Problem 50 - University Calculus: Early Transcendentals

Question

Find the acute angles between the intersecting lines. $$\begin{aligned}&x=2+t, y=4 t+2, z=1+t \quad \text { and }\\\&x=3 t-2, y=-2, z=2-2 t\end{aligned}$$

Step-by-Step Solution

Answer

The acute angle between the intersecting lines is \( \cos^{-1}\left(\frac{1}{\sqrt{18} \times \sqrt{13}}\right) \).

Step 1: Identify the Direction Vectors of the Lines

The first line can be represented in vector form as \( \vec{r_1} = \langle x, y, z \rangle = \langle 2+t, 4t+2, 1+t \rangle \). Hence, the direction vector for the first line is \( \vec{d_1} = \langle 1, 4, 1 \rangle \).The second line is represented as \( \vec{r_2} = \langle x, y, z \rangle = \langle 3t-2, -2, 2-2t \rangle \). The direction vector for the second line is \( \vec{d_2} = \langle 3, 0, -2 \rangle \).

Step 2: Calculate the Dot Product of the Direction Vectors

The dot product of the vectors \( \vec{d_1} = \langle 1, 4, 1 \rangle \) and \( \vec{d_2} = \langle 3, 0, -2 \rangle \) is calculated as follows:\[\vec{d_1} \cdot \vec{d_2} = 1 \times 3 + 4 \times 0 + 1 \times (-2) = 3 + 0 - 2 = 1\]

Step 3: Calculate the Magnitudes of the Direction Vectors

Calculate the magnitude of \( \vec{d_1} = \langle 1, 4, 1 \rangle \):\[\| \vec{d_1} \| = \sqrt{1^2 + 4^2 + 1^2} = \sqrt{1 + 16 + 1} = \sqrt{18}\]Calculate the magnitude of \( \vec{d_2} = \langle 3, 0, -2 \rangle \):\[\| \vec{d_2} \| = \sqrt{3^2 + 0^2 + (-2)^2} = \sqrt{9 + 0 + 4} = \sqrt{13}\]

Step 4: Determine the Cosine of the Angle Between the Direction Vectors

The cosine of the angle \( \theta \) between two vectors \( \vec{d_1} \) and \( \vec{d_2} \) is given by:\[\cos(\theta) = \frac{\vec{d_1} \cdot \vec{d_2}}{\| \vec{d_1} \| \| \vec{d_2} \|} = \frac{1}{\sqrt{18} \times \sqrt{13}}\]

Step 5: Calculate the Acute Angle Between the Lines

To find \( \theta \), we use the inverse cosine function:\[\theta = \cos^{-1}\left(\frac{1}{\sqrt{18} \times \sqrt{13}}\right)\]Since we require the acute angle, we ensure \( \theta \) is between \( 0 \) and \( \frac{\pi}{2} \). Compute \( \theta \) using a calculator to find the angle in degrees or radians.

Key Concepts

Dot Product

The dot product is a fundamental operation in vector algebra. It takes two vectors and returns a single number, also known as a scalar. This value is crucial in determining the relationship between the two vectors.

To find the dot product of vectors, you multiply corresponding components and sum the results. For example, given vectors \( \vec{a} = \langle a_1, a_2, a_3 \rangle \) and \( \vec{b} = \langle b_1, b_2, b_3 \rangle \, \), their dot product is calculated as follows:

• Step 1: Multiply the first components: \( a_1 \cdot b_1 \)
• Step 2: Multiply the second components: \( a_2 \cdot b_2 \)
• Step 3: Multiply the third components: \( a_3 \cdot b_3 \)
• Step 4: Sum the products: \( a_1 \cdot b_1 + a_2 \cdot b_2 + a_3 \cdot b_3 \)

The result is a scalar value which, in the context of angles, helps us determine how aligned two vectors are: a higher dot product indicates a smaller angle, implying greater alignment.

Direction Vectors

Direction vectors provide us a way to express lines in vector form. They indicate the direction a line moves in the 3D space.

In a parametric equation of a line, such as \( x = a + t \cdot b \, \), \( y = c + t \cdot d \, \), \( z = e + t \cdot f \, \, \), the constants \( b, d, f \) form the direction vector \( \vec{d} = \langle b, d, f \rangle \).

Direction vectors are crucial because they form the basis for many calculations in vector algebra including finding angles between lines. They provide us with a compact representation of a line, allowing us to easily compute other essential properties such as angles or intersections.

Magnitude of Vectors

The magnitude of a vector represents its length in space. It is calculated using the Pythagorean theorem in three dimensions.

For a vector \( \vec{v} = \langle x, y, z \rangle \), the magnitude is given by:

• Step 1: Square each component: \( x^2, y^2, z^2 \)
• Step 2: Sum the squares: \( x^2 + y^2 + z^2 \)
• Step 3: Take the square root of the sum: \( \sqrt{x^2 + y^2 + z^2} \)

Knowing the magnitude is key when computing the cosine of the angle between two vectors, as it acts like the denominator in the equation for computing the cosine.

Cosine of Angles

The cosine of the angle between two vectors relates directly to their dot product and their magnitudes. This relationship is given by the formula:
\[ \cos(\theta) = \frac{\vec{a} \cdot \vec{b}}{\|\vec{a}\| \cdot \|\vec{b}\|}\]

Here, \( \theta \) represents the angle between vectors \( \vec{a} \) and \( \vec{b} \, \), whereas \( \vec{a} \cdot \vec{b} \) is the dot product, and \( \|\vec{a}\| \cdot \|\vec{b}\| \) are the products of their magnitudes.

The cosine value helps us determine the nature of the angle:

• If it is positive, the angle is acute.
• If it is zero, the vectors are perpendicular.
• If it is negative, the angle is obtuse.

This formula thus provides a straightforward way to find the angle between vectors when their directions and magnitudes are known.