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Chapter 7: Problem 1

Find \(a+b, a-b, 4 a+5 b, 4 a-5 b,\) and \(\|a\|\) $$\mathbf{a}=\langle 2,-3\rangle, \quad \mathbf{b}=\langle 1,4\rangle$$

Short Answer

Expert verified
1) \(a+b = \langle 3, 1 \rangle\); 2) \(a-b = \langle 1, -7 \rangle\); 3) \(4a+5b = \langle 13, 8 \rangle\); 4) \(4a-5b = \langle 3, -32 \rangle\); 5) \(\|a\| = \sqrt{13}\).

Step by step solution

01

Calculate a+b

To find \(a+b\), add the corresponding components of the vectors \(\mathbf{a}\) and \(\mathbf{b}\). \(\mathbf{a} = \langle 2, -3 \rangle\) and \(\mathbf{b} = \langle 1, 4 \rangle\) so, \[ a+b = \langle 2+1, -3+4 \rangle = \langle 3, 1 \rangle.\]
02

Calculate a-b

To find \(a-b\), subtract the components of \(\mathbf{b}\) from \(\mathbf{a}\).\[ a-b = \langle 2-1, -3-4 \rangle = \langle 1, -7 \rangle.\]
03

Calculate 4a+5b

First, multiply each component of \(\mathbf{a}\) by 4 and each component of \(\mathbf{b}\) by 5, then add the results.\[ 4\mathbf{a} = \langle 8, -12 \rangle, \ 5\mathbf{b} = \langle 5, 20 \rangle\]\[ 4a+5b = \langle 8+5, -12+20 \rangle = \langle 13, 8 \rangle.\]
04

Calculate 4a-5b

Multiply each component of \(\mathbf{a}\) by 4 and each component of \(\mathbf{b}\) by 5, then subtract the second from the first.\[ 4\mathbf{a} = \langle 8, -12 \rangle, \ 5\mathbf{b} = \langle 5, 20 \rangle\]\[ 4a-5b = \langle 8-5, -12-20 \rangle = \langle 3, -32 \rangle.\]
05

Calculate \(\|a\|\)

The magnitude \(\|\mathbf{a}\|\) is calculated using the formula \(\sqrt{x^2 + y^2}\).For \(\mathbf{a} = \langle 2, -3 \rangle\): \[ \|a\| = \sqrt{2^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13}.\]

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Key Concepts

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

Vector Addition
Vector addition involves combining two vectors to create a new vector. When you add vectors, simply add their corresponding components. For example, consider vectors \( \mathbf{a} = \langle 2, -3 \rangle \) and \( \mathbf{b} = \langle 1, 4 \rangle \).
To compute \( \mathbf{a} + \mathbf{b} \), do the following:
  • Add the first components: \(2 + 1 = 3\)
  • Add the second components: \(-3 + 4 = 1\)
The result is a new vector: \( \langle 3, 1 \rangle \).
This method can be visualized as two arrows joined head-to-tail to form a diagonal of a parallelogram.
Vector Subtraction
Vector subtraction is the process of finding the difference between two vectors. This is done by subtracting the corresponding components. For instance, if you have vectors \( \mathbf{a} = \langle 2, -3 \rangle \) and \( \mathbf{b} = \langle 1, 4 \rangle \), then the subtraction \( \mathbf{a} - \mathbf{b} \) is calculated by:
  • Subtract the first components: \(2 - 1 = 1\)
  • Subtract the second components: \(-3 - 4 = -7\)
Thus, \( \mathbf{a} - \mathbf{b} = \langle 1, -7 \rangle \).
This can also be seen as reversing the direction of \( \mathbf{b} \) and then performing vector addition.
Scalar Multiplication
In scalar multiplication, a vector is multiplied by a scalar (a real number), scaling the size of the vector while retaining its direction. Given vector \( \mathbf{a} = \langle 2, -3 \rangle \), multiplying by scalar 4 involves:
  • Multiply the first component: \(4 \cdot 2 = 8\)
  • Multiply the second component: \(4 \cdot (-3) = -12\)
The resulting scaled vector is \( \langle 8, -12 \rangle \).
Similarly, scaling \( \mathbf{b} = \langle 1, 4 \rangle \) by 5 gives \( \langle 5, 20 \rangle \).
After scalar multiplication, you can proceed with vector addition or subtraction as discussed earlier.
Magnitude of a Vector
The magnitude of a vector, also known as its length, is found using the Pythagorean theorem. For vector \( \mathbf{a} = \langle 2, -3 \rangle \), the magnitude \( \|\mathbf{a}\| \) is calculated by:
  • Square each component: \(2^2 = 4\), \((-3)^2 = 9\)
  • Add the squares: \(4 + 9 = 13\)
  • Take the square root: \(\sqrt{13}\)
The magnitude \( \|\mathbf{a}\| \) represents the distance from the origin to the point \( \langle 2, -3 \rangle \) in a coordinate system. This measure is always non-negative and provides a sense of the vector's size or intensity.

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