/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 Find (a) \(3 \mathbf{a}\), (b) \... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 2

Find (a) \(3 \mathbf{a}\), (b) \(\mathbf{a}+\mathbf{b},(\mathbf{c}) \mathbf{a}-\mathbf{b},(\mathbf{d})\|\mathbf{a}+\mathbf{b}\|\), and \((e)\|\mathbf{a}-\mathbf{b}\|\) . \(\mathbf{a}=\langle 1,1\rangle, \mathbf{b}=\langle 2,3\rangle\)

Short Answer

Expert verified
(a) \(\langle 3, 3 \rangle\), (b) \(\langle 3, 4 \rangle\), (c) \(\langle -1, -2 \rangle\), (d) 5, (e) \(\sqrt{5}\).

Step by step solution

01

Calculate 3a

Multiply each component of \( \mathbf{a} \) by 3. We have \( \mathbf{a} = \langle 1, 1 \rangle \), so \( 3 \mathbf{a} = 3 \times \langle 1, 1 \rangle = \langle 3 \times 1, 3 \times 1 \rangle = \langle 3, 3 \rangle \).
02

Calculate a + b

Add each corresponding component of \( \mathbf{a} \) and \( \mathbf{b} \). We have \( \mathbf{a} = \langle 1, 1 \rangle \) and \( \mathbf{b} = \langle 2, 3 \rangle \), so \( \mathbf{a} + \mathbf{b} = \langle 1 + 2, 1 + 3 \rangle = \langle 3, 4 \rangle \).
03

Calculate a - b

Subtract each component of \( \mathbf{b} \) from the corresponding component of \( \mathbf{a} \). So \( \mathbf{a} - \mathbf{b} = \langle 1 - 2, 1 - 3 \rangle = \langle -1, -2 \rangle \).
04

Calculate ||a + b||

Find the magnitude of \( \mathbf{a} + \mathbf{b} \) from Step 2. The formula for the magnitude is \( ||\mathbf{v}|| = \sqrt{x^2 + y^2} \). Thus, \( ||\mathbf{a} + \mathbf{b}|| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \).
05

Calculate ||a - b||

Find the magnitude of \( \mathbf{a} - \mathbf{b} \) from Step 3. Using the magnitude formula, \( ||\mathbf{a} - \mathbf{b}|| = \sqrt{(-1)^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5} \).

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Key Concepts

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

vector addition
Vector addition is a basic operation in vector mathematics that involves combining two vectors to create a new vector. Vector addition is performed by adding the corresponding components of the two vectors. If you have two vectors, \( \mathbf{a} = \langle x_1, y_1 \rangle \) and \( \mathbf{b} = \langle x_2, y_2 \rangle \), their sum is given by:
  • \( \mathbf{a} + \mathbf{b} = \langle x_1 + x_2, y_1 + y_2 \rangle \)
In the original exercise, the addition of vectors \( \mathbf{a} = \langle 1, 1 \rangle \) and \( \mathbf{b} = \langle 2, 3 \rangle \) results in:
  • \( \mathbf{a} + \mathbf{b} = \langle 1 + 2, 1 + 3 \rangle = \langle 3, 4 \rangle \)
This operation is visually represented as moving from the tip of the first vector to the tip of the second, forming a triangle with the vectors.
vector subtraction
Vector subtraction is closely related to vector addition and can be understood as the addition of a negative vector. When subtracting vectors, you are essentially adding the opposite of the vector being subtracted. To subtract vector \( \mathbf{b} \) from \( \mathbf{a} \), we compute:
  • \( \mathbf{a} - \mathbf{b} = \langle x_1 - x_2, y_1 - y_2 \rangle \)
From the given example, with vectors \( \mathbf{a} = \langle 1, 1 \rangle \) and \( \mathbf{b} = \langle 2, 3 \rangle \), the subtraction results in:
  • \( \mathbf{a} - \mathbf{b} = \langle 1 - 2, 1 - 3 \rangle = \langle -1, -2 \rangle \)
Think of this as reversing the direction of vector \( \mathbf{b} \) and then performing vector addition with \( \mathbf{a} \). This results in moving backwards along vector \( \mathbf{b} \).
scalar multiplication
Scalar multiplication involves multiplying each component of a vector by a scalar (a real number). This operation scales the vector, either stretching or shrinking it, without changing its direction.
Consider a vector \( \mathbf{a} = \langle x, y \rangle \). If you multiply \( \mathbf{a} \) by a scalar \( c \), you get:
  • \( c \mathbf{a} = \langle c \times x, c \times y \rangle \)
In the problem, the original vector \( \mathbf{a} = \langle 1, 1 \rangle \) is multiplied by 3:
  • \( 3 \mathbf{a} = \langle 3 \times 1, 3 \times 1 \rangle = \langle 3, 3 \rangle \)
This means that the vector is stretched to three times its original length, pointing in the same direction as \( \mathbf{a} \).
vector magnitude
The magnitude of a vector is a measure of its length. It is found using the Pythagorean theorem in the context of a vector's components. For a vector \( \mathbf{v} = \langle x, y \rangle \), its magnitude \( ||\mathbf{v}|| \) is calculated by:
  • \( ||\mathbf{v}|| = \sqrt{x^2 + y^2} \)
In the original exercise, the magnitude of \( \mathbf{a} + \mathbf{b} = \langle 3, 4 \rangle \) is:
  • \( ||\mathbf{a} + \mathbf{b}|| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \)
Similarly, the magnitude of \( \mathbf{a} - \mathbf{b} = \langle -1, -2 \rangle \) is:
  • \( ||\mathbf{a} - \mathbf{b}|| = \sqrt{(-1)^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5} \)
These calculations give the physical length of the resultant vectors, a crucial step in understanding the scale of vector quantities.

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