3. Equations of Planes

Video coming soon3 worked examples

Theory

A plane is fixed by a normal vector $\mathbf{n} = (a, b, c)$ and a point on it. Its Cartesian equation is:

ax + by + cz = d

where $d$ is found by substituting a known point. If the plane is given by three points, find the normal by taking the cross product of two direction vectors lying in the plane.

Angles are measured through the normals. For two planes with normals $\mathbf{n}_1$ and $\mathbf{n}_2$ , the angle between the planes equals the angle between the normals:

\cos\theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|}

The angle between a line and a plane is the one case that behaves differently. The line's direction $\mathbf{d}$ makes some angle with the normal, but the plane itself lies at $90^\circ$ to that normal — so the angle we want is the complement. Since $\cos(90^\circ - \alpha) = \sin\alpha$ , the formula uses sine:

\sin\alpha = \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}

The Golden Rule: the coefficients $(a, b, c)$ in $ax + by + cz = d$ are the normal vector; find $d$ by substituting a point on the plane. For angles, use $\cos$ between two planes and $\sin$ between a line and a plane.

⚠️ Common Examiner Traps

Normal from coefficients: read the normal directly from the Cartesian equation's coefficients.
Three points: use a cross product of two in-plane vectors to get the normal.
Finding $d$ : substitute any known point into $ax + by + cz$ .

Worked examples

Example 1

Find the Cartesian equation of the plane with normal $\mathbf{n} = (2, -1, 3)$ passing through the point $(1, 0, 4)$ .

Step 1: The equation is $2x - y + 3z = d$ . Substitute the point to find $d$ :

2(1) - (0) + 3(4) = 2 + 12 = 14

Step 2: So the plane is:

2x - y + 3z = 14

Example 2

Find the equation of the plane through the points $A(1, 0, 0)$ , $B(0, 1, 0)$ and $C(0, 0, 1)$ .

Step 1: Two direction vectors in the plane are $\overrightarrow{AB} = (-1, 1, 0)$ and $\overrightarrow{AC} = (-1, 0, 1)$ . The normal is their cross product:

\mathbf{n} = \overrightarrow{AB}\times\overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{vmatrix} = (1,\ 1,\ 1)

Step 2: Using normal $(1,1,1)$ and point $A(1,0,0)$ , the equation $x + y + z = d$ gives $d = 1$ :

x + y + z = 1

Example 3

Find the acute angle between the line $\mathbf{r} = (1,-1,0) + t(2,1,2)$ and the plane $3x - y + 2z = 5$ .

Step 1: Read off the direction of the line and the normal of the plane:

\mathbf{d} = (2,1,2), \qquad \mathbf{n} = (3,-1,2)

Step 2: Compute the scalar product and the two magnitudes:

\mathbf{d}\cdot\mathbf{n} = 6 - 1 + 4 = 9

|\mathbf{d}| = \sqrt{4+1+4} = 3, \qquad |\mathbf{n}| = \sqrt{9+1+4} = \sqrt{14}

Step 3: This is a line and a plane, so use sine — not cosine. The modulus signs ensure the acute angle:

\sin\alpha = \frac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}||\mathbf{n}|} = \frac{9}{3\sqrt{14}} = \frac{3}{\sqrt{14}}

Step 4: Take the inverse sine:

\alpha = \sin^{-1}\!\left(\frac{3}{\sqrt{14}}\right) = \sin^{-1}(0.8018) = 53.3^\circ \ \text{(1 d.p.)}

Had cosine been used by mistake, the answer would have been $36.7^\circ$ — the complement, and a very common lost mark.