Problem 36 - Advanced Engineering Mathematics

Question

The angle between two lines \(\mathscr{L}_{a}\) and \(\mathscr{L}_{b}\) is the angle between their direction vectors a and \(\mathbf{b}\). In Problems, find the angle between the given lines. $$ \frac{x-1}{2}=\frac{y+5}{7}=\frac{z-1}{-1} ; \quad \frac{x+3}{-2}=y-9=\frac{z}{4} $$

Step-by-Step Solution

Answer

\(\theta \approx 92.85^\circ\)

Step 1: Identify Direction Vectors

For the line \(\mathscr{L}_a\) represented by \(\frac{x-1}{2}=\frac{y+5}{7}=\frac{z-1}{-1}\), the direction vector is \(\mathbf{a} = \langle 2, 7, -1 \rangle\). For the line \(\mathscr{L}_b\) given by \(\frac{x+3}{-2}=y-9=\frac{z}{4}\), the direction vector is \(\mathbf{b} = \langle -2, 1, 4 \rangle\).

Step 2: Dot Product of Direction Vectors

Calculate the dot product of the vectors \(\mathbf{a}\) and \(\mathbf{b}\): \(\mathbf{a} \cdot \mathbf{b} = 2 \times (-2) + 7 \times 1 + (-1) \times 4 = -4 + 7 - 4 = -1\).

Step 3: Magnitude of Direction Vectors

Compute the magnitude of each direction vector. For \(\mathbf{a}\): \(|\mathbf{a}| = \sqrt{2^2 + 7^2 + (-1)^2} = \sqrt{4 + 49 + 1} = \sqrt{54}\). For \(\mathbf{b}\): \(|\mathbf{b}| = \sqrt{(-2)^2 + 1^2 + 4^2} = \sqrt{4 + 1 + 16} = \sqrt{21}\).

Step 4: Calculate Cosine of the Angle

Use the dot product and magnitudes to find the cosine of the angle \(\theta\) between the vectors: \(\cos(\theta) = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} = \frac{-1}{\sqrt{54} \cdot \sqrt{21}}\).

Step 5: Determine the Angle

Using the inverse cosine function, find \(\theta\): \(\theta = \cos^{-1}\left(\frac{-1}{\sqrt{54 \times 21}}\right)\). Calculate the result using a calculator to get the angle in degrees or radians.

Key Concepts

Direction Vectors

In vector geometry, a direction vector is a vector that gives the direction of a line in space. For any line in 3D, its direction vector is crucial. It provides an understanding of how the line extends in each coordinate direction. To determine the direction vector from the given line equations:

• Look at the denominators in the equation. These tell you how far the line moves in each axis direction when it steps through one unit of the parameter.
• For example, in the line equation \(\frac{x-1}{2}=\frac{y+5}{7}=\frac{z-1}{-1}\), the direction vector is \( \langle 2, 7, -1 \rangle \), indicating movement 2 units in the x direction, 7 in y, and -1 in z per unit parameter change.
• Similarly, the line \(\frac{x+3}{-2}=y-9=\frac{z}{4}\) has a direction vector \( \langle -2, 1, 4 \rangle \).

Understanding how to extract these vectors is foundational for analyzing any line in a three-dimensional space.

Dot Product

The dot product is a fundamental operation in vector mathematics. It is calculated by multiplying corresponding components of two vectors and then summing the results. For vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \), the dot product is computed as:\[\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3\]The dot product has several uses, one of which is determining the angle between two vectors. If the dot product is zero, the vectors are orthogonal. In the example, the vectors \( \mathbf{a} = \langle 2, 7, -1 \rangle \) and \( \mathbf{b} = \langle -2, 1, 4 \rangle \) have a dot product of -1, calculated as -4 + 7 - 4. This operation helps in understanding the geometric relationship between the lines represented by these vectors.

Angle Between Lines

The angle between two lines in space can be found through their direction vectors. We do this by using the dot product formula and magnitudes of the vectors. The formula for the cosine of the angle \(\theta\) between two vectors \(\mathbf{a}\) and \(\mathbf{b}\) is given by:\[\cos(\theta) = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| \cdot |\mathbf{b}|}\]For the given vectors, the dot product is -1. The magnitudes are \(\sqrt{54}\) for \(\mathbf{a}\) and \(\sqrt{21}\) for \(\mathbf{b}\). The cosine of the angle is then:\[\cos(\theta) = \frac{-1}{\sqrt{54} \times \sqrt{21}}\]By calculating the inverse cosine (\(\cos^{-1}\) of the result), you can find the angle \(\theta\) in degrees or radians. The angle helps describe how the two lines orient with respect to one another in space.

Magnitude of Vectors

The magnitude of a vector, often referred to as its length, is a measure of how long it is in space. For a vector with components \(\langle x, y, z \rangle\), the magnitude is calculated using the formula:\[|\mathbf{a}| = \sqrt{x^2 + y^2 + z^2}\]This is derived from the Pythagorean theorem in three dimensions. It gives a scalar value representing the vector's length. For example:

• Vector \(\mathbf{a} = \langle 2, 7, -1 \rangle \) has a magnitude of \(\sqrt{2^2 + 7^2 + (-1)^2} = \sqrt{54}\).
• Vector \(\mathbf{b} = \langle -2, 1, 4 \rangle \) has a magnitude of \(\sqrt{(-2)^2 + 1^2 + 4^2} = \sqrt{21}\).

Calculating the magnitude is an essential step in many vector calculations, including finding the angle between vectors.