The Plane in Space · MathLopedia

1. The plane through a point, perpendicular to a vector

Through a given point \(M_1\) there are infinitely many planes, but only one perpendicular to a given nonzero vector \(\vec N\). That vector is called a normal vector of the plane.

A point \(M\) lies in this plane exactly when the vector \(\overrightarrow{M_1M}\) lies in the plane, that is, when it is perpendicular to \(\vec N\). Perpendicularity means a zero dot product, and that is the whole equation.

Equation of a plane, 1° The plane through \(M_1(x_1, y_1, z_1)\) perpendicular to the vector \( \vec N = \{A, B, C\} \) has, in vector form, the equation \[ (\vec r - \vec r_1)\cdot\vec N = 0 \qquad\text{or}\qquad \vec r\cdot\vec N = \alpha, \] where \( \vec r = \overrightarrow{OM} \), \( \vec r_1 = \overrightarrow{OM_1} \) and \( \alpha = \vec r_1\cdot\vec N \). In scalar form: \[ A(x - x_1) + B(y - y_1) + C(z - z_1) = 0. \]

(Here \( \vec r - \vec r_1 = \overrightarrow{M_1M} = \{x - x_1, y - y_1, z - z_1\} \), and the scalar form is just the dot product written out.)

Figure 1. The plane through \(M_1\) with normal vector \(\vec N\). For every point \(M\) of the plane, \(\overrightarrow{M_1M} \perp \vec N\). The plane shown cuts the axes at \(a\), \(b\), \(c\) (Section 3).

Example 1 The plane through \(M_1(1, 2, 3)\) with normal vector \( \vec N = \{2, -1, 4\} \): \[ 2(x - 1) - (y - 2) + 4(z - 3) = 0 \quad\Longleftrightarrow\quad 2x - y + 4z - 12 = 0. \]

2. The general equation \(Ax + By + Cz + D = 0\)

Multiplying out the scalar form gives \( Ax + By + Cz + D = 0 \) with \( D = -(Ax_1 + By_1 + Cz_1) = -\alpha \). Conversely, every equation

\[ Ax + By + Cz + D = 0 \qquad (A, B, C \text{ not all zero}) \]

is a plane, and you can read the normal vector straight off the coefficients: \( \vec N = \{A, B, C\} \). This is the most useful fact about planes. (Any point satisfying the equation can serve as \(M_1\).)

Special positions

Equation	Position of the plane	Reason
\(D = 0\)	passes through the origin	\((0, 0, 0)\) satisfies it
\(A = 0\): \(By + Cz + D = 0\)	parallel to the \(x\)-axis	\(\vec N \perp \vec i\)
\(A = B = 0\): \(Cz + D = 0\)	parallel to the \(xy\)-plane (\(z = \text{const}\))	\(\vec N \parallel \vec k\)
\(A = D = 0\): \(By + Cz = 0\)	contains the \(x\)-axis	both of the above

Example 2 · Plane through three points Through \(A(1, 0, 0)\), \(B(0, 2, 0)\), \(C(0, 0, 3)\). The vectors \( \overrightarrow{AB} = \{-1, 2, 0\} \) and \( \overrightarrow{AC} = \{-1, 0, 3\} \) lie in the plane, so their cross product is a normal vector: \[ \vec N = \overrightarrow{AB}\times\overrightarrow{AC} = \{6, 3, 2\}, \qquad 6(x - 1) + 3y + 2z = 0 \ \Longleftrightarrow\ 6x + 3y + 2z - 6 = 0. \]

3. The intercept form

Intercept form, 2° \[ \frac xa + \frac yb + \frac zc = 1, \] where \(a\), \(b\) and \(c\) are the intercepts: the plane cuts the coordinate axes at the points \((a, 0, 0)\), \((0, b, 0)\), \((0, 0, c)\).

Check: putting \(y = z = 0\) gives \(x = a\), and similarly for the others. From the general equation with \(D \ne 0\), divide by \(-D\): \( a = -\frac DA \), \( b = -\frac DB \), \( c = -\frac DC \). The form exists only when \(a, b, c\) are all finite and nonzero, i.e. when the plane does not pass through the origin and is not parallel to any axis (\(A, B, C, D \ne 0\)). The book calls it the canonical form; it is also called the segment form.

Example 3 The plane of Example 2, \( 6x + 3y + 2z = 6 \), divided by \(6\): \( \frac x1 + \frac y2 + \frac z3 = 1 \), the intercepts we started from. The plane \( 2x + 3y - 6z - 12 = 0 \) becomes \( \frac x6 + \frac y4 + \frac{z}{-2} = 1 \): it cuts the \(z\)-axis below the origin.

The intercept form is perfect for sketching: mark the three intercepts and join them. The triangle in Figure 1 is drawn this way.

4. Pencils of planes

Pencil of planes, 3° The planes passing through the line of intersection of the planes \[ A_1x + B_1y + C_1z + D_1 = 0 \quad\text{and}\quad A_2x + B_2y + C_2z + D_2 = 0 \] are given by \[ A_1x + B_1y + C_1z + D_1 + \lambda(A_2x + B_2y + C_2z + D_2) = 0, \qquad \lambda \in \mathbb R. \]

Why: for each \(\lambda\), this is a linear equation, so a plane. And every point of the common line makes both brackets zero, so it satisfies the equation for every \(\lambda\): all these planes contain the line. Like the pages of a book around its spine, they turn around the line as \(\lambda\) changes.

One plane is missing No value of \(\lambda\) gives the second plane itself (it corresponds to "\(\lambda = \infty\)"). The complete pencil is \( \lambda_1(A_1x + \dots + D_1) + \lambda_2(A_2x + \dots + D_2) = 0 \), with \(\lambda_1, \lambda_2\) not both zero. In practice: if the computation for \(\lambda\) leads to a contradiction, check whether the answer is the second plane.

Example 4 Find the plane through the line of intersection of \( x + y + z - 1 = 0 \) and \( 2x - y + 3z = 0 \) that passes through \(P(1, 1, 1)\). Substitute \(P\) into the pencil: \[ (1 + 1 + 1 - 1) + \lambda(2 - 1 + 3) = 0 \ \Rightarrow\ 2 + 4\lambda = 0 \ \Rightarrow\ \lambda = -\tfrac12. \] The plane: \( x + y + z - 1 - \frac12(2x - y + 3z) = 0 \), i.e. \( 3y - z - 2 = 0 \). No need to compute the line at all.

5. The angle between two planes

Two planes meet at the same angle as their normals (rotate the picture so that you look along the line of intersection: the planes become two lines, and their normals are those lines turned by \(90^\circ\)). So the angle comes from the dot product of the normals.

Angle between planes, 4° For planes with normal vectors \( \vec N_1 = \{A_1, B_1, C_1\} \) and \( \vec N_2 = \{A_2, B_2, C_2\} \): \[ \cos\varphi = \pm\frac{\vec N_1\cdot\vec N_2}{|\vec N_1|\,|\vec N_2|} = \pm\frac{A_1A_2 + B_1B_2 + C_1C_2}{\sqrt{A_1^2 + B_1^2 + C_1^2}\,\sqrt{A_2^2 + B_2^2 + C_2^2}}. \]

The \(\pm\) reflects that two planes form two angles, \(\varphi\) and \(180^\circ - \varphi\) (a normal can point to either side). Usually the acute angle is meant, obtained by taking the absolute value.

Two special cases follow immediately:

Parallel planes: the normals are parallel, i.e. proportional: \[ \frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2}. \] (If moreover \( \frac{D_1}{D_2} \) is the same ratio, the planes coincide. A zero in a denominator means the corresponding numerator is zero too: the condition is really \( \vec N_1 = k\vec N_2 \).)
Perpendicular planes: the normals are perpendicular: \[ A_1A_2 + B_1B_2 + C_1C_2 = 0. \]

Example 5 The angle between \( x - y + 2z - 3 = 0 \) and \( 2x + y + z + 1 = 0 \): \[ \cos\varphi = \frac{2 - 1 + 2}{\sqrt6\cdot\sqrt6} = \frac12, \qquad \varphi = 60^\circ. \]

6. The distance from a point to a plane

Distance formula The distance of the point \(M(x_0, y_0, z_0)\) from the plane \( Ax + By + Cz + D = 0 \) is \[ d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}. \]

Derivation. Take any point \(M_1(x_1, y_1, z_1)\) of the plane. The distance is the length of the projection of \(\overrightarrow{M_1M}\) onto the normal direction, i.e. onto the unit vector \( \frac{\vec N}{|\vec N|} \):

\[ d = \frac{|\overrightarrow{M_1M}\cdot\vec N|}{|\vec N|} = \frac{|A(x_0 - x_1) + B(y_0 - y_1) + C(z_0 - z_1)|}{\sqrt{A^2 + B^2 + C^2}}, \]

and since \( Ax_1 + By_1 + Cz_1 = -D \), the numerator is \( |Ax_0 + By_0 + Cz_0 + D| \).

Figure 2. The plane seen edge-on (as a line). The distance \(d\) from \(M\) is the projection of \(\overrightarrow{M_1M}\) onto the direction of \(\vec N\), for any point \(M_1\) of the plane.

A useful by-product: without the absolute value, the expression \( \frac{Ax_0 + By_0 + Cz_0 + D}{\sqrt{A^2 + B^2 + C^2}} \) is a signed distance: positive on the side the normal points to, negative on the other. So plugging a point into the left side of the equation tells you on which side of the plane it lies, and two points are on the same side exactly when they give the same sign.

Example 6 The distance of \(M(1, 2, 3)\) from the plane \( 2x - y + 2z + 4 = 0 \): \[ d = \frac{|2 - 2 + 6 + 4|}{\sqrt{4 + 1 + 4}} = \frac{10}{3}. \]

Example 7 · Two parallel planes \( 2x - y + 2z + 4 = 0 \) and \( 2x - y + 2z - 5 = 0 \) are parallel (same normal). Take any point of the second, e.g. \((0, 1, 3)\), and measure its distance from the first: \( \frac{|0 - 1 + 6 + 4|}{3} = 3 \). In general, for planes \( Ax + By + Cz + D_1 = 0 \) and \( Ax + By + Cz + D_2 = 0 \): \( d = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}} \).

7. Summary

Concept	Formula
Plane through \(M_1\), normal \(\vec N\)	\( (\vec r - \vec r_1)\cdot\vec N = 0 \); \( A(x - x_1) + B(y - y_1) + C(z - z_1) = 0 \)
General equation	\( Ax + By + Cz + D = 0 \), normal \( \vec N = \{A, B, C\} \)
Intercept form	\( \frac xa + \frac yb + \frac zc = 1 \)
Pencil through a line	\( (A_1x + \dots + D_1) + \lambda(A_2x + \dots + D_2) = 0 \)
Angle	\( \cos\varphi = \frac{\|\vec N_1\cdot\vec N_2\|}{\|\vec N_1\|\|\vec N_2\|} \)
Parallel / perpendicular	\( \vec N_1 = k\vec N_2 \) / \( A_1A_2 + B_1B_2 + C_1C_2 = 0 \)
Distance	\( d = \frac{\|Ax_0 + By_0 + Cz_0 + D\|}{\sqrt{A^2 + B^2 + C^2}} \)

8. Practice problems

Try each one before opening the solution. More solved problems from this topic are in Geometry.

Problem 1: Find the plane through the origin parallel to \( 3x - 2y + z = 5 \).

Same normal \(\{3, -2, 1\}\), passing through \((0, 0, 0)\): \( 3x - 2y + z = 0 \).

Problem 2: Find the plane through \(A(1, 1, 1)\) perpendicular to both planes \( x - y = 0 \) and \( y - z = 0 \).

Its normal must be perpendicular to both normals: \( \vec N = \{1, -1, 0\}\times\{0, 1, -1\} = \{1, 1, 1\} \). Plane: \( x + y + z - 3 = 0 \).

Problem 3: For which \(k\) are the planes \( 2x + ky + 3z = 1 \) and \( x - y + kz = 4 \) perpendicular?

\( 2\cdot1 + k\cdot(-1) + 3\cdot k = 0 \Rightarrow 2 + 2k = 0 \Rightarrow k = -1 \).

Problem 4: Find the volume of the tetrahedron cut off from the first octant by \( 6x + 3y + 2z - 6 = 0 \).

Intercepts \(a = 1\), \(b = 2\), \(c = 3\). The tetrahedron has three mutually perpendicular edges \(a, b, c\): \( V = \frac{abc}{6} = 1 \).

Problem 5: Find the foot of the perpendicular from \(M(1, 2, 3)\) to the plane \( 2x - y + 2z + 4 = 0 \).

Move from \(M\) along the normal: \( (1 + 2t, 2 - t, 3 + 2t) \). Substituting into the plane: \( 10 + 9t = 0 \), \( t = -\frac{10}{9} \). Foot: \( \big(-\frac{11}{9}, \frac{28}{9}, \frac79\big) \). Its distance from \(M\) is \( |t|\cdot|\vec N| = \frac{10}{9}\cdot 3 = \frac{10}{3} \), matching Example 6.

Problem 6: Are the points \(P(1, 0, 0)\) and \(Q(0, 0, 2)\) on the same side of the plane \( x + y + z - 1.5 = 0 \)?

\(P\): \(1 - 1.5 = -0.5 < 0\). \(Q\): \(2 - 1.5 = 0.5 > 0\). Different signs: the plane separates them.

Final thoughts

One idea carries the whole chapter: a plane is determined by a point and a normal vector, and "lies in the plane" means "perpendicular to the normal". The general equation hands you the normal for free; parallelism, perpendicularity and angles are statements about normals; and the distance formula is a projection onto the normal. The pencil of planes is a different kind of trick, letting you build planes through a line without ever computing the line. With these tools, most problems about planes come down to a dot product or a cross product.

Related posts: vectors (dot and cross products) and affine transformations of coordinates.