Problem 7 - Calculus: Early Transcendentals

Question

Find the cosine of the angle between \langle-1,-2,-3\rangle and \langle 5,0,2\rangle\(;\) use a calculator if necessary to find the angle.

Step-by-Step Solution

Answer

The cosine of the angle is approximately -0.546.

Step 1: Understand the Problem

We need to find the cosine of the angle between two vectors: \( \langle -1, -2, -3 \rangle \) and \( \langle 5, 0, 2 \rangle \). The formula to find the cosine of the angle \( \theta \) between two vectors \( \mathbf{a} \) and \( \mathbf{b} \) is given by \[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} \]where \( \mathbf{a} \cdot \mathbf{b} \) is the dot product of the vectors, and \( \|\mathbf{a}\| \) and \( \|\mathbf{b}\| \) are the magnitudes of the vectors.

Step 2: Calculate the Dot Product

The dot product \( \mathbf{a} \cdot \mathbf{b} \) can be calculated as:\[ (-1 \times 5) + (-2 \times 0) + (-3 \times 2) = -5 + 0 - 6 = -11 \]

Step 3: Calculate the Magnitude of Each Vector

The magnitude of a vector \( \langle x, y, z \rangle \) is calculated as:\[ \|\mathbf{a}\| = \sqrt{(-1)^2 + (-2)^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \] \[ \|\mathbf{b}\| = \sqrt{(5)^2 + (0)^2 + (2)^2} = \sqrt{25 + 0 + 4} = \sqrt{29} \]

Step 4: Apply the Cosine Formula

Substitute the dot product and the magnitudes into the cosine formula:\[ \cos \theta = \frac{-11}{\sqrt{14} \times \sqrt{29}} \] Simplify the expression:\[ \cos \theta = \frac{-11}{\sqrt{406}} \approx \frac{-11}{20.15} \approx -0.546 \]

Step 5: Interpret the Result

The cosine of the angle between the vectors is approximately \(-0.546\). This indicates that the angle is greater than 90 degrees since the cosine is negative.

Key Concepts

Dot Product

The dot product, also known as the scalar product, is a fundamental operation you can perform on two vectors. It is defined as the sum of the products of each corresponding pair of vector components. Given two vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \):
\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \]
This operation results in a single number, a scalar, which reflects the extent to which the vectors point in the same direction. If the dot product is positive, the vectors tend in the same direction. If it is negative, they are more likely to point in opposite directions.

• Usefulness: Helps in finding the angle between vectors.
• Interpretation: Positive, negative, or zero values indicate the type of angle.

In our exercise, the dot product of vectors \( \langle -1, -2, -3 \rangle \) and \( \langle 5, 0, 2 \rangle \) is \(-11\), hinting at an obtuse angle since it is negative.

Magnitude of Vectors

The magnitude of a vector, also referred to as its length or norm, is a measure of the distance from its tail to its head if it were drawn as an arrow in space. It is calculated using the following formula for a vector \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \):
\[ \|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2 + a_3^2} \]
This formula is derived from the Pythagorean theorem and provides a way to understand how "long" a vector is.

• The magnitude is always non-negative.
• It gives an idea about the size of the vector.

In our calculation, the magnitudes of vector \( \langle -1, -2, -3 \rangle \) and vector \( \langle 5, 0, 2 \rangle \) are \( \sqrt{14} \) and \( \sqrt{29} \), respectively.
These magnitudes are used to compute the cosine of the angle between vectors.

Cosine of Angle Between Vectors

To find the cosine of the angle between two vectors, use the formula:
\[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} \]
This formula states that the cosine of the angle \( \theta \) between vectors \( \mathbf{a} \) and \( \mathbf{b} \) can be found by dividing their dot product by the product of their magnitudes. This expression helps determine the directional relationship between the vectors.

• If \( \cos \theta \) is positive, the angle is acute (less than 90 degrees).
• If \( \cos \theta \) is zero, the vectors are perpendicular.
• If \( \cos \theta \) is negative, the angle is obtuse (greater than 90 degrees).

In our problem, the cosine calculated is approximately \(-0.546\), indicating that the angle between the vectors is obtuse because the cosine value is negative.
This calculations help in various applications such as physics, engineering, and computer graphics, where understanding the spatial relation of vectors is essential.