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Chapter 9: Problem 25

Find \(|\mathbf{u}|,|\mathbf{v}|,|2 \mathbf{u}|,\left|\frac{1}{2} \mathbf{v}\right|,|\mathbf{u}+\mathbf{v}|,|\mathbf{u}-\mathbf{v}|,\) and \(|\mathbf{u}|-|\mathbf{v}|\) $$ \mathbf{u}=\langle 10,-1\rangle, \quad \mathbf{v}=\langle- 2,-2\rangle $$

Short Answer

Expert verified
\(|\mathbf{u}| = \sqrt{101}, |\mathbf{v}| = 2\sqrt{2}, |2\mathbf{u}| = 2\sqrt{101}, |\frac{1}{2}\mathbf{v}| = \sqrt{2}, |\mathbf{u}+\mathbf{v}| = \sqrt{73}, |\mathbf{u}-\mathbf{v}| = \sqrt{145}, |\mathbf{u}| - |\mathbf{v}| = \sqrt{101} - 2\sqrt{2}.\)

Step by step solution

01

Find \(|\mathbf{u}|\)

Calculate the magnitude of \(\mathbf{u}\) by using the formula for the magnitude of a vector: \(|\mathbf{u}| = \sqrt{u_1^2 + u_2^2}\). For \(\mathbf{u} = \langle 10, -1 \rangle\), we have: \[ |\mathbf{u}| = \sqrt{10^2 + (-1)^2} = \sqrt{100 + 1} = \sqrt{101}. \]
02

Find \(|\mathbf{v}|\)

Determine the magnitude of \(\mathbf{v}\) using the same formula: \(|\mathbf{v}| = \sqrt{v_1^2 + v_2^2}\). For \(\mathbf{v} = \langle -2, -2 \rangle\), we calculate: \[ |\mathbf{v}| = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}. \]
03

Find \(|2\mathbf{u}|\)

Multiply \(\mathbf{u}\) by 2 to find \(2\mathbf{u}\) and then determine its magnitude. The magnitude of \(c\mathbf{u}\) for any scalar \(c\) is \(|c| \times |\mathbf{u}|\):\[ |2\mathbf{u}| = 2 \times |\mathbf{u}| = 2\sqrt{101}. \]
04

Find \(\left|\frac{1}{2}\mathbf{v}\right|\)

Divide \(\mathbf{v}\) by 2 to compute \(\frac{1}{2}\mathbf{v}\) and then find its magnitude. Here, we divide \(|\mathbf{v}|\) by 2:\[ \left|\frac{1}{2}\mathbf{v}\right| = \frac{1}{2} \times |\mathbf{v}| = \frac{1}{2} \times 2\sqrt{2} = \sqrt{2}. \]
05

Find \(|\mathbf{u} + \mathbf{v}|\)

First, add vectors \(\mathbf{u}\) and \(\mathbf{v}\): \(\mathbf{u} + \mathbf{v} = \langle 10, -1 \rangle + \langle -2, -2 \rangle = \langle 8, -3 \rangle\). Now, find the magnitude:\[ |\mathbf{u} + \mathbf{v}| = \sqrt{8^2 + (-3)^2} = \sqrt{64 + 9} = \sqrt{73}. \]
06

Find \(|\mathbf{u} - \mathbf{v}|\)

Subtract \(\mathbf{v}\) from \(\mathbf{u}\): \(\mathbf{u} - \mathbf{v} = \langle 10, -1 \rangle - \langle -2, -2 \rangle = \langle 12, 1 \rangle\). Then, calculate the magnitude:\[ |\mathbf{u} - \mathbf{v}| = \sqrt{12^2 + 1^2} = \sqrt{144 + 1} = \sqrt{145}. \]
07

Find \(|\mathbf{u}| - |\mathbf{v}|\)

Subtract the magnitudes of \(\mathbf{v}\) from \(\mathbf{u}\):\[ |\mathbf{u}| - |\mathbf{v}| = \sqrt{101} - 2\sqrt{2}. \]

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Key Concepts

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

Vector Operations
Working with vectors involves performing a variety of operations, which are crucial in many fields like physics, engineering, and computer science. These operations include addition, subtraction, and scalar multiplication.
Let's discuss vector addition and subtraction:
  • Addition: To add vectors, like \(\mathbf{u} = \langle 10, -1 \rangle\) and \(\mathbf{v} = \langle -2, -2 \rangle\), we add their corresponding components. That means adding the first component of \(\mathbf{u}\) to the first component of \(\mathbf{v}\), and the same for the second component. The resultant vector is \(\mathbf{u} + \mathbf{v} = \langle 8, -3 \rangle\).
  • Subtraction: Similarly, to subtract \(\mathbf{v}\) from \(\mathbf{u}\), subtract the corresponding components: \(\mathbf{u} - \mathbf{v} = \langle 10 - (-2), -1 - (-2) \rangle = \langle 12, 1 \rangle\).
These operations form the basis for more complex tasks involving vectors.
Magnitude Calculation
The magnitude of a vector, often referred to as its 'length' or 'norm,' gives a measure of how long the vector is. To find the magnitude, we use the formula:
  • For vector \(\mathbf{u} = \langle u_1, u_2 \rangle\), the magnitude is \(|\mathbf{u}| = \sqrt{u_1^2 + u_2^2}\).
Consider the vector \(\mathbf{u} = \langle 10, -1 \rangle\). Its magnitude is calculated as:\[|\mathbf{u}| = \sqrt{10^2 + (-1)^2} = \sqrt{100 + 1} = \sqrt{101}.\]For another vector \(\mathbf{v} = \langle -2, -2 \rangle\), the calculation is:\[|\mathbf{v}| = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}.\]Thus, the magnitude gives you a scalar indicating the size of the vector, irrespective of its direction.
Scalar Multiplication
Scalar multiplication involves taking a vector and multiplying it by a constant, a scalar. This is useful for scaling the vector's magnitude without changing its direction.
For instance, if you have the vector \(\mathbf{u}\) and you need to find \(2\mathbf{u}\):
  • Multiply each component of \(\mathbf{u}\) by 2. So, \(2\mathbf{u}\) becomes \(\langle 2 \times 10, 2 \times (-1) \rangle = \langle 20, -2 \rangle\).
  • The magnitude of \(2\mathbf{u}\) is \(2\times|\mathbf{u}|\), which equals \(2\sqrt{101}\).
Similarly, dividing \(\mathbf{v}\) by 2 results in \(\frac{1}{2}\mathbf{v}\) having each component halved:
  • The operation gives \(\langle -1, -1 \rangle\), and the magnitude is \(\frac{1}{2}\times|\mathbf{v}|\), or \(\sqrt{2}\).
Scalar multiplication stretches or shrinks the vector's length while keeping its direction the same.

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