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Chapter 9: Problem 33

Two vectors \(u\) and \(v\) are given. Find the angle (expressed in degrees) between \(\mathbf{u}\) and \(v\) $$\mathbf{u}=\langle 2,-2,-1\rangle, \quad \mathbf{v}=\langle 1,2,2\rangle$$

Short Answer

Expert verified
The angle between the vectors is approximately 116.57 degrees.

Step by step solution

01

Calculate the Dot Product

The dot product of two vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \) can be calculated using the formula \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \). For \( \mathbf{u} = \langle 2, -2, -1 \rangle \) and \( \mathbf{v} = \langle 1, 2, 2 \rangle \), the dot product is:\[ \mathbf{u} \cdot \mathbf{v} = (2)(1) + (-2)(2) + (-1)(2) = 2 - 4 - 2 = -4 \]
02

Calculate the Magnitude of Vectors

The magnitude of a vector \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) is given by \( ||\mathbf{a}|| = \sqrt{a_1^2 + a_2^2 + a_3^2} \).- For \( \mathbf{u} \): \[ ||\mathbf{u}|| = \sqrt{2^2 + (-2)^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \]- For \( \mathbf{v} \): \[ ||\mathbf{v}|| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]
03

Use Cosine Formula to Find the Angle

The angle \( \theta \) between two vectors can be found using the formula: \( \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| \, ||\mathbf{v}||} \).Substitute the dot product and magnitudes from above:\[ \cos \theta = \frac{-4}{3 \times 3} = \frac{-4}{9} \]
04

Calculate the Angle in Degrees

To find \( \theta \), take the inverse cosine (or arccos) of \( \frac{-4}{9} \) and then convert the angle from radians to degrees using the conversion factor \( \frac{180}{\pi} \).\[ \theta = \cos^{-1} \left( \frac{-4}{9} \right) \approx 116.57^{\circ} \]

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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 operation in vector algebra that helps in analyzing the relationship between two vectors. Imagine two vectors, \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \). To find their dot product, you use the formula:
\[\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3\]This expression essentially involves multiplying each corresponding component of the vectors and then summing the results.
  • Example: For vectors \( \mathbf{u} = \langle 2, -2, -1 \rangle \) and \( \mathbf{v} = \langle 1, 2, 2 \rangle \), the dot product is calculated as \( (2)(1) + (-2)(2) + (-1)(2) = -4 \).
  • The result tells you something about the orientation of the vectors; when the dot product is zero, the vectors are perpendicular.
Understanding the dot product helps in many fields such as physics and computer graphics where vector directions and magnitudes are key.
Magnitude of Vectors
The magnitude of a vector gives you an idea about its 'length' or 'size' in the space it occupies. For a vector \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \), its magnitude is computed through:
\[||\mathbf{a}|| = \sqrt{a_1^2 + a_2^2 + a_3^2}\]This formula involves squaring each of the vector's components, summing these squares, and then taking the square root of the result.
  • For vector \( \mathbf{u} = \langle 2, -2, -1 \rangle \), the magnitude is calculated as \( \sqrt{4 + 4 + 1} = 3 \).
  • Similarly, for vector \( \mathbf{v} = \langle 1, 2, 2 \rangle \), its magnitude is also \( 3 \).
The magnitude is crucial when normalizing vectors or when calculating angles between vectors.
Angle Between Vectors
Finding the angle between two vectors is a powerful way to understand their spatial relationship. The angle \( \theta \) can be calculated by using the dot product and the magnitudes of the vectors. The formula used is:
\[\cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| \, ||\mathbf{v}||}\]This involves dividing the dot product of the two vectors by the product of their magnitudes.
  • For vectors \( \mathbf{u} = \langle 2, -2, -1 \rangle \) and \( \mathbf{v} = \langle 1, 2, 2 \rangle \), using the previously calculated dot product \( -4 \) and magnitudes \( 3 \), you find \( \cos \theta = \frac{-4}{9} \).
  • To find the actual angle \( \theta \), take the arccosine of \( \frac{-4}{9} \), which gives approximately \( 116.57^{\circ} \).
This calculation helps in applications such as determining alignment in physics or congruence in geometry.

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Most popular questions from this chapter

A constant force \(\mathbf{F}=\langle 2,8\rangle\) moves an object along a straight line from the point \((2,5)\) to the point \((11,13) .\) Find the work done if the distance is measured in feet and the force is measured in pounds.

Find the magnitude and direction (in degrees) of the vector. $$\mathbf{v}=\langle 3,4\rangle$$

Find the horizontal and vertical components of the vector with given length and direction, and write the vector in terms of the vectors i and j. $$|\mathbf{v}|=800, \quad \theta=125^{\circ}$$

Two nonzero vectors are parallel if they point in the same direction or in opposite directions. This means that if two vectors are parallel, one must be a scalar multiple of the other. Determine whether the given vectors \(\mathbf{u}\) and \(\mathbf{v}\) are parallel. If they are, express \(\mathbf{v}\) as a scalar multiple of \(\mathbf{u}\). (a) \(\mathbf{u}=\langle 3,-2,4\rangle, \quad \mathbf{v}=\langle- 6,4,-8\rangle\) (b) \(\mathbf{u}=\langle- 9,-6,12\rangle, \quad \mathbf{v}=\langle 12,8,-16\rangle\) (c) \(\mathbf{u}=\mathbf{i}+\mathbf{j}+\mathbf{k}, \quad \mathbf{v}=2 \mathbf{i}+2 \mathbf{j}-2 \mathbf{k}\)

Equilibrium of Forces The forces \(\mathbf{F}_{1}, \mathbf{F}_{2}, \ldots, \mathbf{F}_{n}\) acting at the same point \(P\) are said to be in equilibrium if the resultant force is zero, that is, if \(\mathbf{F}_{1}+\mathbf{F}_{2}+\cdots+\mathbf{F}_{n}=0 .\) Find (a) the resultant forces acting at \(P,\) and (b) the additional force required (if any) for the forces to be in equilibrium. $$\begin{aligned} &\mathbf{F}_{1}=4 \mathbf{i}-\mathbf{j}, \quad \mathbf{F}_{2}=3 \mathbf{i}-7 \mathbf{j}, \quad \mathbf{F}_{3}=-8 \mathbf{i}+3 \mathbf{j}\\\&\mathbf{F}_{4}=\mathbf{i}+\mathbf{j}\end{aligned}$$
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