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Chapter 6: Problem 46

Finding a Unit Vector In Exercises \(39-48,\) find a unit vector in the direction of the given vector. Verify that the result has a magnitude of 1 . . $$\mathbf{w}=-6 \mathbf{i}$$

Short Answer

Expert verified
The unit vector in the direction of vector \(\mathbf{w}\) is - \mathbf{i}, and its magnitude is indeed 1.

Step by step solution

01

Identify the Given Vector

The given vector is \(\mathbf{w}=-6 \mathbf{i}\).
02

Determine the Magnitude of the Given Vector

The magnitude, denoted by \(||\mathbf{w}||\), is obtained using the formula \(||\mathbf{w}||= \sqrt{w_{1}^2}\), where \(w_{1}\) is the coefficient of \(\mathbf{i}\), which in this case is -6. Therefore, the magnitude of vector \(\mathbf{w}\) is \(||\mathbf{w}||= \sqrt{(-6)^{2}} = 6\).
03

Calculating the Unit Vector

A unit vector, denoted by \(\hat{u}\), is obtained by dividing the given vector by its magnitude. So, the unit vector in the direction of \(\mathbf{w}\) is \(\hat{u}=\mathbf{w}/||\mathbf{w}|| = -6 \mathbf{i} / 6 = - \mathbf{i}\).
04

Verify the Magnitude of the Unit Vector

Finally, let's verify that the magnitude of the unit vector is indeed 1. The magnitude of \(\hat{u}\) is \(||\hat{u}||=\sqrt{u_{1}^{2}}\), where \(u_{1}\) is the coefficient of \(\mathbf{i}\), which in this case is -1. Therefore, \(||\hat{u}||=\sqrt{(-1)^2} = 1\). So, the magnitude of the unit vector \(\hat{u}\) is indeed 1, as it should be.

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Key Concepts

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

Vector Magnitude
The magnitude of a vector gives us its size or length, regardless of direction. Imagine a vector as an arrow. The magnitude is how long that arrow is. In math, it's like measuring the distance of a point from the origin.
To calculate the magnitude of a vector, we use the formula that involves the square root of the sum of the squares of its components. For a simple 1-dimensional vector like \( \mathbf{w} = -6 \mathbf{i} \), we only have one component to consider:

\[ ||\mathbf{w}|| = \sqrt{(-6)^2} = 6 \]
This means our vector has a length of 6. This step is essential because knowing the magnitude helps us in finding the unit vector.
Vector Direction
The direction of a vector is the direction in which the vector points. While magnitude tells us how far or big, direction tells us where. Think of it as where an arrow is pointing on a compass.
In 1-dimensional vectors, like \(-6 \mathbf{i} \), the direction is along the axis of the vector's component. The negative sign indicates the vector points in the opposite direction of the positive axis. This concept becomes even more critical in multi-dimensional vectors, where direction involves angles or multiple axes.
Knowing the direction is crucial as it helps in determining the orientation of the vector, which plays a significant role in physics and engineering applications.
Unit Vector Calculation
A unit vector is a vector with a magnitude of 1. It's like having an arrow with a fixed length pointing in a specific direction. Unit vectors are handy when you need to retain direction but not size.
To find a unit vector in the direction of a given vector, divide the vector by its magnitude. For example, with \( \mathbf{w} = -6 \mathbf{i} \) and magnitude 6:

\[ \hat{u} = \frac{\mathbf{w}}{||\mathbf{w}||} = \frac{-6 \mathbf{i}}{6} = - \mathbf{i} \]
This result, \(- \mathbf{i}\), is our unit vector. It preserves the direction but scales the size down to 1.
Always verify by checking the magnitude of the unit vector. In this case, the magnitude is \(1\), confirming it's a unit vector. Ensuring your unit vector has a magnitude of 1 is vital for accuracy in vector operations.

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Most popular questions from this chapter

Finding Orthogonal Vectors In Exercises \(67-70\) , find two vectors in opposite directions that are orthogonal to the vector u. (There are many correct answers.) $$\mathbf{u}=\langle- 8,3\rangle$$

Resultant Force In Exercises 81 and \(82,\) find the angle between the forces given the magnitude of their resultant. (Hint. Write force 1 as a vector in the direction of the positive \(x\) -axis and force 2 as a vector at an angle \(\theta\) with the positive \(x\) -axis.) \(\begin{array}{ll}{\text { Force } 1} & {\text { Force } 2} & {\text { Resultant Force }} \\ {45 \text { pounds }} & {60 \text { pounds }} & {90 \text { pounds }}\end{array}\)

Graphical Reasoning Consider two forces $$\mathbf{F}_{1}=\langle 10,0\rangle\( and \)\mathbf{F}_{2}=5\langle\cos \theta, \sin \theta\rangle$$ (a) Find \(\left\|\mathbf{F}_{1}+\mathbf{F}_{2}\right\|\) as a function of \(\theta\) (b) Use a graphing utility to graph the function in part \(\quad(\) a) for \(0 \leq \theta<2 \pi\) (c) Use the graph in part (b) to determine the range of the function. What is its maximum, and for what value of \(\theta\) does it occur? What is its minimum, and for what value of \(\theta\) does it occur? (d) Explain why the magnitude of the resultant is never 0.

Finding the Component Form of a Vector In Exercises \(67-74\) , find the component form of \(v\) given its magnitude and the angle it makes with the positive \(x\) -axis. Sketch y. $$\begin{array}{ll}{\text { Magnitude }} & {\text { Angle }} \\\ {\|\mathbf{v}\|=1} & {\theta=45^{\circ}}\end{array}$$

Solving an Equation In Exercises \(97-104,\) use the formula on page 446 to find all solutions of the equation and represent the solutions graphically. $$x^{3}-(1-i)=0$$
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