Problem 3 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the cosine of the angle between \(\mathbf{a}\) and \(\mathbf{b}\) and make a sketch. (a) \(\mathbf{a}=\langle 1,-3\rangle, \mathbf{b}=\langle-1,2\rangle\) (b) \(\mathbf{a}=\langle-1,-2\rangle, \mathbf{b}=\langle 6,0\rangle\) (c) \(\mathbf{a}=\langle 2,-1\rangle, \mathbf{b}=\langle-2,-4\rangle\) (d) \(\mathbf{a}=\langle 4,-7\rangle, \mathbf{b}=\langle-8,10\rangle\)

Step-by-Step Solution

Answer

(a) \( \cos(\theta) = \frac{-7}{5\sqrt{2}} \); (b) \( \cos(\theta) = \frac{-1}{\sqrt{5}} \); (c) \( \cos(\theta) = 0 \); (d) \( \cos(\theta) = \frac{-102}{\sqrt{65}\sqrt{164}} \).

Step 1a: Find Dot Product (a)

Calculate the dot product of \( \mathbf{a} = \langle 1, -3 \rangle \) and \( \mathbf{b} = \langle -1, 2 \rangle \): \[ \mathbf{a} \cdot \mathbf{b} = (1)(-1) + (-3)(2) = -1 - 6 = -7 \]

Step 2a: Find Magnitudes (a)

Compute the magnitudes of vectors \( \mathbf{a} \) and \( \mathbf{b} \): \[ ||\mathbf{a}|| = \sqrt{1^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10} \] \[ ||\mathbf{b}|| = \sqrt{(-1)^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5} \]

Step 3a: Calculate Cosine (a)

Use the formula for cosine between two vectors: \[ \cos(\theta) = \frac{\mathbf{a} \cdot \mathbf{b}}{||\mathbf{a}|| ||\mathbf{b}||} = \frac{-7}{\sqrt{10}\sqrt{5}} = \frac{-7}{\sqrt{50}} = \frac{-7}{5\sqrt{2}} \]

Step 4a: Sketch Vectors (a)

Draw vectors \( \mathbf{a} = \langle 1, -3 \rangle \) and \( \mathbf{b} = \langle -1, 2 \rangle \) on the coordinate plane. Vector \( \mathbf{a} \) points to near the fourth quadrant while \( \mathbf{b} \) points to the second quadrant.

Step 1b: Find Dot Product (b)

Calculate the dot product of \( \mathbf{a} = \langle -1, -2 \rangle \) and \( \mathbf{b} = \langle 6, 0 \rangle \): \[ \mathbf{a} \cdot \mathbf{b} = (-1)(6) + (-2)(0) = -6 \]

Step 2b: Find Magnitudes (b)

Compute the magnitudes of vectors \( \mathbf{a} \) and \( \mathbf{b} \): \[ ||\mathbf{a}|| = \sqrt{(-1)^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5} \] \[ ||\mathbf{b}|| = \sqrt{6^2 + 0^2} = \sqrt{36} = 6 \]

Step 3b: Calculate Cosine (b)

Use the formula for cosine between two vectors: \[ \cos(\theta) = \frac{\mathbf{a} \cdot \mathbf{b}}{||\mathbf{a}|| ||\mathbf{b}||} = \frac{-6}{\sqrt{5} \times 6} = \frac{-6}{6\sqrt{5}} = \frac{-1}{\sqrt{5}} \]

Step 4b: Sketch Vectors (b)

Draw vectors \( \mathbf{a} = \langle -1, -2 \rangle \) and \( \mathbf{b} = \langle 6, 0 \rangle \). Vector \( \mathbf{a} \) points in the third quadrant, and \( \mathbf{b} \) is along the positive x-axis.

Step 1c: Find Dot Product (c)

Calculate the dot product of \( \mathbf{a} = \langle 2, -1 \rangle \) and \( \mathbf{b} = \langle -2, -4 \rangle \): \[ \mathbf{a} \cdot \mathbf{b} = (2)(-2) + (-1)(-4) = -4 + 4 = 0 \]

Step 2c: Find Magnitudes (c)

Compute the magnitudes of vectors \( \mathbf{a} \) and \( \mathbf{b} \): \[ ||\mathbf{a}|| = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \] \[ ||\mathbf{b}|| = \sqrt{(-2)^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \]

Step 3c: Calculate Cosine (c)

Since the dot product is zero, the angle \( \theta \) is 90 degrees, and \( \cos(90^\circ) = 0 \). Thus, vectors are perpendicular.

Step 4c: Sketch Vectors (c)

Draw vectors \( \mathbf{a} = \langle 2, -1 \rangle \) and \( \mathbf{b} = \langle -2, -4 \rangle \) intersecting at right angles, showing they are perpendicular.

Step 1d: Find Dot Product (d)

Calculate the dot product of \( \mathbf{a} = \langle 4, -7 \rangle \) and \( \mathbf{b} = \langle -8, 10 \rangle \): \[ \mathbf{a} \cdot \mathbf{b} = (4)(-8) + (-7)(10) = -32 - 70 = -102 \]

Step 2d: Find Magnitudes (d)

Compute the magnitudes of vectors \( \mathbf{a} \) and \( \mathbf{b} \): \[ ||\mathbf{a}|| = \sqrt{4^2 + (-7)^2} = \sqrt{16 + 49} = \sqrt{65} \] \[ ||\mathbf{b}|| = \sqrt{(-8)^2 + 10^2} = \sqrt{64 + 100} = \sqrt{164} \]

Step 3d: Calculate Cosine (d)

Use the formula for cosine between two vectors: \[ \cos(\theta) = \frac{\mathbf{a} \cdot \mathbf{b}}{||\mathbf{a}|| ||\mathbf{b}||} = \frac{-102}{\sqrt{65}\sqrt{164}} \]

Step 4d: Sketch Vectors (d)

Draw vectors \( \mathbf{a} = \langle 4, -7 \rangle \) and \( \mathbf{b} = \langle -8, 10 \rangle \) on the coordinate plane to visualize the angle between them.

Key Concepts

Dot Product

The dot product is a key tool when working with vectors. It tells us how much one vector extends in the direction of another. Essentially, it's a way to multiply two vectors. To calculate the dot product of two vectors, such as \( \mathbf{a} = \langle x_1, y_1 \rangle \) and \( \mathbf{b} = \langle x_2, y_2 \rangle \), we simply multiply their corresponding components and add the results:

• Multiply the x-components: \( x_1 \times x_2 \)
• Multiply the y-components: \( y_1 \times y_2 \)
• Add these products: \( a \cdot b = x_1 y_1 + x_2 y_2 \)

If the dot product is positive, the angle between the vectors is less than 90 degrees. If it's negative, the vectors point in opposite directions. A dot product of zero means the vectors are perpendicular, indicating they form a 90-degree angle.

Magnitude of Vectors

The magnitude of a vector represents its length or size. Think of it as the vector's distance from the origin in a coordinate system. Calculating magnitude involves the Pythagorean theorem. For a vector \( \mathbf{a} = \langle x, y \rangle \), the magnitude is given by:

• Square each component: \( x^2, y^2 \)
• Add the squared values: \( x^2 + y^2 \)
• Take the square root of the sum: \( ||\mathbf{a}|| = \sqrt{x^2 + y^2} \)

Understanding vector magnitudes is essential when comparing the sizes of different vectors or applying them in physics to determine velocity, force, or displacement. It's also crucial for finding the cosine of the angle between vectors.

Vector Sketch

Sketching vectors helps to visualize their direction and magnitude. For beginners, drawing vectors can clarify their position in a coordinate system and assist in understanding vector operations, like addition or determining angles between vectors. When sketching vectors, follow these steps:

• Start at the origin, or the tail of the vector, in a coordinate system.
• From there, plot the x-component along the x-axis.
• Next, plot the y-component upwards or downwards based on the positive or negative sign.
• Draw an arrow from the origin to this point. The arrow’s tip marks the vector’s head.
• Always indicate direction, as this conveys more than just length.

Sketches offer an intuitive grasp of how vectors relate to each other, important when calculating angles. This visual insight into the vector’s orientation and relative position helps confirm algebraic calculations.

Perpendicular Vectors

Perpendicular vectors meet at a right angle, marking an important geometric relationship. When two vectors are perpendicular, their dot product is zero. This is because they exert no influence on each other in their respective directions. Understanding perpendicular vectors is helpful in fields like graphics, where it's vital to know when lines or planes intersect orthogonally.

• Two vectors \( \mathbf{a} \) and \( \mathbf{b} \) are perpendicular if: \( \mathbf{a} \cdot \mathbf{b} = 0 \)
• This means \( \cos(\theta) = 0 \) since \( \theta = 90^{\circ} \)
• Perpendicular vectors are common in coordinate systems, simplifying vector component analysis.

Perpendicularity simplifies many calculations and is fundamental in creating coordinate axes and planes. Recognizing when vectors are perpendicular, through their dot product, is foundational in both geometry and algebra.