Chapter 1: Problem 3
Find all unit vectors in \(\mathbb{R}^{2}\) that are orthogonal to \(\mathbf{x}=(1,2)\).
Short Answer
Expert verified
The unit vectors orthogonal to \((1,2)\) are \(\left(-\frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}}\right)\) and \(\left(\frac{2}{\sqrt{5}}, -\frac{1}{\sqrt{5}}\right)\).
Step by step solution
01
Understanding Orthogonal Vectors
Two vectors are orthogonal if their dot product is zero. This means that for two vectors \(\mathbf{a} = (a_1, a_2)\) and \(\mathbf{b} = (b_1, b_2)\), they are orthogonal if \(a_1b_1 + a_2b_2 = 0\). We are looking for unit vectors in \(\mathbb{R}^2\) that are orthogonal to \(\mathbf{x} = (1, 2)\).
02
Set Up the Orthogonality Condition
Let \(\mathbf{y} = (y_1, y_2)\) be a vector that is orthogonal to \(\mathbf{x}\). The condition for orthogonality is:\[1 \cdot y_1 + 2 \cdot y_2 = 0\]which simplifies to \(y_1 + 2y_2 = 0\).
03
Solve the Equation for One Parameter
From the equation \(y_1 + 2y_2 = 0\), we can express \(y_1\) in terms of \(y_2\): \(y_1 = -2y_2\). This means any vector orthogonal to \(\mathbf{x}\) can be written in the form \((-2t, t)\) for some parameter \(t\).
04
Add the Unit Vector Condition
A unit vector \(\mathbf{y}\) has a magnitude (or length) of 1, i.e., \(\sqrt{y_1^2 + y_2^2} = 1\). Substituting \(y_1 = -2y_2\), we have:\[\sqrt{(-2t)^2 + t^2} = 1\]This simplifies to:\[\sqrt{4t^2 + t^2} = \sqrt{5t^2} = 1\]
05
Solve for Parameter t
We solve \(\sqrt{5t^2} = 1\) by first squaring both sides to get \(5t^2 = 1\), leading to \(t^2 = \frac{1}{5}\). Hence \(t = \pm \frac{1}{\sqrt{5}}\).
06
Find the Unit Vectors
Substituting back \(t = \frac{1}{\sqrt{5}}\) and \(t = -\frac{1}{\sqrt{5}}\) into \((-2t, t)\), we get the unit vectors:\[\mathbf{y}_1 = \left(-2\left(\frac{1}{\sqrt{5}}\right), \frac{1}{\sqrt{5}}\right) = \left(-\frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}}\right)\] and \[\mathbf{y}_2 = \left(-2\left(-\frac{1}{\sqrt{5}}\right), -\frac{1}{\sqrt{5}}\right) = \left(\frac{2}{\sqrt{5}}, -\frac{1}{\sqrt{5}}\right)\].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Unit Vectors
Unit vectors are fundamental in understanding the direction of vectors. A unit vector has a magnitude, or length, of exactly one, making it useful for indicating directions without scaling the vector's magnitude. Any nonzero vector \( \mathbf{v} \) can be converted into a unit vector by dividing it by its magnitude. For example, if \( \mathbf{v} = (v_1, v_2) \), the unit vector \( \mathbf{u} \) in the same direction is given by:\[ \mathbf{u} = \frac{\mathbf{v}}{\|\mathbf{v}\|} \]where \( \|\mathbf{v}\| \) is the magnitude of \( \mathbf{v} \). Unit vectors are often used in normalizing vectors, ensuring they merely represent direction when needed so you can focus on angles and orientations, such as when working with orthogonal vectors.
- They aid in computation by simplifying equations, as the unit vector itself signifies mere direction.
- In coordinate systems, standard unit vectors \( \mathbf{i} \) and \( \mathbf{j} \) in 2D or \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) in 3D, are used as basis vectors with each having a magnitude of 1.
Dot Product
The dot product, also known as the scalar product, is a powerful tool in vector mathematics, especially when determining orthogonality. The dot product of two vectors \( \mathbf{a} = (a_1, a_2) \) and \( \mathbf{b} = (b_1, b_2) \) is given by:\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \]This operation results in a single scalar value rather than a vector. A key property of the dot product is that it equals zero when the vectors are orthogonal, meaning they are perpendicular to each other.
- The dot product helps in projecting one vector onto another, significant in physics and engineering.
- It's also used to find the angle \( \theta \) between two vectors by the formula:\[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\|\|\mathbf{b}\|} \]
- Orthogonal vectors have \( \theta = 90^\circ \), which validates their dot product being zero.
Magnitude of Vectors
Understanding the magnitude of vectors is crucial as it represents the length or size of the vector. For a vector \( \mathbf{v} = (v_1, v_2) \), its magnitude is computed using the formula:\[ \|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2} \]This formula is derived from the Pythagorean theorem, and it gives a sense of the vector's "reach" in space.
- The magnitude is always a non-negative value, showing the extent of displacement or force represented by the vector.
- In physical contexts, magnitudes can represent measurable quantities like distance, speed, or force.
- Unit vectors have a magnitude of 1, which is deeply relevant for specifying vector direction without influence from its length.