Question

Orthogonal unit vectors in \(\mathbb{R}^{2}\) 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 I and J are orthogonal unit vectors.

Short answer

Answer: Yes, the given vectors I and J are orthogonal unit vectors.

Step-by-step solution

Calculate the dot product

Compute the dot product of the given vectors \(\mathbf{I}\) and \(\mathbf{J}\) to verify if they are orthogonal. Recall that the dot product of two vectors \(\mathbf{A}=\langle a_1, a_2\rangle\) and \(\mathbf{B}=\langle b_1, b_2\rangle\) is given by: $$\mathbf{A} \cdot \mathbf{B} = a_1b_1 + a_2b_2$$ Calculate the dot product of \(\mathbf{I}\) and \(\mathbf{J}\): $$\mathbf{I} \cdot \mathbf{J} = \frac{1}{\sqrt{2}} \cdot \frac{-1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}$$

Simplify the dot product

Simplify the expression obtained in step 1: $$\mathbf{I} \cdot \mathbf{J} = \frac{-1}{2} + \frac{1}{2} = 0$$ Since the dot product is equal to 0, the vectors \(\mathbf{I}\) and \(\mathbf{J}\) are orthogonal.

Calculate the magnitudes of I and J

Determine the magnitudes of \(\mathbf{I}\) and \(\mathbf{J}\). Recall that the magnitude of a vector \(\mathbf{A}=\langle a_1, a_2\rangle\) is given by: $$||\mathbf{A}|| = \sqrt{a_1^2 + a_2^2}$$ Calculate the magnitudes of \(\mathbf{I}\) and \(\mathbf{J}\): $$||\mathbf{I}|| = \sqrt{\left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2}$$ $$||\mathbf{J}|| = \sqrt{\left(\frac{-1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2}$$

Simplify the magnitudes of I and J

Simplify the expressions obtained in step 3: $$||\mathbf{I}|| = \sqrt{\frac{1}{2} + \frac{1}{2}} = \sqrt{1} = 1$$ $$||\mathbf{J}|| = \sqrt{\frac{1}{2} + \frac{1}{2}} = \sqrt{1} = 1$$ Since both magnitudes are equal to 1, the vectors \(\mathbf{I}\) and \(\mathbf{J}\) are unit vectors. In conclusion, the given vectors \(\mathbf{I}\) and \(\mathbf{J}\) are orthogonal unit vectors.

Key concepts

Dot Product

The dot product is a fundamental operation in vector algebra. It helps determine the relationship between two vectors, particularly whether they are orthogonal (i.e., perpendicular).
The dot product of two vectors \(\mathbf{A}= \langle a_1, a_2 \rangle\) and \(\mathbf{B}= \langle b_1, b_2 \rangle\) is calculated using the formula:

• \( \mathbf{A} \cdot \mathbf{B} = a_1b_1 + a_2b_2 \).

In our exercise, the vectors \(\mathbf{I}\) and \(\mathbf{J}\) had their dot product calculated:

• \( \mathbf{I} \cdot \mathbf{J} = \frac{1}{\sqrt{2}} \cdot \frac{-1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = 0 \).

This result signifies that the vectors \(\mathbf{I}\) and \(\mathbf{J}\) are orthogonal because their dot product equals zero. Understanding this property is crucial when working with vectors, as orthogonal vectors are often desired in applications involving projections and coordinate systems.

Magnitude of a Vector

The magnitude of a vector, often referred to as its length or norm, measures how long a vector is. For a vector \(\mathbf{A} = \langle a_1, a_2 \rangle\), the magnitude is found using the formula:

• \( ||\mathbf{A}|| = \sqrt{a_1^2 + a_2^2} \).

In the problem at hand, the magnitudes of vectors \(\mathbf{I}\) and \(\mathbf{J}\) were calculated as:

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

Because both magnitudes are equal to 1, these vectors are unit vectors, which have particular importance in defining direction without implying magnitude, often used in physics and engineering.

Unit Vector

Unit vectors are vectors that have a magnitude of exactly 1. They are useful in many mathematical and practical applications because they provide direction while maintaining a standard length. In our exercise, after computing the magnitudes, we confirmed that both \(\mathbf{I}\) and \(\mathbf{J}\) are unit vectors because their magnitudes are both 1.

• This characteristic allows them to be used directly as directions along certain axes or in defining bases for vector spaces.
• In any dimension, a unit vector is found by dividing each component of a vector by its magnitude.

For instance, if \(\mathbf{A}\) has a magnitude of \(|\mathbf{A}||\), the unit vector \(\mathbf{A}^\text{unit}\) is \(\mathbf{A}/||\mathbf{A}||\).
This process is known as normalization. Unit vectors simplify calculations in vector mathematics, allowing clear representation of direction without the complicating factor of different magnitudes.