Problem 23 - Calculus: Early Transcendentals

Question

Find the cosine of the angle between the planes \(x+y+z=2\) and \(x+2 y+3 z=8\).

Step-by-Step Solution

Answer

The cosine of the angle between the planes is \(\frac{\sqrt{42}}{7}\).

Step 1: Identify the normal vectors of the planes

The normal vector of a plane given by the equation \(Ax + By + Cz = D\) is \(\mathbf{n} = \langle A, B, C \rangle\). For plane 1, \(x+y+z=2\), the normal vector \(\mathbf{n}_1 = \langle 1, 1, 1 \rangle\). For plane 2, \(x+2y+3z=8\), the normal vector \(\mathbf{n}_2 = \langle 1, 2, 3 \rangle\).

Step 2: Use the dot product to find the cosine of the angle between the normals

To find the cosine of the angle \(\theta\) between two vectors \(\mathbf{a}\) and \(\mathbf{b}\), use the formula \(\cos(\theta) = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|}\). Here \(\mathbf{n}_1 \cdot \mathbf{n}_2 = 1 \cdot 1 + 1 \cdot 2 + 1 \cdot 3 = 6\).

Step 3: Compute the magnitudes of the normal vectors

Find \(\|\mathbf{n}_1\| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}\) and \(\|\mathbf{n}_2\| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}\).

Step 4: Solve for the cosine of the angle

Substitute the dot product and magnitudes into the cosine formula: \(\cos(\theta) = \frac{6}{\sqrt{3} \cdot \sqrt{14}} = \frac{6}{\sqrt{42}}\). Simplify to get \(\cos(\theta) = \frac{6}{\sqrt{42}} = \frac{6\sqrt{42}}{42} = \frac{\sqrt{42}}{7}\).

Key Concepts

Normal Vector

In plane geometry, a normal vector is essential when dealing with planes. A normal vector is a vector that is perpendicular (or orthogonal) to a surface—in this case, a plane.

If a plane is described by the equation \(Ax + By + Cz = D\), the normal vector \(\mathbf{n}\) can be easily found and is represented by \(\langle A, B, C \rangle\). This vector tells us the direction that is perpendicular to the surface of the plane.

In the example given, we have two planes:

• For the equation \(x+y+z=2\), the normal vector is \(\langle 1, 1, 1 \rangle\).
• For the equation \(x+2y+3z=8\), the normal vector is \(\langle 1, 2, 3 \rangle\).

Understanding how to determine the normal vector is critical for solving problems in plane geometry, especially when calculating angles between planes.

Dot Product

The dot product, also known as the scalar product, is a way to multiply two vectors to obtain a scalar. It's used extensively in geometry and physics to solve problems involving angles and distances.

When we have two vectors \(\mathbf{a} = \langle a_1, a_2, a_3 \rangle\) and \(\mathbf{b} = \langle b_1, b_2, b_3 \rangle\), their dot product is computed as: \[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \]
In our problem, the dot product between the normal vectors \(\mathbf{n}_1 = \langle 1, 1, 1 \rangle\) and \(\mathbf{n}_2 = \langle 1, 2, 3 \rangle\) is 6, calculated as:

• \(1 \cdot 1\)
• + \(1 \cdot 2\)
• + \(1 \cdot 3\)
• = 6

This dot product helps determine the cosine of the angle between the two normal vectors.

Magnitude of a Vector

The magnitude of a vector, often called its length, is an essential property that tells us how long the vector is in terms of units. Calculating the magnitude of a vector \(\mathbf{v} = \langle v_1, v_2, v_3 \rangle\) can be done using the formula:

\[ \|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2 + v_3^2} \]

In this case, we need to find the magnitudes of the normal vectors for the planes:

• For \(\mathbf{n}_1 = \langle 1, 1, 1 \rangle\), the magnitude is \(\sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}\).
• For \(\mathbf{n}_2 = \langle 1, 2, 3 \rangle\), the magnitude is \(\sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}\).

Knowing the magnitude is pivotal when calculating angles between vectors as it forms part of the formula for finding the cosine of angles.

Cosine of Angle Between Vectors

To find the cosine of the angle between two vectors, we use a formula that relates their dot product and magnitudes. The general equation is:

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

This equation finds the angle between two vectors by dividing their dot product by the product of their magnitudes. In our example, the dot product calculated was 6, the magnitudes were \(\sqrt{3}\) and \(\sqrt{14}\) respectively.

Substituting these values into the formula gives:

• \(\cos(\theta) = \frac{6}{\sqrt{3} \cdot \sqrt{14}} = \frac{6}{\sqrt{42}}\)
• Simplify further to \(\cos(\theta) = \frac{6\sqrt{42}}{42} = \frac{\sqrt{42}}{7}\)

This method allows us to calculate the cosine of the angle between any two intersecting planes, giving insight into their spatial relation.