Problem 48 - University Calculus: Early Transcendentals

Question

Find the angles between the planes. $$5 x+y-z=10, \quad x-2 y+3 z=-1$$

Step-by-Step Solution

Answer

The planes are perpendicular, meaning the angle between them is \(90^\circ\).

Step 1: Identify the Normal Vectors

To find the angle between the two planes, first identify their normal vectors from the equation coefficients. The normal vector of the first plane \(5x + y - z = 10\) is \(\mathbf{n_1} = (5, 1, -1)\), and for the second plane \(x - 2y + 3z = -1\), the normal vector is \(\mathbf{n_2} = (1, -2, 3)\).

Step 2: Use the Dot Product Formula

The cosine of the angle \(\theta\) between the normal vectors \(\mathbf{n_1}\) and \(\mathbf{n_2}\) is given by the dot product formula: \(\cos \theta = \frac{\mathbf{n_1} \cdot \mathbf{n_2}}{\|\mathbf{n_1}\| \|\mathbf{n_2}\|}\). Compute the dot product \(\mathbf{n_1} \cdot \mathbf{n_2}\).

Step 3: Calculate the Dot Product

Calculate the dot product: \(\mathbf{n_1} \cdot \mathbf{n_2} = 5 \times 1 + 1 \times (-2) + (-1) \times 3 = 5 - 2 - 3 = 0\).

Step 4: Compute Magnitudes of Vectors

Calculate the magnitudes of the normal vectors: \(\|\mathbf{n_1}\| = \sqrt{5^2 + 1^2 + (-1)^2} = \sqrt{27}\) and \(\|\mathbf{n_2}\| = \sqrt{1^2 + (-2)^2 + 3^2} = \sqrt{14}\).

Step 5: Determine the Cosine Value

Since the dot product is 0, \(\cos \theta = \frac{0}{\sqrt{27} \cdot \sqrt{14}} = 0\). Thus, \(\theta = \cos^{-1}(0)\).

Step 6: Conclusion About the Angle

The angle \(\theta\), where \(\cos \theta = 0\), is \(90^\circ\). This means the planes are perpendicular to each other.

Key Concepts

Dot Product

The dot product is a crucial concept when working with vectors, especially in determining angles between them. It serves as a way to multiply two vectors, resulting in a scalar. For vectors \( \mathbf{a} = (a_1, a_2, a_3) \) and \( \mathbf{b} = (b_1, b_2, b_3) \), the dot product is calculated as:

• \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \)

This formula sums the products of their corresponding components.
In the context of this exercise, the dot product helps us find the angle between the normals of the planes. By determining the dot product, you can further compute the cosine of the angle, thanks to the relationship between dot products and angles.

Normal Vectors

Normal vectors are vectors that are perpendicular to a surface. In the case of planes, they are perpendicular to the entirety of the plane. This means they make the ideal candidates for analyzing angles between planes.
Every plane in three-dimensional space can be represented by a normal vector, attributed to it through the equation of the plane itself. For instance:

• Given \(5x + y - z = 10\), the normal vector is \( \mathbf{n_1} = (5, 1, -1) \).
• Similarly, for \(x - 2y + 3z = -1\), the normal vector is \( \mathbf{n_2} = (1, -2, 3) \).

These vectors are derived from the coefficients of \(x, y, \) and \(z\). By using these normal vectors, we can determine the orientation of the planes and their angles relative to one another.

Cosine of Angle

The cosine of the angle between two vectors can be found using the dot product and the magnitudes of the vectors. This relationship is expressed as:

• \( \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} \)

Where \(\mathbf{a}\) and \(\mathbf{b}\) are the vectors, and \(\|\mathbf{a}\|\) and \(\|\mathbf{b}\|\) are their magnitudes. This formula elegantly ties together several fundamental concepts in vector mathematics, allowing us to calculate angles efficiently.
In this problem:

• The dot product was calculated to be zero \((\mathbf{n_1} \cdot \mathbf{n_2} = 0)\).
• When the dot product is zero, it implies that \(\cos \theta = 0\), resulting in \(\theta = 90^\circ\).

Thus, showing that the planes are indeed perpendicular, all through the magic of the cosine formula.