Problem 52 - University Calculus: Early Transcendentals

Question

Find the acute angles between the lines and planes. $$x=2, y=3+2 t, z=1-2 t ; \quad x-y+z=0$$

Step-by-Step Solution

Answer

The acute angle is \(\arccos \left( \frac{\sqrt{6}}{3} \right) \).

Step 1: Identify the line's direction vector

Given the line equation in parametric form: \( x = 2, y = 3 + 2t, z = 1 - 2t \).The direction vector of the line is derived from the coefficients of the parameter \( t \) in the equations for \( y \) and \( z \). Thus, the direction vector of the line is \( \mathbf{d} = \langle 0, 2, -2 \rangle \).

Step 2: Identify the plane's normal vector

For the plane given by the equation: \( x - y + z = 0 \),the normal vector is found from the coefficients of \( x, y, \) and \( z \) in the plane equation. Therefore, the normal vector is \( \mathbf{n} = \langle 1, -1, 1 \rangle \).

Step 3: Compute the dot product of direction and normal vectors

The dot product is calculated as follows:\( \mathbf{d} \cdot \mathbf{n} = \langle 0, 2, -2 \rangle \cdot \langle 1, -1, 1 \rangle = (0 \times 1) + (2 \times -1) + (-2 \times 1) = -2 - 2 = -4 \).

Step 4: Compute magnitudes of the vectors

Calculate the magnitude of \( \mathbf{d} \):\[ \|\mathbf{d}\| = \sqrt{0^2 + 2^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2} \].Calculate the magnitude of \( \mathbf{n} \):\[ \|\mathbf{n}\| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3} \].

Step 5: Find the cosine of the angle

The cosine of the angle \( \theta \) between the line and the plane is found using the formula:\( \cos \theta = \frac{\mathbf{d} \cdot \mathbf{n}}{\|\mathbf{d}\| \times \|\mathbf{n}\|} \).Substituting the dot product and magnitudes:\( \cos \theta = \frac{-4}{(2\sqrt{2}) \times \sqrt{3}} = \frac{-4}{2\sqrt{6}} = \frac{-2}{\sqrt{6}} = -\frac{\sqrt{6}}{3} \).

Step 6: Derive the acute angle using the arccos function

The acute angle \( \phi \) is given by:\( \phi = 90^\circ - \arccos(|\cos \theta|) \).Calculate:\( \phi = \arccos \left( \frac{\sqrt{6}}{3} \right) \).

Key Concepts

Direction Vector of a Line

When examining a line in three-dimensional space described with a parameter, the direction vector plays a crucial role. It essentially tells us in which direction the line is moving.
For a line expressed in the parametric form like \( x = a, y = b + \ t imes c, z = m + \ t imes n \), the direction vector is given by the coefficients of \( t \) in the equations for \( y \) and \( z \).

In our example, the line is given by \( x = 2, y = 3 + 2t, z = 1 - 2t \).

• The direction vector thus becomes \( \mathbf{d} = \langle 0, 2, -2 \rangle \).

This vector \( \mathbf{d} \) indicates that as \( t \) increases, \( y \) increases by 2 units and \( z \) decreases by 2 units. The direction vector is essential for analyzing how the line interacts with planes, other lines, or spaces.

Normal Vector of a Plane

To understand the orientation of a plane in space, the normal vector is valuable. This vector is orthogonal, or perpendicular, to the plane, defining its spatial orientation.
A plane described by an equation like \( ax + by + cz = d \), has a normal vector determined directly from the coefficients \( a, b, \text{ and } c \).

Let's consider the equation of our plane, \( x - y + z = 0 \).

• Here, the normal vector becomes \( \mathbf{n} = \langle 1, -1, 1 \rangle \).

This normal vector is fundamental for tasks like finding angles between planes and lines, determining perpendicularity, and many more geometric operations. It serves as an essential tool in understanding and working with the plane.

Dot Product and Magnitude

The dot product is a mathematical operation that takes two equal-length sequences of numbers and returns a single number. It's incredibly helpful in finding angles between vectors.
The dot product of two vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \) is found using the formula:

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

In our scenario:

• \( \mathbf{d} \cdot \mathbf{n} = \langle 0, 2, -2 \rangle \cdot \langle 1, -1, 1 \rangle = -4 \).

The magnitude, or length, of a vector \( \mathbf{v} = \langle x, y, z \rangle \) is calculated as:

• \( \|\mathbf{v}\| = \sqrt{x^2 + y^2 + z^2} \).

It provides the length of the vector in space, such as:

• For \( \mathbf{d} \), \( \|\mathbf{d}\| = 2\sqrt{2} \).
• For \( \mathbf{n} \), \( \|\mathbf{n}\| = \sqrt{3} \).

The combination of the dot product and magnitudes allows us to compute angles between vectors efficiently.

Cosine of an Angle

The cosine of the angle between two vectors is a measure of how much one vector extends along the direction of the other, revealing the angle they form in space.
To find the cosine of the angle \( \theta \) between a line and a plane, we use the formula:
\[ \cos \theta = \frac{\mathbf{d} \cdot \mathbf{n}}{\|\mathbf{d}\| \times \|\mathbf{n}\|} \]
For our vectors:

• Substitute \( \mathbf{d} \cdot \mathbf{n} = -4 \), \( \|\mathbf{d}\| = 2\sqrt{2} \), and \( \|\mathbf{n}\| = \sqrt{3} \).
• Thus, \( \cos \theta = \frac{-4}{2\sqrt{6}} = -\frac{\sqrt{6}}{3} \).

This gives us the cosine of the angle formed between our line and plane, directing us closer to understanding their spatial relationship.

Arccos Function

The arccos (or inverse cosine) function is instrumental in converting a cosine value back into the corresponding angle, especially when dealing with vectors and angles in vector spaces.
In our case, the arccos function helps find the angle \( \phi \) after getting the cosine of an angle. For an acute angle, remember to consider the absolute value because angles need to be between 0 and \( 90^\circ \) in this context.
The formula for the acute angle \( \phi \) is:
\[ \phi = \arccos \left( \left| \cos \theta \right| \right) \]
The calculation proceeds as:

• \( \phi = \arccos \left( \frac{\sqrt{6}}{3} \right) \)

This result delivers the acute angle between the given line and plane, giving us a tangible measure of their geometrical interrelation.