Question

Find the angle between the given planes. $$2 x+y-2 z=3 \text { and } 3 x-6 y-2 z=4$$

Short answer

The angle between the planes is \(\theta = \cos^{-1}\left(\frac{4}{21}\right)\).

Step-by-step solution

Identify the Normal Vectors of Each Plane

The equation of a plane in the form \(Ax + By + Cz = D\) has a normal vector \(\begin{bmatrix} A \ B \ C \end{bmatrix}\). For the first plane, \(2x + y - 2z = 3\), the normal vector is \(\begin{bmatrix} 2 \ 1 \ -2 \end{bmatrix}\). For the second plane, \(3x - 6y - 2z = 4\), the normal vector is \(\begin{bmatrix} 3 \ -6 \ -2 \end{bmatrix}\).

Use the Dot Product to Find the Cosine of the Angle

The dot product of two vectors \(\mathbf{a} = \begin{bmatrix} a_1 \ a_2 \ a_3 \end{bmatrix}\) and \(\mathbf{b} = \begin{bmatrix} b_1 \ b_2 \ b_3 \end{bmatrix}\) is given by \(\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3\). The cosine of the angle \(\theta\) between them can be found using \[\cos \theta = \dfrac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \cdot \|\mathbf{b}\|}\]. For the vectors \(\begin{bmatrix} 2 \ 1 \ -2 \end{bmatrix}\) and \(\begin{bmatrix} 3 \ -6 \ -2 \end{bmatrix}\), the dot product is \((2 \cdot 3) + (1 \cdot -6) + (-2 \cdot -2) = 6 - 6 + 4 = 4\).

Calculate the Magnitudes of the Vectors

The magnitude of a vector \(\mathbf{a}\) is given by \[\|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2 + a_3^2}\]. For \(\begin{bmatrix} 2 \ 1 \ -2 \end{bmatrix}\), the magnitude is \(\sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3\). For \(\begin{bmatrix} 3 \ -6 \ -2 \end{bmatrix}\), the magnitude is \(\sqrt{3^2 + (-6)^2 + (-2)^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7\).

Find the Cosine of the Angle

Using the dot product and the magnitudes calculated, the cosine of the angle between the two planes is found by \[\cos \theta = \frac{4}{3 \cdot 7} = \frac{4}{21}\].

Calculate the Angle Using the Inverse Cosine Function

The angle \(\theta\) can be computed by taking the inverse cosine of \(\frac{4}{21}\). Therefore, \(\theta = \cos^{-1}\left(\frac{4}{21}\right)\).

Key concepts

normal vector

In geometry, a normal vector is a vector that is perpendicular to a surface or a plane. This concept is crucial for understanding the orientation of a plane in 3D space.
A plane given by the equation \(Ax + By + Cz = D\) has a normal vector \(\begin{bmatrix} A \ B \ C \ end{bmatrix}\).
For instance, in the exercise, the plane equations are \(2x + y - 2z = 3\) and \(3x - 6y - 2z = 4\). Their respective normal vectors are \(\begin{bmatrix} 2 \ 1 \ -2 \ end{bmatrix}\) and \(\begin{bmatrix} 3 \ -6 \ -2 \ end{bmatrix}\).
Finding the normal vector helps in identifying the orientation and properties of the plane, which is essential when calculating the angle between two planes.

dot product

The dot product is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number.
The dot product of two vectors \(\mathbf{a} = \begin{bmatrix} a_1 \ a_2 \ a_3 \ end{bmatrix}\) and \(\mathbf{b} = \begin{bmatrix} b_1 \ b_2 \ b_3 \ end{bmatrix}\) is given by:
\(\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3\).
In the exercise, the normal vectors were \(\begin{bmatrix} 2 \ 1 \ -2 \ end{bmatrix}\) and \(\begin{bmatrix} 3 \ -6 \ -2 \ end{bmatrix}\).
The dot product of these vectors is:
\((2 \cdot 3) + (1 \cdot -6) + (-2 \cdot -2) = 6 - 6 + 4 = 4\).
The dot product is essential to find the cosine of the angle between two vectors.

vector magnitude

The magnitude of a vector represents its length and is calculated using the Euclidean norm.
For a vector \(\mathbf{a} = \begin{bmatrix} a_1 \ a_2 \ a_3 \ end{bmatrix}\), the magnitude \(\|\mathbf{a}\|\) is computed as:
\(\|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2 + a_3^2}\).
For example, in the exercise, the magnitudes of the normal vectors \(\begin{bmatrix} 2 \ 1 \ -2 \ end{bmatrix}\) and \(\begin{bmatrix} 3 \ -6 \ -2 \ end{bmatrix}\) were calculated as:
\(\|\begin{bmatrix} 2 \ 1 \ -2 \ end{bmatrix}\| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{9} = 3\),\(
\|\begin{bmatrix} 3 \ -6 \ -2 \ end{bmatrix}\| = \sqrt{3^2 + (-6)^2 + (-2)^2} = \sqrt{49} = 7\).
The vector magnitude is a key component in normalizing vectors and calculating angles between them.

cosine

Cosine (abbreviated as \(\cos\)) is a trigonometric function that, in the context of vectors, helps to find the angle between two vectors.
Using the dot product and magnitudes of vectors, the cosine of the angle \(\theta\) between two vectors \(\mathbf{a}\) and \(\mathbf{b}\) is calculated by:
\(\cos \theta = \dfrac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \cdot \|\mathbf{b}\|}\).
From the exercise, the dot product of normal vectors is 4, and their magnitudes are 3 and 7, respectively.
Thus, the cosine of the angle is:
\(\cos \theta = \dfrac{4}{3 \cdot 7} = \dfrac{4}{21}\).
Cosine values range from -1 to 1 and are particularly useful for understanding angles and calculating angles between vectors.

inverse cosine

The inverse cosine function, denoted as \(\cos^{-1}\) or \(\arccos\), is used to find the angle when the cosine value is known.
Given that \(\cos \theta = \dfrac{4}{21}\), the angle \(\theta\) can be determined by:
\(\theta = \cos^{-1}\left(\dfrac{4}{21}\right)\).
The inverse cosine function returns the angle in radians within the range [0,\pi], and it is particularly useful in various areas of mathematics and physics where angle measurements are required.
In this exercise, taking the inverse cosine of \(\dfrac{4}{21}\) yields the angle between the planes.