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Chapter 9: Problem 7

Find \((a) u \cdot v\) and \((b)\) the angle between \(u\) and \(v\) to the nearest degree. $$\mathbf{u}=\langle 2,7\rangle, \quad \mathbf{v}=\langle 3,1\rangle$$

Short Answer

Expert verified
(a) \(u \cdot v = 13\); (b) angle \(\approx 53\) degrees.

Step by step solution

01

Dot Product Formula

The dot product of two vectors \(\mathbf{u} = \langle u_1, u_2 \rangle\) and \(\mathbf{v} = \langle v_1, v_2 \rangle\) is given by the formula: \(u \cdot v = u_1 \cdot v_1 + u_2 \cdot v_2\).
02

Calculate Dot Product

Using the vectors \(\mathbf{u} = \langle 2,7 \rangle\) and \(\mathbf{v} = \langle 3,1 \rangle\), we compute the dot product: \(u \cdot v = 2 \cdot 3 + 7 \cdot 1 = 6 + 7 = 13\).
03

Magnitude of Vectors

To find the angle between the vectors, we first calculate their magnitudes. The magnitude of a vector \(\mathbf{u} = \langle u_1, u_2 \rangle\) is given by \(|\mathbf{u}| = \sqrt{u_1^2 + u_2^2}\). For \(\mathbf{u}\), \(|\mathbf{u}| = \sqrt{2^2 + 7^2} = \sqrt{4 + 49} = \sqrt{53}\). For \(\mathbf{v}\), \(|\mathbf{v}| = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10}\).
04

Angle Formula

The cosine of the angle \(\theta\) between two vectors is given by the formula: \(\cos \theta = \frac{u \cdot v}{|\mathbf{u}| |\mathbf{v}|}\).
05

Calculate Angle

Substitute the known values into the angle formula: \(\cos \theta = \frac{13}{\sqrt{53} \cdot \sqrt{10}} = \frac{13}{\sqrt{530}}\). Calculate \(\theta\) using \(\theta = \cos^{-1}\left(\frac{13}{\sqrt{530}}\right)\), which approximately equals 53 degrees.

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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 key operation when working with vectors. It combines two vectors, yielding a scalar, rather than another vector. Formally, the dot product between two vectors, say \(\mathbf{u} = \langle u_1, u_2 \rangle\) and \(\mathbf{v} = \langle v_1, v_2 \rangle\), is computed as:
  • \(u \cdot v = u_1 \cdot v_1 + u_2 \cdot v_2\).
In the given problem, for \(\mathbf{u} = \langle 2, 7 \rangle\) and \(\mathbf{v} = \langle 3, 1 \rangle\), calculating the dot product is straightforward:
  • First, multiply the corresponding components: \(2 \cdot 3\) and \(7 \cdot 1\).
  • Then, add these results to obtain the dot product: \(2 \cdot 3 + 7 \cdot 1 = 6 + 7 = 13\).
This scalar result provides valuable information for further calculations, such as determining angles between the vectors.
Magnitude of Vectors
The magnitude of a vector is essentially the "length" of the vector, providing an idea of its size in a geometric sense. For a vector \(\mathbf{u} = \langle u_1, u_2 \rangle\), its magnitude is calculated using the Pythagorean theorem:
  • \(|\mathbf{u}| = \sqrt{u_1^2 + u_2^2}\).
Applying this formula to our vectors:
  • For \(\mathbf{u} = \langle 2, 7 \rangle\), calculate \(|\mathbf{u}| = \sqrt{2^2 + 7^2} = \sqrt{4 + 49} = \sqrt{53}\).
  • For \(\mathbf{v} = \langle 3, 1 \rangle\), compute \(|\mathbf{v}| = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10}\).
Understanding vector magnitudes is crucial, especially when determining the angle between vectors or normalizing vectors.
Angle Between Vectors
The angle between vectors provides insight into their directional relationship. To find the angle \(\theta\) between vectors \(\mathbf{u}\) and \(\mathbf{v}\), you use the cosine formula:
  • \(\cos \theta = \frac{u \cdot v}{|\mathbf{u}| \times |\mathbf{v}|}\).
For our specific vectors:
  • The dot product is \(13\), as calculated.
  • The magnitudes are \(\sqrt{53}\) and \(\sqrt{10}\).
Substitute these values into the cosine formula:
  • \(\cos \theta = \frac{13}{\sqrt{53} \times \sqrt{10}} = \frac{13}{\sqrt{530}}\).
To find \(\theta\), compute the inverse cosine:
  • \(\theta = \cos^{-1}\left(\frac{13}{\sqrt{530}}\right)\).
This approximately gives an angle of 53 degrees. The angle helps in understanding how aligned or perpendicular the vectors are to each other.

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Most popular questions from this chapter

Find the magnitude and direction (in degrees) of the vector. $$\mathbf{v}=\mathbf{i}+\sqrt{3} \mathbf{j}$$

Find the work done by the force \(\mathbf{F}\) in moving an object from \(P\) to \(Q\). $$\mathbf{F}=4 \mathbf{i}-5 \mathbf{j}: \quad P(0,0), Q(3,8)$$

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. $$\begin{aligned} &\mathbf{F}_{1}=4 \mathbf{i}-\mathbf{j}, \quad \mathbf{F}_{2}=3 \mathbf{i}-7 \mathbf{j}, \quad \mathbf{F}_{3}=-8 \mathbf{i}+3 \mathbf{j}\\\&\mathbf{F}_{4}=\mathbf{i}+\mathbf{j}\end{aligned}$$

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. $$\mathbf{F}_{1}=\langle 2,5\rangle, \quad \mathbf{F}_{2}=\langle 3,-8\rangle$$

An object located at the origin in a three-dimensional coordinate system is held in equilibrium by four forces. One has magnitude 7 Ib and points in the direction of the positive \(x\) -axis, so it is represented by the vector 7 i. The second has magnitude 24 Ib and points in the direction of the positive \(y\) -axis. The third has magnitude 25 Ib and points in the direction of the negative z-axis. (a) Use the fact that the four forces are in equilibrium (that is, their sum is 0 ) to find the fourth force. Express it in terms of the unit vectors \(\mathbf{i}, \mathbf{j},\) and \(\mathbf{k}\) (b) What is the magnitude of the fourth force?
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