Lesson 03

Plotting and Sketching in R³

Algebra tells you what a line or a plane is; a sketch tells you what it looks like. This lesson is about translating equations into accurate three-dimensional drawings — and reading a drawing back into an equation — using one consistent, shared convention for how the axes sit on the page.

By the end of this lesson

  • Draw and label a consistent set of three-dimensional coordinate axes.
  • Plot a point in R³ using its shadow on the \(xy\)-plane.
  • Sketch a plane using its three intercepts.
  • Recognise and sketch a plane that is missing one variable.

Section 1

Setting up the axes

Before drawing anything, we fix a convention, so that every sketch in this course looks the same and can be read the same way: the \(z\)-axis points straight up, the \(x\)-axis points down and toward the viewer, and the \(y\)-axis points to the right. Together the \(x\)- and \(y\)-axes form the \(xy\)-plane — the “floor” that the \(z\)-axis rises out of.

That convention is a right-handed frame, and it is worth knowing the test: point the fingers of your right hand along the positive \(x\)-axis, curl them toward the positive \(y\)-axis, and your thumb points along positive \(z\). It matters because the cross product obeys the same rule. Draw a left-handed frame by accident and every normal vector you compute points into the page when it should point out of it, so every picture you draw afterwards disagrees with your algebra.

A right-handed set of coordinate axes Three coordinate axes drawn from a common origin and ticked at the integers: x pointing toward the viewer, y to the right, and z upward, with the xy-plane outlined. x y z the xy-plane toward you to the right upward
fig. 3.1The standard sketching convention used throughout this course: \(z\) upward, \(x\) toward you, \(y\) to the right, with the \(xy\)-plane forming the floor of the picture. The three positive axes divide space into eight octants; the one drawn here, with all three coordinates positive, is the first.

Section 2

Plotting a point in R³

To plot a point like \(P(2,3,4)\), walk out along each axis in turn: \(2\) units along \(x\), then \(3\) units parallel to \(y\), then \(4\) units straight up parallel to \(z\). Marking the point directly below \(P\) on the \(xy\)-plane — its shadow — makes the sketch far easier to read, since it shows exactly where \(P\) sits relative to the floor of the picture.

The three signs of the coordinates also tell you immediately which of the eight octants the point lies in. For \(P(2,3,4)\) all three coordinates are positive, so \(P\) lies in the octant where \(x>0,\ y>0,\ z>0\). A negative coordinate simply reverses one of the three walks: \((3,-2,4)\) means forward, left, and up.

Plotting the point (2, 3, 4) in three-space Ticked coordinate axes with dashed construction lines forming a box: two units along x, then three units parallel to y, then four units upward parallel to z, arriving at the marked point P at (2, 3, 4). The open circles are the intermediate corners. x y z P(2, 3, 4) (2, 3, 0) (2, 0, 0)
fig. 3.2Plotting \(P(2,3,4)\): walk \(2\) along \(x\), then \(3\) parallel to \(y\), then \(4\) parallel to \(z\). The open circle at \((2,3,0)\) on the \(xy\)-plane is the point’s shadow.
Plot any point, then drag to rotate Interactive

Type a point, then drag the figure to see the projection box from another angle. Rotating is the fastest way to convince yourself which octant a point with a negative coordinate is in.

Section 3

Sketching a plane from its intercepts

The fastest way to sketch a plane \(Ax + By + Cz + D = 0\) — with \(A,B,C,D\) all non-zero — is to find where it crosses each axis. Setting \(y = z = 0\) gives the \(x\)-intercept, setting \(x = z = 0\) gives the \(y\)-intercept, and setting \(x = y = 0\) gives the \(z\)-intercept. Plotting those three points and joining them with straight edges gives a triangular sketch: the visible corner of the plane.

\[\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1\]
(3.1)

Dividing the equation through by \(-D\) puts it in intercept form (3.1), where \(a,\,b,\,c\) are exactly the three intercepts — a useful check that they were computed correctly. The triangle is only a window onto the plane, which carries on past all three edges for ever; say so in your answer.

The intercept triangle of the plane 3x + 2y + 6z = 12 Ticked coordinate axes with a shaded triangle joining the three axis intercepts at (4, 0, 0), (0, 6, 0) and (0, 0, 2). Where the axes pass behind the triangle they are drawn faint and dashed. The triangle is a window onto a plane that continues beyond it. x y z (4, 0, 0) (0, 6, 0) (0, 0, 2)
fig. 3.3The plane \(3x + 2y + 6z - 12 = 0\), sketched from its three intercepts \((4,0,0)\), \((0,6,0)\) and \((0,0,2)\), matching the intercept form \(\tfrac{x}{4} + \tfrac{y}{6} + \tfrac{z}{2} = 1\).
Intercept-triangle builder Interactive

Type any coefficients and watch the intercepts and the triangle appear, including the awkward cases where a coefficient is zero, or where \(D=0\) and all three intercepts collapse to the origin.

Section 4

Planes missing a variable

When one variable is missing from a plane’s equation — for example \(2x + 5y = 10\), with no \(z\) term — that variable is completely unrestricted, and the plane runs parallel to its axis. Here \(z\) can take any value at all, so the plane stretches infinitely up and down, parallel to the \(z\)-axis.

To sketch this kind of plane, first draw the line \(2x + 5y = 10\) in the \(xy\)-plane using its two intercepts, then extend that line straight upward and downward to show the plane sweeping parallel to the missing axis. That ground line is the plane’s trace, and two traces are usually enough to place a plane convincingly when there is no intercept triangle to draw.

The plane 2x + 5y = 10 drawn in three-space Ticked coordinate axes with a vertical sheet standing on the line from (5, 0, 0) to (0, 2, 0) in the xy-plane and running parallel to the z-axis, because the equation contains no z term. That ground line is its trace. x y z (5, 0, 0) (0, 2, 0) 2x + 5y = 10
fig. 3.4The plane \(2x + 5y = 10\), missing a \(z\) term, sketched as the line through \((5,0,0)\) and \((0,2,0)\) extended parallel to the \(z\)-axis.

Practice

Worked examples

Example 1 · routine

Sketch the plane \(4x + 3y + 2z - 12 = 0\) using its intercepts.

  1. Set \(y = z = 0\): \(4x = 12\), so \(x = 3\) — the \(x\)-intercept is \((3,0,0)\).
  2. Set \(x = z = 0\): \(3y = 12\), so \(y = 4\) — the \(y\)-intercept is \((0,4,0)\).
  3. Set \(x = y = 0\): \(2z = 12\), so \(z = 6\) — the \(z\)-intercept is \((0,0,6)\).
Plot \((3,0,0)\), \((0,4,0)\) and \((0,0,6)\) and join them to form the triangular sketch of the plane, noting that the plane continues beyond all three edges. Intercept form checks it: \(\tfrac{x}{3} + \tfrac{y}{4} + \tfrac{z}{6} = 1\).
Example 2 · demanding

Describe and sketch the plane \(x - 2z = 6\). Explain why it has only two intercepts, not three.

  1. Notice that \(y\) is missing from the equation, so \(y\) is unrestricted — the plane runs parallel to the \(y\)-axis.
  2. Set \(z = 0\): \(x = 6\), giving the \(x\)-intercept \((6,0,0)\). Set \(x = 0\): \(-2z = 6\), so \(z = -3\), giving the \(z\)-intercept \((0,0,-3)\).
  3. There is no \(y\)-intercept, because setting \(x = z = 0\) gives \(0 = 6\), which is never true. The plane never crosses the \(y\)-axis, which is exactly what running parallel to it means.
Draw the line through \((6,0,0)\) and \((0,0,-3)\) in the \(xz\)-plane, then extend it parallel to the \(y\)-axis in both directions.

Watch out

Common mistakes

  • Forgetting to mark the point’s shadow on the \(xy\)-plane when plotting in R³, which leaves the sketch hard to read. Always drop a dashed line to the \(xy\)-plane and mark the shadow point — it anchors the picture and makes the depth clear.
  • Assuming every plane has exactly three intercepts. A plane missing a variable is parallel to that variable’s axis and has only two. Check for a missing term before hunting for a third intercept that does not exist — and when \(D = 0\), all three collapse to the origin and you should sketch traces instead.