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Chapter 7: Problem 54

Vector \(\mathbf{F}\) is in standard position and makes an angle of \(315^{\circ}\) with the positive \(x\)-axis. Its magnitude is 30 . Write \(\mathbf{F}\) in component form \(\langle a, b\rangle\) and in vector component form \(a \mathbf{i}+b \mathbf{j}\).

Short Answer

Expert verified
\(\mathbf{F} = \langle 15\sqrt{2}, -15\sqrt{2} \rangle = 15\sqrt{2} \mathbf{i} - 15\sqrt{2} \mathbf{j}\)

Step by step solution

01

Understand the Problem

We need to express the vector \(\mathbf{F}\), which has a magnitude of 30 and makes an angle of \(315^{\circ}\) with the positive \(x\)-axis, in component form \(\langle a, b \rangle\) and in vector component form \(a \mathbf{i} + b \mathbf{j}\).
02

Calculate the Components Using Trigonometry

The components \(a\) and \(b\) of a vector with magnitude \(r\) and direction angle \(\theta\) can be found using the formulas \(a = r \cos \theta\) and \(b = r \sin \theta\). Here, \(r = 30\) and \(\theta = 315^{\circ}\).
03

Compute the x-component (a)

Calculate the \(x\)-component using the cosine of \(\theta\). \[ a = 30 \cos 315^{\circ} = 30 \times \frac{\sqrt{2}}{2} = 15\sqrt{2} \].
04

Compute the y-component (b)

Calculate the \(y\)-component using the sine of \(\theta\). \[ b = 30 \sin 315^{\circ} = 30 \times \left(-\frac{\sqrt{2}}{2}\right) = -15\sqrt{2} \].
05

Write the Vector in Component Form

Combine the components into the component form \(\langle a, b \rangle\). The vector is \(\langle 15\sqrt{2}, -15\sqrt{2} \rangle\).
06

Write the Vector in Vector Component Form

The vector can also be expressed in terms of the unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) as \(15\sqrt{2} \mathbf{i} - 15\sqrt{2} \mathbf{j}\).

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Key Concepts

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

Trigonometry and Vectors
Trigonometry plays a key role in understanding vectors. When we talk about a vector in the plane, we're typically interested in two main quantities: its magnitude, or length, and its direction, which is usually measured as an angle from the positive x-axis.
To find components of a vector, trigonometry is essential.
  • The cosine of the angle helps us determine the x-component, as it measures how much of the vector points horizontally.
  • The sine of the angle gives us the y-component, showing how much of the vector points vertically.

In our exercise, with an angle of \(315^{\circ}\), we first need to recognize this angle is in the fourth quadrant (because it is between \(270^{\circ}\) and \(360^{\circ}\)). Here, the cosine is positive, and the sine is negative. This tells us that the x-component is positive, and the y-component, being negative, points downward.
Remembering these aspects of trigonometry helps immensely in analyzing vectors in component form.
Vector Magnitude
The magnitude of a vector is a measure of how long it is, much like the length of a line segment. We denote it as \(r\) and given its value, we can think of it as the vector's "speed" or "strength" regardless of direction.
Magnitude is always a non-negative value.
To calculate a vector's components using its magnitude and direction angle:
  • The x-component, \(a\), is computed by multiplying the magnitude \(r\) by the cosine of the angle \(\theta\): \[ a = r \cos \theta \].
  • The y-component, \(b\), is found by multiplying the magnitude by the sine of the angle: \[ b = r \sin \theta \].

In our example, the magnitude of the vector \( \mathbf{F} \) is 30, and we follow these formulas using \(\theta = 315^{\circ}\) to find the exact values for the x and y components. Magnitude is crucial as it scales the components proportionally to keep the same directional balance dictated by the angle.
Direction Angle of a Vector
The direction angle of a vector indicates its orientation in the plane. Measured from the positive x-axis, this angle helps us understand where the vector points.
Angles are often presented in degrees, and an angle can help us categorize which quadrant the vector lands in:
  • First quadrant: 0掳 to 90掳
  • Second quadrant: 90掳 to 180掳
  • Third quadrant: 180掳 to 270掳
  • Fourth quadrant: 270掳 to 360掳 (or -90掳 from positive y-axis going clockwise)
In our exercise, an angle of \(315^{\circ}\) positions the vector in the fourth quadrant. This has significant implications on the signs of its components:
  • In the fourth quadrant, the x-component (cosine) remains positive while the y-component (sine) turns negative.
Such understanding aids in correctly calculating and assigning the sign to each component, making sure the vector's overall direction stays accurate when transitioning from magnitude and angle to standard component forms.

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Most popular questions from this chapter

For each pair of vectors, find \(\mathbf{U} \cdot \mathbf{V}\). \(\mathbf{U}=5 \mathbf{i}+3 \mathbf{j} . \mathbf{V}=-5 \mathbf{i}+3 \mathbf{j}\)

Draw vectors representing the course of a ship that travels $$ 25 \text { miles on a course with heading } 225^{\circ} $$

For Problems 37 through 42, use your knowledge of bearing, heading, and true course to sketch a diagram that will help you solve each problem. Bearing and Distance Two ships leave the harbor at the same time. One ship is traveling at 15 miles per hour on a course with a bearing of \(\mathrm{S} \mathrm{} 13^{\circ} \mathrm{W}\), while the other is traveling at 11 miles per hour on a course with a bearing of \(\mathrm{N} 75^{\circ} \mathrm{E}\). How far apart are they after 3 hours?

Which pair of vectors are perpendicular? a. \(3 \mathbf{i}+4 \mathbf{j}\) and \(8 \mathbf{i}-6 \mathbf{j}\) b. \(3 \mathbf{i}+2 \mathbf{j}\) and \(2 \mathbf{i}+3 \mathbf{j}\) c. \(\mathbf{i}+5 \mathbf{j}\) and \(\mathbf{i}-5 \mathbf{j}\) d. \(2 \mathbf{i}-5 \mathbf{j}\) and \(-7 \mathbf{i}-3 \mathbf{j}\)

Show that each pair of vectors is perpendicular. In general, show that the vectors \(\mathbf{V}=a \mathbf{i}+b \mathbf{j}\) and \(\mathbf{W}=-b \mathbf{i}+a \mathbf{j}\) are always perpendicular. Assume \(a\) and \(b\) are not both equal to zero.
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