Worked examples · Sheet 1

Equations of planes

Theory used: Plane through a point orthogonal to a vector Cartesian equation of the plane Plane through three points Parallelism and perpendicularity

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

Write the equation of the plane passing through $P_{0} = (1, - 2, 3)$ and parallel to the plane $π : 2 x + y - z + 5 = 0$ .

n = (2, 1, - 1)

2 (x - 1) + (y + 2) - (z - 3) = 0

2 x - 2 + y + 2 - z + 3 = 0 ⟹ 2 x + y - z + 3 = 0

Two parallel planes have parallel normal vectors, so the required plane has the same normal vector as $π$ , which is read off the coefficients of $x$ , $y$ , $z$ : $n = (2, 1, - 1)$ . By the Cartesian equation of the plane through a point with a given normal vector we write $a (x - x_{0}) + b (y - y_{0}) + c (z - z_{0}) = 0$ with $(a, b, c) = (2, 1, - 1)$ and $(x_{0}, y_{0}, z_{0}) = (1, - 2, 3)$ . Expanding the products gives the plane $2 x + y - z + 3 = 0$ . As a check, the point $P_{0}$ satisfies the equation, since $2 - 2 - 3 + 3 = 0$ . The plane found is distinct from $π$ , because the constant term $3$ is different from $5$ .

Exercise 2

Write the equation of the plane passing through the points $A = (1, 1, 0)$ , $B = (2, 0, 1)$ , $C = (0, 1, 2)$ .

A B = (1, - 1, 1), A C = (- 1, 0, 2)

A B \times A C = i 1 - 1 j - 1 0 k 12 = (- 2 - 0, - 1 - 2, 0 - 1) = (- 2, - 3, - 1)

n = (2, 3, 1)

2 (x - 1) + 3 (y - 1) + z = 0 ⟹ 2 x + 3 y + z - 5 = 0

The vectors $A B$ and $A C$ are parallel to the plane and are not parallel to each other, since their components are not proportional; therefore the three points are not collinear and the plane through three points is unique. The cross product $A B \times A C$ is orthogonal to both, so it is a normal vector to the plane. The result $(- 2, - 3, - 1)$ can be replaced by its opposite $(2, 3, 1)$ , because the normal vector is defined up to a nonzero factor. Writing the equation through the point $A$ gives the plane $2 x + 3 y + z - 5 = 0$ . As a check, $B$ and $C$ satisfy the equation, since $4 + 0 + 1 - 5 = 0$ and $0 + 3 + 2 - 5 = 0$ .

Exercise 3

Write the equation of the plane that contains the line $r : x = 1 + t$ , $y = 2 - t$ , $z = 2 t$ and the point $P = (2, 1, 1)$ .

A = (1, 2, 0) \in r, v = (1, - 1, 2), A P = (1, - 1, 1)

n = v \times A P = i 11 j - 1 - 1 k 21 = (- 1 + 2, 2 - 1, - 1 + 1) = (1, 1, 0)

(x - 1) + (y - 2) = 0 ⟹ x + y - 3 = 0

First we check that $P$ does not belong to $r$ . The first two equations with $x = 2$ , $y = 1$ give $t = 1$ , while the third with $z = 1$ gives $t = \frac{1}{2}$ ; hence $P \in / r$ and the plane containing $r$ and $P$ is unique. The plane contains the point $A$ of $r$ , obtained for $t = 0$ , and is parallel both to the direction vector $v$ and to the vector $A P$ . These two vectors are not parallel, therefore their cross product $(1, 1, 0)$ is a normal vector. The equation through $A$ gives the plane $x + y - 3 = 0$ . As a check, the generic point of $r$ satisfies the equation, since $(1 + t) + (2 - t) - 3 = 0$ for every $t$ , and $P$ satisfies it as well, since $2 + 1 - 3 = 0$ . The variable $z$ does not appear in the equation, so the plane is parallel to the $z$ -axis.

Exercise 4

Write the equation of the plane passing through $P = (1, 0, - 1)$ and perpendicular to the two planes $π_{1} : x + y - z = 2$ and $π_{2} : 2 x - y + z = 0$ .

n_{1} = (1, 1, - 1), n_{2} = (2, - 1, 1)

n_{1} \times n_{2} = i 12 j 1 - 1 k - 1 1 = (1 - 1, - 2 - 1, - 1 - 2) = (0, - 3, - 3)

n = (0, 1, 1)

y + (z + 1) = 0 ⟹ y + z + 1 = 0

Two planes are perpendicular if and only if their normal vectors are orthogonal, so the normal vector $n$ of the required plane must be orthogonal both to $n_{1}$ and to $n_{2}$ . The vectors $n_{1}$ and $n_{2}$ are not parallel, therefore the vectors orthogonal to both are the multiples of the cross product $n_{1} \times n_{2} = (0, - 3, - 3)$ , and one can choose $n = (0, 1, 1)$ . The plane through $P$ with this normal vector is $y + z + 1 = 0$ . As a check, $n \cdot n_{1} = 0 + 1 - 1 = 0$ and $n \cdot n_{2} = 0 - 1 + 1 = 0$ , and the point $P$ satisfies the equation, since $0 - 1 + 1 = 0$ . The plane found is perpendicular to the line of intersection of $π_{1}$ and $π_{2}$ , which has direction vector precisely $n_{1} \times n_{2}$ .