Chapter 1: Problem 7
Compute \(\|\mathbf{u}\|,\|\mathbf{v}\|,\) and \(\mathbf{u} \cdot \mathbf{v}\) for the given vectors in \(\mathbb{R}^{3}\). $$\mathbf{u}=2 \mathbf{j}-\mathbf{i}, \mathbf{v}=-\mathbf{j}+\mathbf{i}$$
Short Answer
Expert verified
\( \|\mathbf{u}\| = \sqrt{5}, \|\mathbf{v}\| = \sqrt{2}, \mathbf{u} \cdot \mathbf{v} = -3 \).
Step by step solution
01
Understand the Vectors
Identify the components of the vectors \( \mathbf{u} \) and \( \mathbf{v} \). The vector \( \mathbf{u} \) is given as \( -\mathbf{i} + 2\mathbf{j} \), which corresponds to the components \((-1, 2, 0)\) in \(\mathbb{R}^3\). Similarly, \( \mathbf{v} \) is \( \mathbf{i} - \mathbf{j} \) corresponding to \((1, -1, 0)\).
02
Compute the Magnitude of \( \mathbf{u} \)
The magnitude (or norm) of a vector \( \mathbf{a} = (a_1, a_2, a_3) \) is given by the formula: \( \|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2 + a_3^2} \). For \( \mathbf{u} = (-1, 2, 0) \), calculate: \[ \|\mathbf{u}\| = \sqrt{(-1)^2 + 2^2 + 0^2} = \sqrt{1 + 4 + 0} = \sqrt{5}. \]
03
Compute the Magnitude of \( \mathbf{v} \)
Using the same formula for the magnitude, apply it to \( \mathbf{v} = (1, -1, 0) \): \[ \|\mathbf{v}\| = \sqrt{1^2 + (-1)^2 + 0^2} = \sqrt{1 + 1 + 0} = \sqrt{2}. \]
04
Compute the Dot Product \( \mathbf{u} \cdot \mathbf{v} \)
The dot product of two vectors \( \mathbf{a} = (a_1, a_2, a_3) \) and \( \mathbf{b} = (b_1, b_2, b_3) \) is given by \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \). With \( \mathbf{u} = (-1, 2, 0) \) and \( \mathbf{v} = (1, -1, 0) \), compute: \[ \mathbf{u} \cdot \mathbf{v} = (-1)(1) + (2)(-1) + (0)(0) = -1 - 2 + 0 = -3. \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
vector magnitude
The vector magnitude, also known as the norm of a vector, is a measure of its length. In mathematics, if you have a vector \(\mathbf{a} = (a_1, a_2, a_3)\) in three-dimensional space \(\mathbb{R}^3\), its magnitude is calculated using the formula:\[\|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2 + a_3^2}.\]This formula essentially extends the Pythagorean theorem into three dimensions, where each component \(a_1\), \(a_2\), and \(a_3\) is squared, summed, and then the square root of the total is taken.
For example, consider the vector \(\mathbf{u} = (-1, 2, 0)\). To find its magnitude, apply the formula:
For example, consider the vector \(\mathbf{u} = (-1, 2, 0)\). To find its magnitude, apply the formula:
- Square each component: \((-1)^2 = 1\), \(2^2 = 4\), \(0^2 = 0\).
- Sum the squares: \(1 + 4 + 0 = 5\).
- Take the square root: \(\sqrt{5}\).
dot product
The dot product is a way to multiply two vectors, resulting in a scalar. This scalar describes how much of one vector lies in the direction of another. For vectors \(\mathbf{a} = (a_1, a_2, a_3)\) and \(\mathbf{b} = (b_1, b_2, b_3)\) in \(\mathbb{R}^3\), the formula is:\[\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3.\]The dot product involves multiplying corresponding components of the vectors and then summing these products.
Take the vectors \(\mathbf{u} = (-1, 2, 0)\) and \(\mathbf{v} = (1, -1, 0)\) as examples:
Take the vectors \(\mathbf{u} = (-1, 2, 0)\) and \(\mathbf{v} = (1, -1, 0)\) as examples:
- Multiply the first components: \((-1) \cdot 1 = -1\).
- Multiply the second components: \(2 \cdot (-1) = -2\).
- Multiply the third components: \(0 \cdot 0 = 0\).
- Sum the results: \(-1 - 2 + 0 = -3\).
vectors in R^3
Vectors in \(\mathbb{R}^3\) are fundamental in understanding three-dimensional space. Each vector in \(\mathbb{R}^3\) is identified by three components, often associated with the \(x\), \(y\), and \(z\) axes. These components are typically denoted as \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\) for unit vectors along these axes, respectively.For instance, look at the vectors \(\mathbf{u} = 2 \mathbf{j} - \mathbf{i}\) and \(\mathbf{v} = -\mathbf{j} + \mathbf{i}\).
When expressed as coordinates, they become \((-1, 2, 0)\) and \((1, -1, 0)\). The first vector
When expressed as coordinates, they become \((-1, 2, 0)\) and \((1, -1, 0)\). The first vector
- has no \(z\)-component (0) which means it lies in the \(xy\)-plane.
- has a negative \(x\)-component and a positive \(y\)-component, indicating its direction is towards both negative \(x\) and positive \(y\).
- also lies in the \(xy\)-plane.
- has a positive \(x\)-component and a negative \(y\)-component, indicating its direction is towards positive \(x\) and negative \(y\).