Question

\(\mathbb{R}^{2}\) Consider the vectors \(\mathbf{I}=\langle 1 / \sqrt{2}, 1 / \sqrt{2}\rangle\) and \(\mathbf{J}=\langle-1 / \sqrt{2}, 1 / \sqrt{2}\rangle\). Show that \(\mathbf{I}\) and \(\mathbf{J}\) are orthogonal unit vectors.

Short answer

Answer: Yes, vectors \(\mathbf{I}\) and \(\mathbf{J}\) are orthogonal unit vectors since their dot product is zero and their magnitudes are both equal to 1.

Step-by-step solution

Calculate the dot product of \(\mathbf{I}\) and \(\mathbf{J}\)

To calculate the dot product of two-dimensional vectors, we simply multiply the corresponding components and add the products. The dot product of vectors \(\mathbf{I}\) and \(\mathbf{J}\) is: \(\mathbf{I} \cdot \mathbf{J} = \frac{1}{\sqrt{2}} \cdot \left(-\frac{1}{\sqrt{2}}\right) + \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}\)

Simplify the expression

By simplifying the expression above, we can find the value of the dot product of \(\mathbf{I}\) and \(\mathbf{J}\): \(\mathbf{I} \cdot \mathbf{J} = -\frac{1}{2} + \frac{1}{2} = 0\) Since the dot product is zero, vectors \(\mathbf{I}\) and \(\mathbf{J}\) are orthogonal.

Calculate the magnitude of each vector

To determine if a vector is a unit vector, we should calculate its magnitude. The magnitude of a two-dimensional vector is given by the equation: \(||\mathbf{V}|| = \sqrt{v_1^2 + v_2^2}\) Let's calculate the magnitudes of vectors \(\mathbf{I}\) and \(\mathbf{J}\). For \(\mathbf{I}\): \(||\mathbf{I}|| = \sqrt{\left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2}\) For \(\mathbf{J}\): \(||\mathbf{J}|| = \sqrt{\left(-\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2}\)

Simplify the expressions

By simplifying the expressions in Step 3, we can find the magnitudes of vectors \(\mathbf{I}\) and \(\mathbf{J}\): For \(\mathbf{I}\): \(||\mathbf{I}|| = \sqrt{\frac{1}{2} + \frac{1}{2}} = \sqrt{1} = 1\) For \(\mathbf{J}\): \(||\mathbf{J}|| = \sqrt{\frac{1}{2} + \frac{1}{2}} = \sqrt{1} = 1\) Since both vectors have a magnitude of 1, they are unit vectors. #Conclusion#Vectors \(\mathbf{I}\) and \(\mathbf{J}\) are orthogonal unit vectors, as we demonstrated that their dot product is zero and their magnitudes are both equal to 1.

Key concepts

Unit Vectors

A unit vector is a vector that has a magnitude of 1. These vectors are important in mathematics and physics because they provide a sense of direction without scaling. To verify if a vector is a unit vector, simply calculate its magnitude and check if it equals 1.
In our exercise, we are given the vectors \( \mathbf{I}=\langle \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\rangle \) and \( \mathbf{J}=\langle-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\rangle \).
The magnitudes of these vectors were calculated as follows:

• For \( \mathbf{I} \), \( ||\mathbf{I}|| = \sqrt{\left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2} = 1 \)
• For \( \mathbf{J} \), \( ||\mathbf{J}|| = \sqrt{\left(-\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2} = 1 \)

Since both magnitudes are 1, \( \mathbf{I} \) and \( \mathbf{J} \) are indeed unit vectors. They maintain their direction but have no effect on the scale of operations, making them ideal for defining directions.

Dot Product

The dot product is a fundamental operation in vector algebra. It tells us how much of one vector goes in the direction of another, effectively quantifying the overlap between the two vectors. Notably, vectors are orthogonal if their dot product is zero, meaning they are at right angles to each other.
To find the dot product of two vectors \( \mathbf{A} = \langle a_1, a_2 \rangle \) and \( \mathbf{B} = \langle b_1, b_2 \rangle \), we multiply corresponding components and add the results:
\[ \mathbf{A} \cdot \mathbf{B} = a_1 b_1 + a_2 b_2 \]
In our exercise, for vectors \( \mathbf{I} \) and \( \mathbf{J} \), the calculation was:

• \( \frac{1}{\sqrt{2}} \times \left(-\frac{1}{\sqrt{2}}\right) + \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = -\frac{1}{2} + \frac{1}{2} = 0 \)

Since the dot product is zero, these vectors are orthogonal, which matches the definition perfectly.

Magnitude of Vectors

The magnitude of a vector, often referred to as its "length," is a measure of how long the vector is. As such, it's a positive scalar value. Knowing the magnitude is essential when working with vectors as it helps in understanding their scale or size.
You can calculate the magnitude of a two-dimensional vector \( \mathbf{V} = \langle v_1, v_2 \rangle \) with the formula:
\[ ||\mathbf{V}|| = \sqrt{v_1^2 + v_2^2} \]
In the exercise, for vectors \( \mathbf{I} \) and \( \mathbf{J} \), the magnitude calculations showed:

• For \( \mathbf{I} \): \( ||\mathbf{I}|| = \sqrt{\left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2} = 1 \)
• For \( \mathbf{J} \): \( ||\mathbf{J}|| = \sqrt{\left(-\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2} = 1 \)

Both magnitudes are 1, confirming that \( \mathbf{I} \) and \( \mathbf{J} \) are unit vectors, but also demonstrating how the magnitude ensures vectors retain their true scale.