Question

Show that the vectors \(\mathbf{a}=\mathbf{i}-\mathbf{j}, \mathbf{b}=\mathbf{i}+\mathbf{j}, \quad\) and \(\mathbf{c}=2 \mathbf{k}\) are mutually orthogonal, that is, each pair of vectors is orthogonal.

Short answer

The vectors are mutually orthogonal; each pair has a dot product of zero.

Step-by-step solution

Understand Orthogonality

Two vectors are orthogonal if their dot product is zero. For vectors \(\mathbf{v}\) and \(\mathbf{w}\), the condition is \(\mathbf{v} \cdot \mathbf{w} = 0\). We will test this condition for each pair of vectors (\(\mathbf{a}, \mathbf{b}\)), (\(\mathbf{b}, \mathbf{c}\)), and (\(\mathbf{c}, \mathbf{a}\)).

Calculate \(\mathbf{a} \cdot \mathbf{b}\)

The dot product \(\mathbf{a} \cdot \mathbf{b}\) is given by: \((\mathbf{i} - \mathbf{j}) \cdot (\mathbf{i} + \mathbf{j})\). Using the distributive property, this becomes: \(\mathbf{i} \cdot \mathbf{i} + \mathbf{i} \cdot \mathbf{j} - \mathbf{j} \cdot \mathbf{i} - \mathbf{j} \cdot \mathbf{j}\). Since \(\mathbf{i} \cdot \mathbf{i} = 1\), \(\mathbf{j} \cdot \mathbf{j} = 1\), and \(\mathbf{i} \cdot \mathbf{j} = \mathbf{j} \cdot \mathbf{i} = 0\), the resulting dot product is \(1 - 1 = 0\).

Calculate \(\mathbf{b} \cdot \mathbf{c}\)

The dot product \(\mathbf{b} \cdot \mathbf{c}\) is \((\mathbf{i} + \mathbf{j}) \cdot 2\mathbf{k}\). This results in: \(2 \mathbf{i} \cdot \mathbf{k} + 2 \mathbf{j} \cdot \mathbf{k}\). Because both \(\mathbf{i} \cdot \mathbf{k}\) and \(\mathbf{j} \cdot \mathbf{k}\) are 0, the dot product is \(0 + 0 = 0\).

Calculate \(\mathbf{c} \cdot \mathbf{a}\)

The dot product \(\mathbf{c} \cdot \mathbf{a}\) is \(2\mathbf{k} \cdot (\mathbf{i} - \mathbf{j})\). This results in: \(2\mathbf{k} \cdot \mathbf{i} - 2\mathbf{k} \cdot \mathbf{j}\). Since \(\mathbf{k} \cdot \mathbf{i}\) and \(\mathbf{k} \cdot \mathbf{j}\) are both 0, the dot product is \(0 - 0 = 0\).

Conclusion

Since all dot products \(\mathbf{a} \cdot \mathbf{b}\), \(\mathbf{b} \cdot \mathbf{c}\), and \(\mathbf{c} \cdot \mathbf{a}\) are zero, the vectors \(\mathbf{a}\), \(\mathbf{b}\), and \(\mathbf{c}\) are mutually orthogonal.

Key concepts

Dot Product

The dot product, also known as the scalar product, is a fundamental operation in vector mathematics. It provides a way to multiply two vectors, resulting in a scalar (a single number), instead of another vector. The dot product is calculated by multiplying corresponding components of the two vectors and then summing these products.
For example, for vectors \(\mathbf{v} = \langle v_1, v_2, v_3 \rangle\) and \(\mathbf{w} = \langle w_1, w_2, w_3 \rangle\), the dot product is given by:\[ \mathbf{v} \cdot \mathbf{w} = v_1w_1 + v_2w_2 + v_3w_3 \]This operation is commutative, meaning the order of multiplication doesn't affect the result.
Understanding the dot product is essential for assessing two vectors' orthogonality, which leads us to the next concept.

Orthogonality

Orthogonality is a concept that describes the perpendicular relationship between vectors. Two vectors are orthogonal if their dot product equals zero. This geometric interpretation implies that the vectors meet at a right angle.
In the exercise given, each pair of vectors (\(\mathbf{a}, \mathbf{b}\)), (\(\mathbf{b}, \mathbf{c}\)), and (\(\mathbf{c}, \mathbf{a}\)) are assessed for orthogonality by evaluating their dot products:

• \(\mathbf{a} \cdot \mathbf{b} = 0\)
• \(\mathbf{b} \cdot \mathbf{c} = 0\)
• \(\mathbf{c} \cdot \mathbf{a} = 0\)

In all cases, the dot product is zero, proving the vectors are mutually orthogonal. This establishes that each vector is perpendicular to every other vector in the set.

Vector Calculations

Calculating the dot product and determining orthogonality involve several vector operations. Let's break down how these calculations are performed in the context of the provided vectors \(\mathbf{a} = \mathbf{i} - \mathbf{j}\), \(\mathbf{b} = \mathbf{i} + \mathbf{j}\), and \(\mathbf{c} = 2\mathbf{k}\).
Start by understanding the basis vectors \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\), which represent the unit vectors along the X, Y, and Z axes. These basis vectors have a dot product of one with themselves (e.g., \(\mathbf{i} \cdot \mathbf{i} = 1\)) and zero with each other (e.g., \(\mathbf{i} \cdot \mathbf{j} = 0\)).
To verify the orthogonality, compute the dot products using distributive properties:

• For \(\mathbf{a} \cdot \mathbf{b}\): Expand to \((\mathbf{i} - \mathbf{j}) \cdot (\mathbf{i} + \mathbf{j})\). Use distributive property: \(\mathbf{i} \cdot \mathbf{i} + \mathbf{i} \cdot \mathbf{j} - \mathbf{j} \cdot \mathbf{i} - \mathbf{j} \cdot \mathbf{j}\), resulting in \(1 - 1 = 0\).
• For \(\mathbf{b} \cdot \mathbf{c}\): Expands to \((\mathbf{i} + \mathbf{j}) \cdot 2\mathbf{k}\). Perform multiplication: \(2\mathbf{i} \cdot \mathbf{k} + 2\mathbf{j} \cdot \mathbf{k}\), leading to \(0 + 0 = 0\).
• For \(\mathbf{c} \cdot \mathbf{a}\): Expands to \(2\mathbf{k} \cdot (\mathbf{i} - \mathbf{j})\). Use distributive property: \(2\mathbf{k} \cdot \mathbf{i} - 2\mathbf{k} \cdot \mathbf{j}\), resulting in \(0 - 0 = 0\).

By following these computation steps, students can confidently verify that the vectors are mutually orthogonal.