - 2 = 3\). The planes meet at the single point \((1, 2, 3)\). Check the second: \(2 - 2 + 3 = 3\) ✓.

6. Three planes meeting nowhere. Solve \(x + y = 1\), \(y + z = 1\), \(x - z = 2\).

  • Subtract the second from the first: \(x - z = 0\). But the third says \(x - z = 2\). Impossible: no solution.
  • No two normals, \((1, 1, 0)\), \((0, 1, 1)\), \((1, 0, -1)\), are multiples of each other, so no two planes are parallel. Each pair meets in a line, and the three lines are parallel: the triangular tent.

Try it

  1. Write parametric equations for the line through \((2, -1, 0)\) with direction \((1, 4, -2)\). Find the point with \(t = 3\).
  2. Is \((4, 5, 1)\) on the line \(\vec{r} = (1, -1, 3) + t(1, 2, -1)\)?
  3. Find \((2, 0, 1) \times (1, 3, -1)\) and check it is perpendicular to both.
  4. Find the area of the triangle with corners \(O(0, 0, 0)\), \(A(2, 0, 1)\) and \(B(1, 3, -1)\).
  5. Find the equation of the plane through \((1, 2, 3)\) with normal \((2, -1, 4)\).
  6. Where does the line \(x = 1 + t\), \(y = 2t\), \(z = 3 - t\) meet the plane \(x + y + z = 10\)?
  7. Solve \(x + y + z = 2\), \(x - y + 2z = 6\), \(2x + y - z = -1\).
  8. Show that the lines \(\vec{r} = (1, 0, 0) + t(1, 1, 0)\) and \(\vec{r} = (0, 1, 1) + s(1, -1, 0)\) are skew.
Answers
  1. \(x = 2 + t\), \(y = -1 + 4t\), \(z = -2t\). At \(t = 3\): \((5, 11, -6)\).
  2. \(x = 4\) needs \(t = 3\). Then \(y = -1 + 6 = 5\) ✓ but \(z = 3 - 3 = 0\), not 1. No.
  3. \((0 \cdot (-1) - 1 \cdot 3,\; 1 \cdot 1 - 2 \cdot (-1),\; 2 \cdot 3 - 0 \cdot 1) = (-3, 3, 6)\). Checks: \((2, 0, 1) \cdot (-3, 3, 6) = -6 + 0 + 6 = 0\) ✓ and \((1, 3, -1) \cdot (-3, 3, 6) = -3 + 9 - 6 = 0\) ✓.
  4. Half the length of \(\overrightarrow{OA} \times \overrightarrow{OB} = (-3, 3, 6)\) from question 3: \(\tfrac12\sqrt{9 + 9 + 36} = \tfrac12\sqrt{54} = \tfrac{3\sqrt6}{2} \approx 3.67\).
  5. \(2x - y + 4z = d\) with \(d = 2 - 2 + 12 = 12\): \(2x - y + 4z = 12\).
  6. \((1 + t) + 2t + (3 - t) = 10\), so \(4 + 2t = 10\), so \(t = 3\). The point is \((4, 6, 0)\), and \(4 + 6 + 0 = 10\) ✓.
  7. Remove \(z\) twice. The first plus the third: \(3x + 2y = 1\). The second plus twice the third: \((x - y + 2z) + (4x + 2y - 2z) = 6 - 2\), so \(5x + y = 4\). From that, \(y = 4 - 5x\); then \(3x + 2(4 - 5x) = 1\), so \(-7x = -7\), so \(x = 1\) and \(y = -1\). The first equation gives \(z = 2 - 1 + 1 = 2\). The planes meet at \((1, -1, 2)\). Check the second equation: \(1 + 1 + 4 = 6\) ✓.
  8. Directions \((1, 1, 0)\) and \((1, -1, 0)\) are not multiples, so the lines are not parallel. To meet, we need \(1 + t = s\), \(t = 1 - s\) and \(0 = 1\) (the \(z\)-coordinates). The last is impossible, so they never meet. Not parallel and not meeting: skew. (One line lies in the floor \(z = 0\), the other in the level \(z = 1\), like the room in the opening question.)

Common mistake

Using the same parameter for two lines. When you set two lines equal, give them different letters (\(t\) and \(s\)). Using \(t\) for both asks whether the two moving points are in the same place at the same time, which is a different question.

Mixing up the normal and the direction. In \(ax + by + cz = d\), the vector \((a, b, c)\) is perpendicular to the plane, not along it. A line along the direction \((a, b, c)\) crosses the plane straight through.

Where it leads