Problem 49 - Thomas Calculus

Question

Use a calculator to find the acute angles between the planes in Exercises \(49-52\) 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 is approximately 1.82 radians.

Step 1: Identify Normal Vectors

Each plane is defined by its normal vector. For the plane equation \(2x + 2y + 2z = 3\), the normal vector is \( \mathbf{n_1} = \langle 2, 2, 2 \rangle \). For the plane equation \(2x - 2y - z = 5\), the normal vector is \( \mathbf{n_2} = \langle 2, -2, -1 \rangle \).

Step 2: Find the Dot Product

Calculate the dot product of the normal vectors \( \mathbf{n_1} \) and \( \mathbf{n_2} \):\[ \mathbf{n_1} \cdot \mathbf{n_2} = (2 \times 2) + (2 \times -2) + (2 \times -1) = 4 - 4 - 2 = -2 \]

Step 3: Find the Magnitudes

Calculate the magnitude of each normal vector. For \( \mathbf{n_1} = \langle 2, 2, 2 \rangle \):\[ ||\mathbf{n_1}|| = \sqrt{2^2 + 2^2 + 2^2} = \sqrt{12} = 2\sqrt{3} \]For \( \mathbf{n_2} = \langle 2, -2, -1 \rangle \):\[ ||\mathbf{n_2}|| = \sqrt{2^2 + (-2)^2 + (-1)^2} = \sqrt{9} = 3 \]

Step 4: Calculate the Cosine of the Angle

Use the formula for the cosine of the angle \(\theta\) between two vectors:\[ \cos(\theta) = \frac{\mathbf{n_1} \cdot \mathbf{n_2}}{||\mathbf{n_1}|| \times ||\mathbf{n_2}||} \]Substitute the values:\[ \cos(\theta) = \frac{-2}{2\sqrt{3} \times 3} = \frac{-2}{6\sqrt{3}} = \frac{-1}{3\sqrt{3}} \]

Step 5: Find the Angle

To find the angle, take the inverse cosine (arc cosine) of \(\cos(\theta)\):\[ \theta = \cos^{-1}\left(\frac{-1}{3\sqrt{3}}\right) \]Use a calculator to find the angle in radians to the nearest hundredth.

Key Concepts

Dot Product

The dot product, also known as the scalar product, is a way to multiply two vectors together and get a single number, called a scalar. This operation tells us how aligned two vectors are.

• To calculate the dot product, you multiply the corresponding components of the vectors and sum them up.
• In the context of finding the angle between planes, you use the dot product of their normal vectors.
• For example, if you have two normal vectors \( \mathbf{n_1} = \langle a_1, b_1, c_1 \rangle \) and \( \mathbf{n_2} = \langle a_2, b_2, c_2 \rangle \), the dot product is computed as: \( \mathbf{n_1} \cdot \mathbf{n_2} = a_1a_2 + b_1b_2 + c_1c_2 \).

In our exercise, the dot product of the normal vectors \( \langle 2, 2, 2 \rangle \) and \( \langle 2, -2, -1 \rangle \) was calculated as \(-2\). This step is crucial for determining the angle between the two planes.

Normal Vectors

Normal vectors are perpendicular to the surface they define, in this case, a plane. They play a fundamental role in geometry and physics, especially when dealing with surfaces.

• Each plane has an equation of the form \( ax + by + cz = d \), and the normal vector is \( \langle a, b, c \rangle \).
• Normal vectors dictate the orientation or tilt of the plane in three-dimensional space.

In this problem, our normal vectors are given as \( \langle 2, 2, 2 \rangle \) and \( \langle 2, -2, -1 \rangle \). These vectors are derived from the coefficients of the variables \( x, y, z \) in the plane equations. The angle between these normal vectors helps us understand the angle between the two planes themselves.

Inverse Cosine

The inverse cosine function, often written as \( \cos^{-1} \), is the reverse process of finding the cosine of an angle. Here, it is used to find the actual angle from the cosine value.

• Once you have the cosine of an angle, use \( \cos^{-1} \) to determine the angle itself.
• In the context of this exercise, you determine the angle between two normal vectors, which gives the angle between two planes.

After finding \( \cos(\theta) = \frac{-1}{3\sqrt{3}} \), we use \( \cos^{-1} \left( \frac{-1}{3\sqrt{3}} \right) \) to find the angle \( \theta \). Calculators are typically necessary for this step since it results in an angle in radians.

Magnitude of a Vector

The magnitude of a vector, represented by \( ||\mathbf{v}|| \), is like the vector's length or size. It tells us how long the vector is in space.

• The formula to find the magnitude of a vector \( \mathbf{v} = \langle x, y, z \rangle \) is: \( ||\mathbf{v}|| = \sqrt{x^2 + y^2 + z^2} \).
• A vector with a larger magnitude is longer, which in real-world terms might mean greater force or distance.

In our exercise, the magnitudes of \( \langle 2, 2, 2 \rangle \) and \( \langle 2, -2, -1 \rangle \) are calculated as \( 2\sqrt{3} \) and \( 3 \) respectively. Knowing the magnitudes of the normal vectors is required for computing the cosine of the angle between the planes.