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Chapter 1: Problem 79

For the following exercises, use the given vectors to find the quantities.\(\mathbf{a}=9 \mathbf{i}-2 \mathbf{j}, \mathbf{b}=-3 \mathbf{i}+\mathbf{j}\) \(\quad\) a. \(3 \mathbf{a}+\mathbf{b}\) b. \(|\mathbf{a}|\) \(\quad\) c. \(\mathbf{a} \times|\mathbf{b} \times| \mathbf{a}\) \(\quad\) d. \(\mathbf{b} \times \mid \mathbf{a}\)

Short Answer

Expert verified
a) \(24\mathbf{i} - 5\mathbf{j}\). b) \(\sqrt{85}\). c) \(\mathbf{a} \times \sqrt{850}\). d) \(-3\sqrt{85}\mathbf{i} + \sqrt{85}\mathbf{j}\).

Step by step solution

01

Find the expression for 3a

To find \( 3\mathbf{a} \), multiply each component of vector \( \mathbf{a} = 9\mathbf{i} - 2\mathbf{j} \) by 3. This gives: \( 3\mathbf{a} = 3(9\mathbf{i} - 2\mathbf{j}) = 27\mathbf{i} - 6\mathbf{j} \).
02

Add vector b to 3a

Now add vector \( \mathbf{b} = -3\mathbf{i} + \mathbf{j} \) to \( 3\mathbf{a} \) obtained in Step 1: \( 3\mathbf{a} + \mathbf{b} = (27\mathbf{i} - 6\mathbf{j}) + (-3\mathbf{i} + \mathbf{j}) = (27 + (-3))\mathbf{i} + (-6 + 1)\mathbf{j} = 24\mathbf{i} - 5\mathbf{j} \).
03

Calculate the magnitude of vector a

The magnitude of a vector \( \mathbf{a} = 9\mathbf{i} - 2\mathbf{j} \) is given by \( |\mathbf{a}| = \sqrt{9^2 + (-2)^2} = \sqrt{81 + 4} = \sqrt{85} \).
04

Understand the expression c

Expression c is given as \( \mathbf{a} \times |\mathbf{b} \times | \mathbf{a} \). First, calculate \( |\mathbf{b}| \), then substitute into \( 2 |\mathbf{a}| \).
05

Calculate the magnitude of vector b and c

Magnitude of \( \mathbf{b} \): \( |\mathbf{b}| = \sqrt{(-3)^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10} \). Calculate, \(|\mathbf{b}| \times |\mathbf{a}| = \sqrt{10} \times \sqrt{85} = \sqrt{850}\).
06

Calculate vector a times the magnitude result

Expression c becomes \( \mathbf{a} \times \sqrt{850} \), which results in a product scaling: \( (9\mathbf{i} - 2\mathbf{j}) \times \sqrt{850} \). When traversing 0 plane no cross product exists.
07

Calculate vector b time the magnitude result

For expression d: \( \mathbf{b} \times |\mathbf{a}| = (-3\mathbf{i} + \mathbf{j}) \times \sqrt{85} \) results in \( -3\sqrt{85}\mathbf{i} + \sqrt{85}\mathbf{j} \).
08

Put together the results from above expressions

From each calcualtion, we obtain \((24\mathbf{i} - 5\mathbf{j}), |\mathbf{a}| = \sqrt{85}, \mathbf{a} \times \sqrt{850}, \), and \( -3\sqrt{85}\mathbf{i} + \sqrt{85}\mathbf{j}\)
09

Short Answer

Short Answer: a) \(24\mathbf{i} - 5\mathbf{j}\). b) \(\sqrt{85}\). c) \(\mathbf{a} eq \sqrt{850} \). d) No cross product under general assumptions.

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Key Concepts

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

Vector Addition
Adding vectors is similar to adding numbers, though vectors have a direction as well as a magnitude. When we perform **vector addition**, we combine corresponding components of the vectors. This means you add the components in the same direction together.
  • Imagine two vectors, \( \mathbf{a} = 9 \mathbf{i} - 2 \mathbf{j} \) and \( \mathbf{b} = -3 \mathbf{i} + \mathbf{j} \).
  • If you multiply \( \mathbf{a} \) by 3, you get \( 3\mathbf{a} = 27 \mathbf{i} - 6 \mathbf{j} \).
  • To add vector \( \mathbf{b} \) to \( 3\mathbf{a} \), you combine their corresponding components: \((27 - 3) \mathbf{i} + (-6 + 1) \mathbf{j} = 24 \mathbf{i} - 5 \mathbf{j} \).
This result represents a new vector with a different direction and magnitude. It's like walking 24 steps east and then 5 steps south.
Vector Magnitude
The **magnitude of a vector** tells us how long the vector is, which is a key part in understanding its overall effect when combined with other vectors. The magnitude uses the Pythagorean theorem since it's like finding the hypotenuse of a right triangle.
  • For \( \mathbf{a} = 9\mathbf{i} - 2\mathbf{j} \), its magnitude is given by finding the square root of the sum of its components squared: \(|\mathbf{a}| = \sqrt{9^2 + (-2)^2} = \sqrt{81 + 4} = \sqrt{85} \).
This operation allows you to quantify how large the vector is regardless of its direction. In practical terms, it's like measuring the distance of a straight path an object might travel.
Cross Product
The **cross product** differs from addition as it results in a vector that is perpendicular to both original vectors. While the basics are easily understood, the benefits and applications are profound, especially in physics and engineering.
  • Though not applicable in a 2D plane crossing the vector itself or another identical vector, the result helps determine orientation in a 3D system.
  • Consider \( \mathbf{b} \times |\mathbf{a}| = (-3\mathbf{i} + \mathbf{j}) \times \sqrt{85} \), creating a vector: \(-3\sqrt{85}\mathbf{i} + \sqrt{85}\mathbf{j} \).
The cross product is pivotal in finding torque or rotational vectors, which helps explain how forces affect rotating bodies. In simpler terms, it's like determining the spin of a swung baseball bat, where direction and force define the spin.

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Most popular questions from this chapter

For the following exercises, find the traces for the surfaces in planes \(x=k, y=k\), and \(z=k .\) Then, describe and draw the surfaces\(x^{2}=y^{2}+z^{2}\)

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