This note covers where lines and planes meet, and how to measure the angle between a line and a plane or between two planes. These skills let you answer questions such as where a cable pierces a floor or how steeply two roof panels meet.
Key ideas
Intersection of a line and a plane. Write the line in parametric form, \(x=x_0+tb_1\), \(y=y_0+tb_2\), \(z=z_0+tb_3\), and substitute into the Cartesian equation of the plane. This gives an equation in \(t\) only. There are three outcomes:
- One value of \(t\): the line meets the plane at a single point. Put that \(t\) back into the line to get the point.
- An equation that is never true (such as \(0=4\)): the line is parallel to the plane and does not meet it.
- An equation that is always true (such as \(0=0\)): the line lies in the plane.
A quick test for parallel: the line direction \(\mathbf{b}\) is perpendicular to the normal \(\mathbf{n}\), so \(\mathbf{b}\cdot\mathbf{n}=0\). Then check one point of the line to decide between "no intersection" and "line in plane".
Angle between a line and a plane. The angle \(\theta\) between a line and a plane is measured from the line to the plane itself. The normal is at \(90^\circ\) to the plane, so the angle between the line and the normal is \(90^\circ-\theta\). This swaps cosine for sine: \[ \sin\theta=\frac{|\mathbf{b}\cdot\mathbf{n}|}{|\mathbf{b}||\mathbf{n}|}. \]
Angle between two planes. The angle between two planes equals the angle between their normals. Use the acute angle: \[ \cos\theta=\frac{|\mathbf{n}_1\cdot\mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|}. \]
Intersection of two planes. Two planes are either parallel (parallel normals, no common point unless they are the same plane) or they meet in a straight line. That line has direction \(\mathbf{n}_1\times\mathbf{n}_2\), because it lies in both planes and so is perpendicular to both normals. To find a point on it, choose a convenient value for one coordinate (often \(z=0\)) and solve the two equations for the other two.
Intersection of three planes. Treat the three Cartesian equations as a system of three linear equations in \(x\), \(y\), \(z\). Solve it with your GDC's simultaneous equation solver or by elimination. The geometry matches the algebra:
- A unique solution: the three planes meet at one point.
- Infinitely many solutions: the planes share a line (like pages meeting at the spine of a book), or all three are the same plane.
- No solution: the planes have no common point. Either two are parallel, or they form a triangular prism shape with each pair meeting in a line.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| Substitute line into plane | Turns "where do they meet?" into one equation in \(t\). |
| \(\mathbf{b}\cdot\mathbf{n}=0\) | Line is parallel to the plane (or lies in it). |
| \(\sin\theta=\dfrac{|\mathbf{b}\cdot\mathbf{n}|}{|\mathbf{b}||\mathbf{n}|}\) | Angle \(\theta\) between a line with direction \(\mathbf{b}\) and a plane with normal \(\mathbf{n}\). |
| \(\cos\theta=\dfrac{|\mathbf{n}_1\cdot\mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|}\) | Acute angle between two planes with normals \(\mathbf{n}_1\), \(\mathbf{n}_2\). |
| \(\mathbf{n}_1\times\mathbf{n}_2\) | Direction of the line where two non-parallel planes meet. |
| System of three equations | Solve for the intersection of three planes: one point, a line, or nothing. |
Worked examples
Example 1 A laser beam follows the line \(\mathbf{r}=\begin{pmatrix}1\\0\\2\end{pmatrix}+t\begin{pmatrix}2\\1\\-1\end{pmatrix}\) and strikes a mirror lying in the plane \(3x-y+2z=11\). (a) Find the point where the beam meets the mirror. (b) Find the angle between the beam and the mirror.
(a) The line has \(x=1+2t\), \(y=t\), \(z=2-t\). Substitute into the plane:
\(3(1+2t)-t+2(2-t)=11\), so \(3+6t-t+4-2t=11\), so \(7+3t=11\), so \(t=\tfrac43\).
Then \(x=1+\tfrac83=\tfrac{11}{3}\), \(y=\tfrac43\), \(z=2-\tfrac43=\tfrac23\).
Check in the plane: \(3\cdot\tfrac{11}{3}-\tfrac43+2\cdot\tfrac23=11-\tfrac43+\tfrac43=11\). Correct.
(b) \(\mathbf{b}=\begin{pmatrix}2\\1\\-1\end{pmatrix}\), \(\mathbf{n}=\begin{pmatrix}3\\-1\\2\end{pmatrix}\).
\(\mathbf{b}\cdot\mathbf{n}=6-1-2=3\), \(|\mathbf{b}|=\sqrt6\), \(|\mathbf{n}|=\sqrt{14}\).
\(\sin\theta=\dfrac{3}{\sqrt{6}\sqrt{14}}=\dfrac{3}{\sqrt{84}}=0.3273\), so \(\theta=\sin^{-1}(0.3273)=19.1^\circ\).
Answer: (a) \(\left(\tfrac{11}{3},\tfrac43,\tfrac23\right)\); (b) \(19.1^\circ\)
Example 2 Two roof panels lie in the planes \(\Pi_1: x+2y-2z=5\) and \(\Pi_2: 2x-y+2z=3\). (a) Find the acute angle between the panels. (b) Find a vector equation of the line where they meet.
(a) \(\mathbf{n}_1=\begin{pmatrix}1\\2\\-2\end{pmatrix}\), \(\mathbf{n}_2=\begin{pmatrix}2\\-1\\2\end{pmatrix}\).
\(\mathbf{n}_1\cdot\mathbf{n}_2=2-2-4=-4\), \(|\mathbf{n}_1|=\sqrt{1+4+4}=3\), \(|\mathbf{n}_2|=\sqrt{4+1+4}=3\).
\(\cos\theta=\dfrac{|-4|}{3\times3}=\dfrac49\), so \(\theta=63.6^\circ\).
(b) Direction: \(\mathbf{n}_1\times\mathbf{n}_2=\begin{pmatrix}(2)(2)-(-2)(-1)\\ (-2)(2)-(1)(2)\\ (1)(-1)-(2)(2)\end{pmatrix}=\begin{pmatrix}2\\-6\\-5\end{pmatrix}\).
A point on both planes: put \(z=0\). Then \(x+2y=5\) and \(2x-y=3\). From the second, \(y=2x-3\). Substituting, \(x+4x-6=5\), so \(x=\tfrac{11}{5}\) and \(y=\tfrac75\).
Check in \(\Pi_2\): \(2\cdot\tfrac{11}{5}-\tfrac75=\tfrac{15}{5}=3\). Correct.
Answer: (a) \(63.6^\circ\); (b) \(\mathbf{r}=\begin{pmatrix}11/5\\7/5\\0\end{pmatrix}+s\begin{pmatrix}2\\-6\\-5\end{pmatrix}\)
Example 3 Find the intersection of the planes \(x+y+z=6\), \(2x-y+z=3\) and \(x-2y+3z=6\).
This is a system of three linear equations. Use the simultaneous equation solver on your GDC with the coefficient matrix \(\begin{pmatrix}1&1&1\\2&-1&1\\1&-2&3\end{pmatrix}\) and constants \(6,3,6\).
The solver gives a single solution, \(x=1\), \(y=2\), \(z=3\).
Check in each equation: \(1+2+3=6\); \(2-2+3=3\); \(1-4+9=6\). All correct.
A unique solution means the three planes meet at one point.
Answer: the single point \((1,2,3)\)
Common errors
- Common error: using cosine in the line-and-plane angle formula. The formula with the normal gives sine, because the normal is perpendicular to the plane.
- Common error: reporting an obtuse angle between planes or a negative \(\sin\theta\). Use the modulus of the scalar product so that the angle is acute.
- Common error: using a direction vector of the line where a normal is needed, or the other way round. Direction goes with the line, normal goes with the plane.
- Common error: concluding "no intersection" as soon as \(\mathbf{b}\cdot\mathbf{n}=0\). The line may lie in the plane, so test a point of the line.
- Common error: forgetting to substitute \(t\) back into the line to get coordinates, and giving only the value of \(t\).
Exam tip: Always put the intersection point back into the plane equation (or equations) to confirm it. A wrong sign shows up immediately.
Check yourself
- Find where the line \(\mathbf{r}=\begin{pmatrix}0\\1\\2\end{pmatrix}+t\begin{pmatrix}1\\2\\1\end{pmatrix}\) meets the plane \(x+y+z=9\).
Answer
\(t+(1+2t)+(2+t)=9\), so \(3+4t=9\) and \(t=1.5\). The point is \((1.5,\,4,\,3.5)\). Check: \(1.5+4+3.5=9\). - Does the line \(\mathbf{r}=\begin{pmatrix}1\\1\\0\end{pmatrix}+t\begin{pmatrix}1\\-1\\2\end{pmatrix}\) meet the plane \(x+y=5\)?
Answer
\(\mathbf{b}\cdot\mathbf{n}=(1)(1)+(-1)(1)+(2)(0)=0\), so the line is parallel to the plane. The point \((1,1,0)\) gives \(1+1=2\ne5\), so it is not in the plane. There is no intersection. - Find the acute angle between the planes \(x+y+z=4\) and \(z=0\).
Answer
Normals \(\begin{pmatrix}1\\1\\1\end{pmatrix}\) and \(\begin{pmatrix}0\\0\\1\end{pmatrix}\). \(\cos\theta=\dfrac{1}{\sqrt3\times1}=0.5774\), so \(\theta=54.7^\circ\). - Find the angle between the line with direction \(\begin{pmatrix}1\\2\\2\end{pmatrix}\) and the plane \(z=0\).
Answer
\(\mathbf{n}=\begin{pmatrix}0\\0\\1\end{pmatrix}\). \(\mathbf{b}\cdot\mathbf{n}=2\), \(|\mathbf{b}|=3\), \(|\mathbf{n}|=1\). \(\sin\theta=\tfrac23\), so \(\theta=41.8^\circ\).