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Chapter 7: Problem 28

Find the angle \(\theta\left(0^{\circ} \leq \theta \leq 180\right.\); round to the nearest degree) between each pair of vectors. $$ \langle-5,-5 \sqrt{3}\rangle \text { and }\langle 2,-\sqrt{2}\rangle $$

Short Answer

Expert verified
The angle \( \theta \) is approximately \( 109^\circ \).

Step by step solution

01

Write the Formula for Dot Product

The dot product of two vectors \( \langle a_1, b_1 \rangle \) and \( \langle a_2, b_2 \rangle \) is given by: \[ a_1 \cdot a_2 + b_1 \cdot b_2 \] In this problem, the vectors are \( \langle -5, -5\sqrt{3} \rangle \) and \( \langle 2, -\sqrt{2} \rangle \).
02

Calculate the Dot Product

Compute the dot product of the vectors using the formula from Step 1:\[ -5 \cdot 2 + (-5\sqrt{3}) \cdot (-\sqrt{2}) \] Which simplifies to:\[ -10 + 5\sqrt{3} \cdot \sqrt{2} \]\[ -10 + 5\sqrt{6} \]
03

Find the Magnitudes

The magnitude of a vector \( \langle a, b \rangle \) is found using the formula:\[ \sqrt{a^2 + b^2} \]For \( \langle -5, -5\sqrt{3} \rangle \), the magnitude is:\[ \sqrt{(-5)^2 + (-5\sqrt{3})^2} = \sqrt{25 + 75} = \sqrt{100} = 10 \]For \( \langle 2, -\sqrt{2} \rangle \), the magnitude is:\[ \sqrt{2^2 + (-\sqrt{2})^2} = \sqrt{4 + 2} = \sqrt{6} \]
04

Use the Dot Product Formula to Find \( \theta \)

The formula for the angle \( \theta \) between two vectors using the dot product is:\[ \cos \theta = \frac{\text{dot product}}{\text{magnitude of first vector} \times \text{magnitude of second vector}} \]This becomes:\[ \cos \theta = \frac{-10 + 5\sqrt{6}}{10 \cdot \sqrt{6}} \]
05

Calculate \( \theta \) Using Inverse Cosine

Solve for \( \theta \) by applying the inverse cosine function:\[ \theta = \cos^{-1}\left( \frac{-10 + 5\sqrt{6}}{10\sqrt{6}} \right) \]Computing this gives:\[ \theta \approx 109.47^\circ \]
06

Round \( \theta \) to the Nearest Degree

Finally, round the angle to the nearest degree:\[ \theta \approx 109^\circ \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dot Product
The dot product is a fundamental concept in vector mathematics that helps determine the relationship between two vectors. In this exercise, the dot product of two vectors \( \langle a_1, b_1 \rangle \) and \( \langle a_2, b_2 \rangle \) is calculated using the formula:
  • Dot product = \( a_1 \cdot a_2 + b_1 \cdot b_2 \)
The resulting value gives us insight into the nature of the angle between the vectors. If the dot product is positive, the angle is acute. If negative, the angle is obtuse, and if zero, the vectors are perpendicular. In our exercise, we compute:
  • \( -5 \cdot 2 + (-5\sqrt{3}) \cdot (-\sqrt{2}) \)
  • Which simplifies to \( -10 + 5\sqrt{6} \)
The computed dot product is crucial for the next steps where we will calculate the angle between the vectors.
Magnitude of a Vector
The magnitude of a vector is like its length. It helps quantify the vector's size in a geometrical sense, essential for calculating angles between vectors. For any vector \( \langle a, b \rangle \), the formula for magnitude is:
  • Magnitude = \( \sqrt{a^2 + b^2} \)
In this problem, we determine the magnitudes of the vectors \( \langle -5, -5\sqrt{3} \rangle \) and \( \langle 2, -\sqrt{2} \rangle \):
  • \( \sqrt{(-5)^2 + (-5\sqrt{3})^2} = 10 \)
  • \( \sqrt{2^2 + (-\sqrt{2})^2} = \sqrt{6} \)
The magnitude gives a scalar representation of the vectors, which we use alongside the dot product to find the angle between the vectors.
Inverse Cosine Function
The inverse cosine function, or arccos, is used to find the angle between two vectors once you've calculated the cosine value. This function helps to convert a cosine value back into an angle, providing a direct relationship between the dot product and the magnitude of the vectors. The formula applied here is:
  • \( \cos \theta = \frac{\text{dot product}}{\text{magnitude of first vector} \times \text{magnitude of second vector}} \)
Using this formula, the expression becomes:
  • \( \cos \theta = \frac{-10 + 5\sqrt{6}}{10 \cdot \sqrt{6}} \)
Applying the inverse cosine function:
  • \( \theta = \cos^{-1}\left( \frac{-10 + 5\sqrt{6}}{10\sqrt{6}} \right) \)
This calculation gives \( \theta \approx 109.47^\circ \). The inverse cosine function is invaluable for finding the precise angle between vectors, rounding it to \( 109^\circ \) in this case.

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Most popular questions from this chapter

A box weighing 500 pounds is held in place on an inclined plane that has an angle of \(30^{\circ}\) with the ground. What force is required to hold it in place?

In Exercises 63-66, determine whether each statement is true or false. If the dot product of two nonzero vectors is equal to zero, then the vectors must be perpendicular.

Find the angle \(\theta\left(0^{\circ} \leq \theta \leq 180\right.\); round to the nearest degree) between each pair of vectors. $$ \langle-6,2\rangle \text { and }\langle-3,8\rangle $$

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