Question

\(\mathbb{R}^{3}\) Consider the vectors \(\mathbf{I}=\langle 1 / 2,1 / 2,1 / \sqrt{2}), \mathbf{J}=\langle-1 / \sqrt{2}, 1 / \sqrt{2}, 0\rangle,\) and \(\mathbf{K}=\langle 1 / 2,1 / 2,-1 / \sqrt{2}\rangle\) a. Sketch I, J, and K and show that they are unit vectors. b. Show that \(\mathbf{I}, \mathbf{J},\) and \(\mathbf{K}\) are pairwise orthogonal. c. Express the vector \langle 1,0,0\rangle in terms of \(\mathbf{I}, \mathbf{J},\) and \(\mathbf{K}\).

Short answer

Answer: The linear combination of I, J, and K that expresses the vector ⟨1,0,0⟩ is (2/3)I - (1/√2)J + (1/3)K.

Step-by-step solution

Sketching the vectors and showing they are unit vectors

To sketch the vectors \(\mathbf{I}, \mathbf{J},\) and \(\mathbf{K}\), we can represent them as directed line segments in a coordinate system in \(\mathbb{R}^{3}\). To show that they are unit vectors, we must verify that their magnitudes are equal to 1. - Magnitude of I: \(|\mathbf{I}| = \sqrt{(\frac{1}{2})^{2}+(\frac{1}{2})^{2}+(\frac{1}{\sqrt{2}})^{2}} = \sqrt{\frac{1}{4}+\frac{1}{4}+1} = 1\) - Magnitude of J: \(|\mathbf{J}| = \sqrt{(-\frac{1}{\sqrt{2}})^{2}+(\frac{1}{\sqrt{2}})^{2}+0^{2}} = \sqrt{1} = 1\) - Magnitude of K: \(|\mathbf{K}| = \sqrt{(\frac{1}{2})^{2}+(\frac{1}{2})^{2}+(-\frac{1}{\sqrt{2}})^{2}} = \sqrt{\frac{1}{4}+\frac{1}{4}+1} = 1\) As all three vectors have a magnitude of 1, they are unit vectors.

Showing that they are pairwise orthogonal

To show that the vectors \(\mathbf{I}, \mathbf{J},\) and \(\mathbf{K}\) are pairwise orthogonal, we need to calculate the dot product for each pair of vectors and verify that it is equal to 0. - Dot product of I and J: \(\mathbf{I}\cdot\mathbf{J} = \frac{1}{2}(-\frac{1}{\sqrt{2}})+\frac{1}{2}\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\cdot 0 = 0\) - Dot product of I and K: \(\mathbf{I}\cdot\mathbf{K} = \frac{1}{2}\frac{1}{2}+\frac{1}{2}\frac{1}{2}+\frac{1}{\sqrt{2}}(-\frac{1}{\sqrt{2}}) = 0\) - Dot product of J and K: \(\mathbf{J}\cdot\mathbf{K} = -\frac{1}{\sqrt{2}}\frac{1}{2}+\frac{1}{\sqrt{2}}\frac{1}{2}+0\cdot(-\frac{1}{\sqrt{2}}) = 0\) As the dot product for each pair of vectors is equal to 0, they are pairwise orthogonal.

Expressing the vector \langle 1,0,0 \rangle in terms of I, J, and K

We want to express the vector \langle 1,0,0 \rangle as a linear combination of the vectors \(\mathbf{I}, \mathbf{J},\) and \(\mathbf{K}\). Let this combination be represented as: $$\alpha\mathbf{I} + \beta\mathbf{J} + \gamma\mathbf{K} = \langle 1, 0, 0 \rangle$$ Expanding and equating the components gives us a system of linear equations: $$ \begin{cases} \frac{1}{2}\alpha-\frac{1}{\sqrt{2}}\beta+\frac{1}{2}\gamma=1 \\ \frac{1}{2}\alpha+\frac{1}{\sqrt{2}}\beta+\frac{1}{2}\gamma=0 \\ \frac{1}{\sqrt{2}}\alpha+0\beta-\frac{1}{\sqrt{2}}\gamma=0 \end{cases} $$ Solving this system of linear equations, we get: $$\alpha=\frac{2}{3}, \quad \beta=-\frac{1}{\sqrt{2}}, \quad \gamma=\frac{1}{3}$$ So, \(\langle 1,0,0 \rangle\) can be expressed in terms of I, J, and K as: $$\langle 1,0,0 \rangle = \frac{2}{3}\mathbf{I}-\frac{1}{\sqrt{2}}\mathbf{J}+\frac{1}{3}\mathbf{K}$$

Key concepts

Understanding Unit Vectors

A unit vector is a vector with a magnitude (or length) of exactly one unit. It is often used to indicate direction without considering magnitude. In \(\mathbb{R}^{3}\), unit vectors help define orientations along specific axes.
To verify whether a given vector is a unit vector, compute its magnitude using the formula:
\[|\mathbf{v}| = \sqrt{x^2 + y^2 + z^2}\]
where \(x, y,\) and \(z\) are the components of the vector. If the result is 1, the vector is a unit vector.
In our example, the vectors \(\mathbf{I}, \mathbf{J},\) and \(\mathbf{K}\) were shown to have magnitudes of 1, confirming they are indeed unit vectors. Understanding this concept is important for operations like normalizing vectors in various applications, such as physics and computer graphics.

Exploring Orthogonal Vectors

Orthogonal vectors are vectors that meet at right angles. In terms of vector algebra, two vectors are orthogonal if their dot product equals zero. The dot product \(\mathbf{a}\cdot\mathbf{b}\) is computed as:

• Multiply corresponding components of the vectors.
• Sum these products.
  \[\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 = 0\]
  

This zero result indicates perpendicularity in three-dimensional space.
In this exercise, dot products for every pair of the vectors \(\mathbf{I}, \mathbf{J},\) and \(\mathbf{K}\) were calculated to be zero, verifying their orthogonality. Understanding orthogonal vectors is crucial because they are independent of each other and form the basis for vector spaces in areas like machine learning and signal processing.

Linear Combinations of Vectors

Linear combinations involve expressing a vector as a sum of scalar multiples of other vectors. For a vector to be expressed as a linear combination, you need coefficients that are scalars multiplying each vector in the system. This expression follows the form:
\[\mathbf{v} = \alpha\mathbf{u} + \beta\mathbf{v} + \gamma\mathbf{w}\]
where \(\alpha, \beta, \gamma\) are scalars, and \(\mathbf{u}, \mathbf{v}, \mathbf{w}\) are vectors.
In the given exercise, the vector \(\langle 1, 0, 0 \rangle\) was written as a linear combination of \(\mathbf{I}, \mathbf{J},\) and \(\mathbf{K}\). Solving the system of equations allows you to find coefficients that combine these vectors to form the target vector.
This understanding is fundamental in various fields, such as computer graphics and numerical methods, where transformations and alterations of data are frequently required.