Problem 70 - A Graphical Approach to Precalculus with Limits

Question

Find the angle between each pair of vectors. $$\langle 6,8\rangle,\langle- 4,3\rangle$$

Step-by-Step Solution

Answer

The angle between the vectors is 90 degrees.

Step 1: Understand the Formula

To find the angle between two vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \), we use the formula: \[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} \]where \( \theta \) is the angle between the vectors, \( \mathbf{a} \cdot \mathbf{b} \) is the dot product, and \( \|\mathbf{a}\| \), \( \|\mathbf{b}\| \) are the magnitudes of the vectors.

Step 2: Calculate the Dot Product

The dot product of \( \mathbf{a} = \langle 6,8 \rangle \) and \( \mathbf{b} = \langle -4,3 \rangle \) is calculated as follows:\( \mathbf{a} \cdot \mathbf{b} = (6)(-4) + (8)(3) = -24 + 24 = 0 \).

Step 3: Find Magnitudes of Vectors

Calculate the magnitude of each vector using the formula: \( \|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2} \).For \( \mathbf{a} \):\( \|\mathbf{a}\| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \).For \( \mathbf{b} \):\( \|\mathbf{b}\| = \sqrt{(-4)^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \).

Step 4: Use the Formula to Find \( \cos \theta \)

Plug the values into the formula:\[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} = \frac{0}{10 \times 5} = 0 \]

Step 5: Determine the Angle \( \theta \)

Since \( \cos \theta = 0 \), it implies that \( \theta = 90^\circ \) because the cosine of 90 degrees is zero. Thus, the angle between the vectors is 90 degrees.

Key Concepts

Dot Product

To understand the concept of the dot product, think of it as a way to multiply two vectors and get a scalar (a single number) instead of another vector. The dot product helps measure how much one vector goes in the direction of another. It is calculated using the formula:

• If you have two vectors, say \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \), the dot product \( \mathbf{a} \cdot \mathbf{b} \) is found by multiplying the corresponding components of the vectors and then adding them up.
• Mathematically, it's \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \).

In our example with vectors \( \langle 6, 8 \rangle \) and \( \langle -4, 3 \rangle \):

• Calculate the dot product by multiplying: \( 6 \times (-4) + 8 \times 3 \).
• This equals \( -24 + 24 = 0 \).

The dot product being zero is significant here. It tells us that the vectors are perpendicular, meaning they form a right angle (90 degrees) with each other. This is a crucial property because a zero dot product indicates orthogonality in Euclidean space.

Magnitude of Vectors

Next, let's understand what the magnitude of a vector is. You can think of magnitude as the length or size of the vector. It's how long the arrow representing the vector is, if you visualize vectors as arrows pointing in space.

• The magnitude is calculated using the formula: \( \|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2} \).
• This is derived from the Pythagorean theorem, treating the vector components as the sides of a right triangle.

For the vector \( \mathbf{a} = \langle 6, 8 \rangle \):

• The magnitude is \( \|\mathbf{a}\| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \).

Similarly, for \( \mathbf{b} = \langle -4, 3 \rangle \):

• The magnitude is \( \|\mathbf{b}\| = \sqrt{(-4)^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \).

Calculating the magnitude gives us the essential values needed to apply the formula that finds the angle between two vectors.

Cosine Formula

The cosine formula for finding the angle \( \theta \) between two vectors leverages the dot product and the magnitudes of the vectors. The formula is:

• \( \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} \)

This formula can be understood in the following way:

• The numerator, \( \mathbf{a} \cdot \mathbf{b} \), gives a sense of how much the vectors "point" in the same direction.
• The denominator, \( \|\mathbf{a}\| \|\mathbf{b}\| \), scales the dot product by the sizes of the vectors.
• Essentially, it adjusts for differing vector lengths.

In our specific situation:

• The dot product \( \mathbf{a} \cdot \mathbf{b} \) is 0.
• The magnitudes multiplied together are \( 10 \times 5 = 50 \).
• Thus, \( \cos \theta = \frac{0}{50} = 0 \).

When the cosine of an angle is zero, it indicates a 90-degree angle. This result corroborates what we noted with the zero dot product—confirming that the vectors are indeed perpendicular to each other. Understanding this relationship is pivotal in various applications, such as physics and engineering, where vector angles dictate mechanisms and forces.