Problem 53 - University Calculus: Early Transcendentals

Question

Use a calculator to find the acute angles between the planes to the nearest hundredth of a radian. $$2 x+2 y+2 z=3, \quad 2 x-2 y-z=5$$

Step-by-Step Solution

Answer

The acute angle between the planes is approximately 1.82 radians.

Step 1: Identify Normal Vectors

First, we need to identify the normal vectors of the given planes. For the plane given by the equation \(2x + 2y + 2z = 3\), the normal vector \(\vec{n_1}\) is \((2, 2, 2)\). For the plane \(2x - 2y - z = 5\), the normal vector \(\vec{n_2}\) is \((2, -2, -1)\).

Step 2: Use the Dot Product Formula

The cosine of the angle \(\theta\) between two vectors \(\vec{a}\) and \(\vec{b}\) can be calculated using the dot product formula, \(\cos \theta = \frac{\vec{a} \cdot \vec{b}}{\|\vec{a}\| \|\vec{b}\|}\). We will use this formula with vectors \(\vec{n_1} = (2, 2, 2)\) and \(\vec{n_2} = (2, -2, -1)\).

Step 3: Calculate Dot Product and Magnitudes

Calculate the dot product \(\vec{n_1} \cdot \vec{n_2}\) which is \(2*2 + 2*(-2) + 2*(-1) = 4 - 4 - 2 = -2\). Next, compute the magnitudes: \(\|\vec{n_1}\| = \sqrt{2^2 + 2^2 + 2^2} = \sqrt{12} = 2\sqrt{3}\), and \(\|\vec{n_2}\| = \sqrt{2^2 + (-2)^2 + (-1)^2} = \sqrt{9} = 3\).

Step 4: Find Cosine of the Angle

Substitute the values into the dot product formula: \(\cos \theta = \frac{-2}{2\sqrt{3} \times 3} = \frac{-2}{6\sqrt{3}} = -\frac{1}{3\sqrt{3}}\).

Step 5: Compute Angle in Radians

Solve for \(\theta\) by using the inverse cosine: \(\theta = \cos^{-1}(-\frac{1}{3\sqrt{3}})\). Use a calculator to find the angle in radians. The angle to the nearest hundredth is approximately 1.82 radians.

Key Concepts

Normal Vectors

In geometry, the idea of a normal vector is crucial when discussing planes. A normal vector is a vector that is perpendicular to a plane. In the context of an equation of a plane, the coefficients before each variable in the equation usually form the normal vector. For example, consider the plane described by the equation \(2x + 2y + 2z = 3\). Here, the normal vector is \( \vec{n_1} = (2, 2, 2) \). Similarly, for the plane \(2x - 2y - z = 5\), the normal vector is \( \vec{n_2} = (2, -2, -1) \).

These normal vectors are essential because the angle between any two planes can be found by determining the angle between their normal vectors. This is a fundamental step as the plane's orientation in space is directly related to the direction of its normal vector.

Dot Product Formula

The dot product is a fundamental operation in vector algebra and is particularly useful for finding angles between vectors. It involves multiplying corresponding components of two vectors and summing the results. The dot product of two vectors \(\vec{a} = (a_1, a_2, a_3)\) and \(\vec{b} = (b_1, b_2, b_3)\) is given by:

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

By using the dot product, one can conveniently calculate the cosine of the angle between two vectors if the magnitudes of the vectors are also known.

In our specific exercise, the dot product of the normal vectors \(\vec{n_1} = (2, 2, 2)\) and \(\vec{n_2} = (2, -2, -1)\) is calculated as follows:

• \( \vec{n_1} \cdot \vec{n_2} = 2 \times 2 + 2 \times (-2) + 2 \times (-1) = -2 \)

This result is integral to finding the angle between the planes.

Magnitude of Vectors

The magnitude, or length, of a vector is a measure of its size. For a vector \(\vec{v} = (v_1, v_2, v_3)\), its magnitude is computed using the formula:

• \(\|\vec{v}\| = \sqrt{v_1^2 + v_2^2 + v_3^2}\)

This formula applies the Pythagorean theorem in a three-dimensional space to determine the vector's length. Magnitudes are necessary for computing the angle between vectors, as they appear in the denominator of the formula for the cosine of the angle.

In this exercise:

• \(\|\vec{n_1}\| = \sqrt{2^2 + 2^2 + 2^2} = \sqrt{12} = 2\sqrt{3}\)
• \(\|\vec{n_2}\| = \sqrt{2^2 + (-2)^2 + (-1)^2} = \sqrt{9} = 3\)

Accurate calculation of these magnitudes ensures precise determination of the angle between the planes.

Inverse Cosine

The inverse cosine function, denoted as \(\cos^{-1}\), is used to find the angle whose cosine is a given number. In our exercise, once we find the cosine of the angle between the normal vectors using the dot product, the next step is to apply the inverse cosine to find the actual angle in radians.

The formula for finding the angle \(\theta\) between two vectors \(\vec{a}\) and \(\vec{b}\) is:

• \(\theta = \cos^{-1}\left( \frac{\vec{a} \cdot \vec{b}}{\|\vec{a}\| \|\vec{b}\|} \right)\)

In this specific problem, the dot product is \(-2\) and the magnitudes are \(2\sqrt{3}\) and \(3\), leading to:

• \(\cos \theta = -\frac{1}{3\sqrt{3}}\)

Applying the inverse cosine function, \( \theta = \cos^{-1}\left(-\frac{1}{3\sqrt{3}} \right) \), we find that the angle is approximately 1.82 radians. Using a calculator accurately here is crucial, as it ensures that the angle is expressed to the nearest hundredth of a radian.