Lesson 03

Plotting Points and Sketching Planes in R³

Everything in this topic is easier once you can draw it. A sketch does not have to be beautiful — it has to be right about the things that matter: which way the axes point, where the plane crosses them, and whether the object you are describing is a point, a line or a surface.

By the end of this lesson

  • Plot a point in a right-handed frame using its projection box.
  • Sketch a plane from its three intercepts.
  • Recognise, from a missing variable, that a plane is parallel to an axis or to a coordinate plane.
  • Recover an equation from a sketch, using intercept form.

The right-handed frame

R³ uses a right-handed frame: \(x\) toward you, \(y\) to the right, \(z\) upward. The name comes from 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. If you draw a left-handed frame by accident, every normal vector you compute will point into the page instead of out of it, and every picture you draw afterwards will quietly disagree with your algebra.

On paper the frame is drawn in projection, so the axes cannot all be at right angles on the page. The convention is to draw \(z\) vertically, \(y\) off to the right, and \(x\) coming down toward the viewer. Everything on this site is drawn from that one viewpoint.

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 frame every figure on this site uses. The three positive axes divide space into eight octants; the one shown, with all three coordinates positive, is the first.

Plotting a point

To plot \(P(2,3,4)\), walk the coordinates in order rather than trying to see the point directly. Go two units along \(x\); from there, three units parallel to \(y\); from there, four units straight up. Draw the three legs dashed and you have built the point's projection box.

The box is worth drawing every time. It shows the reader — and reminds you — that the point really is where you claim, and it makes the shadow of the point on the \(xy\)-plane, at \((2,3,0)\), visible. That shadow reappears in Lesson 05 as the foot of a perpendicular.

A negative coordinate simply reverses one of the walks. Reading the signs off a plotted point is a good habit: \((3,-2,4)\) 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. 3.2Two along \(x\), three parallel to \(y\), four upward. The open circles are the intermediate corners; the filled one is the point.
Plot any point — 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.

Intercepts and the intercept triangle

To sketch a plane from its Cartesian equation, find where it crosses each axis. On the \(x\)-axis both other coordinates are zero, so set \(y=z=0\) and solve; repeat for the other two. Joining the three intercepts gives a triangle that is a window onto the plane — the plane itself continues past every edge, for ever.

For \(3x+2y+6z=12\): setting \(y=z=0\) gives \(x=4\); setting \(x=z=0\) gives \(y=6\); setting \(x=y=0\) gives \(z=2\). Notice how much easier this is than solving for a general point: the intercepts are the three places where two of the three terms vanish.

In general, from \(Ax+By+Cz+D=0\) the intercepts are \(\left(-\tfrac{D}{A},0,0\right),\ \left(0,-\tfrac{D}{B},0\right),\ \left(0,0,-\tfrac{D}{C}\right)\) — whenever the corresponding coefficient is non-zero.

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 intercept triangle of \(3x+2y+6z=12\). The plane extends beyond all three edges; say so in your answer.
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.

Intercept form

When a plane has three genuine intercepts \(a\), \(b\) and \(c\), its equation can be written in a form that displays them:

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

Check it against the numbers above: dividing \(3x+2y+6z=12\) through by 12 gives \(\tfrac{x}{4}+\tfrac{y}{6}+\tfrac{z}{2}=1\), and the denominators are exactly the intercepts.

This is the form to use when a question runs the other way — when you are handed a sketch, or three intercepts, and asked for the equation. Write the intercepts into the denominators, then clear the fractions.

Missing variables, and traces

Special cases are worth memorising: a missing variable means the plane is parallel to that axis. The equation \(2x+5y=10\) in R³ contains no \(z\), which means \(z\) is free to be anything: whatever height you choose, the same condition on \(x\) and \(y\) applies. The plane is a vertical wall standing on the line \(2x+5y=10\) in the \(xy\)-plane.

Two missing variables go further still: equations like \(z=3\) are planes parallel to a whole coordinate plane — here, three units above the \(xy\)-plane and parallel to it.

When there is no intercept triangle to draw, sketch traces instead. A trace is what you get by setting one variable to zero: the curve — here always a line — where the plane meets a coordinate plane. Setting \(z=0\) in \(Ax+By+Cz+D=0\) leaves \(Ax+By+D=0\), a line in the \(xy\)-plane. Two traces are usually enough to place a plane convincingly.

The awkward case is \(D=0\). Then all three intercepts collapse to the origin, the triangle has no area, and traces are the only option: \(x+y+z=0\) passes through the origin, and its trace in the \(xy\)-plane is the line \(y=-x\).

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. 3.4No \(z\) in the equation means no \(z\)-intercept and a plane parallel to the \(z\)-axis. The gold line is its trace in the \(xy\)-plane.

Practice

Worked examples

Example 1 · routine

Plot \(P(2,3,4)\)

Show solution
  1. Move 2 units along the positive \(x\)-axis, reaching \((2,0,0)\).
  2. From there move 3 units parallel to the \(y\)-axis, reaching \((2,3,0)\) — the point's shadow on the \(xy\)-plane.
  3. From there move 4 units straight up, reaching \(P(2,3,4)\).
  4. Draw all three legs dashed and complete the box, so the reader can see the height.
Example 2 · routine

Sketch \(3x+2y+6z=12\)

Show solution
  1. \(y=z=0 \Rightarrow 3x=12 \Rightarrow (4,0,0)\)
  2. \(x=z=0 \Rightarrow 2y=12 \Rightarrow (0,6,0)\)
  3. \(x=y=0 \Rightarrow 6z=12 \Rightarrow (0,0,2)\)
  4. Plot the three intercepts, join them, and shade the triangle lightly. State that the plane continues beyond the triangle in every direction.
Example 3 · standard

Sketch \(2x+5y=10\) in R³

Show solution
  1. There is no \(z\) term, so setting \(x=y=0\) gives \(0=10\), which is impossible: there is no \(z\)-intercept.
  2. That is not an error. It says \(z\) is unconstrained, so the plane runs parallel to the \(z\)-axis.
  3. Find the trace in the \(xy\)-plane by setting \(z=0\): the line through \((5,0,0)\) and \((0,2,0)\).
  4. Sketch that line, then draw the vertical wall standing on it.
Example 4 · demanding

Recover the equation from a sketch

A plane cuts the axes at \((2,0,0)\), \((0,-3,0)\) and \((0,0,4)\). Find its Cartesian equation.

Show solution
  1. Write the intercepts straight into intercept form — signs included: \(\dfrac{x}{2} + \dfrac{y}{-3} + \dfrac{z}{4} = 1\).
  2. Multiply through by the lowest common denominator, 12: \(6x - 4y + 3z = 12\).
  3. Check all three intercepts: \((2,0,0):\ 12\ \checkmark\), \((0,-3,0):\ 12\ \checkmark\), \((0,0,4):\ 12\ \checkmark\)
  4. As a bonus, the normal can now be read off directly: \(\vv{n} = (6,-4,3)\).

Watch out

Three ways to lose a mark here
  • A missing intercept is information, not a mistake. When \(x=y=0\) produces \(0=10\), do not go back and hunt for an arithmetic slip — write down that the plane is parallel to the \(z\)-axis.
  • When \(D=0\), there is no triangle. All three intercepts sit at the origin. Sketch two traces instead, and say that the plane passes through the origin.
  • The triangle is a window, not the plane. A sketch that stops at the three edges with no comment invites the reader to think the plane is a bounded patch. One sentence fixes it.