Problem 24 - Calculus: Early Transcendentals

Question

Find the cosine of the angle between the planes \(x-y+2 z=2\) and \(3 x-2 y+z=5\).

Step-by-Step Solution

Answer

The cosine of the angle is \(\frac{\sqrt{21}}{6}\).

Step 1: Identify Normal Vectors of the Planes

The general equation of a plane is given by \(Ax + By + Cz = D\), where \( (A, B, C) \) is the normal vector of the plane. For the plane \(x - y + 2z = 2\), the normal vector is \((1, -1, 2)\). For the plane \(3x - 2y + z = 5\), the normal vector is \((3, -2, 1)\).

Step 2: Calculate the Dot Product of the Normal Vectors

The dot product of two vectors \( \mathbf{a} = (a_1, a_2, a_3) \) and \( \mathbf{b} = (b_1, b_2, b_3) \) is given by \(a_1b_1 + a_2b_2 + a_3b_3\). For our normal vectors, the dot product is: \(1 \cdot 3 + (-1) \cdot (-2) + 2 \cdot 1 = 3 + 2 + 2 = 7\).

Step 3: Determine the Magnitudes of the Normal Vectors

The magnitude of a vector \((a, b, c)\) is \(\sqrt{a^2 + b^2 + c^2}\). Calculate the magnitudes of the normal vectors:1. For \((1, -1, 2)\): \(\sqrt{1^2 + (-1)^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}\).2. For \((3, -2, 1)\): \(\sqrt{3^2 + (-2)^2 + 1^2} = \sqrt{9 + 4 + 1} = \sqrt{14}\).

Step 4: Apply the Cosine Formula for Vectors

The cosine of the angle \(\theta\) between two vectors \(\mathbf{a}\) and \(\mathbf{b}\) is given by:\[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} \]Substitute the values obtained:\[ \cos \theta = \frac{7}{\sqrt{6} \cdot \sqrt{14}} \]

Step 5: Simplify the Expression for \(\cos \theta\)

Calculate the product \(\sqrt{6} \cdot \sqrt{14}\):\(\sqrt{6 \times 14} = \sqrt{84}\).Hence, \[ \cos \theta = \frac{7}{\sqrt{84}} \].Simplify \(\sqrt{84}\):\(\sqrt{84} = \sqrt{4 \cdot 21} = 2\sqrt{21}\). Then, \[ \cos \theta = \frac{7}{2\sqrt{21}} \].

Step 6: Rationalize the Denominator

To rationalize, multiply the numerator and the denominator by \(\sqrt{21}\):\[ \cos \theta = \frac{7 \cdot \sqrt{21}}{2 \cdot 21} = \frac{7 \sqrt{21}}{42} = \frac{\sqrt{21}}{6} \].

Key Concepts

dot product

In mathematics, the dot product is a way to multiply two vectors to get a scalar result, which is a single number. It is a fundamental tool used to find angles between vectors. When you have two vectors, say \( \mathbf{a} = (a_1, a_2, a_3) \) and \( \mathbf{b} = (b_1, b_2, b_3) \), the dot product is calculated as follows:

• Multiply corresponding components of the vectors: \( a_1 b_1, a_2 b_2, \) and \( a_3 b_3 \).
• Add these products together: \( a_1 b_1 + a_2 b_2 + a_3 b_3 \).

For instance, using the normal vectors from our planes, the dot product is \( 1 \cdot 3 + (-1) \cdot (-2) + 2 \cdot 1 = 7 \).
This useful calculation helps to determine if vectors are orthogonal (perpendicular), as two vectors are orthogonal if and only if their dot product is zero.

normal vector

A normal vector is a vector that is perpendicular to a surface or a plane. In the context of a plane given by the equation \( Ax + By + Cz = D \), the normal vector is \((A, B, C)\). It describes the orientation of the plane in three-dimensional space.
To find the normal vector for a specific plane, simply look at the coefficients in front of the variables in the plane's equation.

• For the plane \( x - y + 2z = 2 \), the normal vector is \((1, -1, 2)\).
• For the plane \( 3x - 2y + z = 5 \), the normal vector is \((3, -2, 1)\).

Normal vectors play a crucial role when calculating the angle between planes. They are used to find the direction along which one surface is perpendicular to another.

magnitude of a vector

The magnitude of a vector represents its length or size. It is computed similarly to finding the length of a line segment using the distance formula. For a vector \( \mathbf{v} = (a, b, c) \), the magnitude \( \|\mathbf{v}\| \) is given by the formula:\[\|\mathbf{v}\| = \sqrt{a^2 + b^2 + c^2}\]To find the magnitude of the normal vectors from our planes, we proceed as follows:

• For \((1, -1, 2)\): \( \sqrt{1^2 + (-1)^2 + 2^2} = \sqrt{6} \).
• For \((3, -2, 1)\): \( \sqrt{3^2 + (-2)^2 + 1^2} = \sqrt{14} \).

Magnitude is crucial when normalizing vectors to compare their direction rather than their scale, and when using them to compute angles between vectors using the cosine formula.

cosine of angle

The cosine of the angle between two vectors is determined using their dot product and their magnitudes. The formula is:\[\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|}\]This formula takes into account both the directions and lengths of the vectors. In the case of our planes, substituting our calculated dot product and magnitudes gives:
\[\cos \theta = \frac{7}{\sqrt{6} \cdot \sqrt{14}}\]
By simplifying this expression, including rationalizing the denominator, we reach:\[\cos \theta = \frac{\sqrt{21}}{6}\]This final value tells us the cosine of the angle between the normal vectors of the two planes and hence the angle between the planes themselves.
Understanding how cosine relates to angle allows us to determine how "tilted" one plane is relative to another.