Question

Let \(\mathbf{v}\) and \(\mathbf{w}\) denote two nonzero vectors. Show that the vector \(\mathbf{v}-\alpha \mathbf{w}\) is orthogonal to \(\mathbf{w}\) if \(\alpha=\frac{\mathbf{v} \cdot \mathbf{w}}{\|\mathbf{w}\|^{2}}\)

Short answer

If \(\text{alpha} = \frac{\text{v} \cdot \text{w}}{\|\text{w}\|^2}\), then \(\text{v} - \alpha \text{w}\) is orthogonal to \(\text{w}\).

Step-by-step solution

- Understand orthogonality

Two vectors \(\text{u}\) and \(\text{v}\) are orthogonal if their dot product is zero, i.e., \(\text{u} \cdot \text{v} = 0\).

- Consider the vector \(\text{v} - \alpha \text{w}\)

Rewrite the condition for orthogonality considering the vector \(\text{v}\) and \(\text{v} - \alpha \text{w}\). We want to show that \((\text{v} - \alpha \text{w}) \cdot \text{w} = 0\).

- Substitute the value of \(\text{alpha}\)

Substitute \(\text{alpha} = \frac{\text{v} \cdot \text{w}}{\|\text{w}\|^2}\) into the vector expression: \(\text{v} - \left( \frac{\text{v} \cdot \text{w}}{\|\text{w}\|^2} \right)\text{w}\).

- Compute the dot product

Compute the dot product of \(\text{v} - \left( \frac{\text{v} \cdot \text{w}}{\|\text{w}\|^2} \right)\text{w}\) with \(\text{w}\):

- Dot product expansion

Expand the expression: \((\text{v} - \alpha \text{w}) \cdot \text{w} = \text{v} \cdot \text{w} - \left( \frac{\text{v} \cdot \text{w}}{\|\text{w}\|^2} \right) \text{w} \cdot \text{w}\).

- Simplify the terms

Since \(\text{w} \cdot \text{w} = \|\text{w}\|^2\), the expression becomes \(\text{v} \cdot \text{w} - \left( \frac{\text{v} \cdot \text{w}}{\|\text{w}\|^2} \right)\|\text{w}\|^2\).

- Cancel the terms

Cancel the terms: \(\text{v} \cdot \text{w} - \left( \text{v} \cdot \text{w} \right) = 0\).

- Conclusion

Since the dot product is zero, \(\text{v} - \left( \frac{\text{v} \cdot \text{w}}{\|\text{w}\|^2} \right)\text{w}\) is orthogonal to \(\text{w}\).

Key concepts

Dot Product

Understanding the dot product is essential when dealing with vector orthogonality. The dot product of two vectors \(\text{u}\) and \(\text{v}\), denoted as \(\text{u} \cdot \text{v}\), is the product of their magnitudes and the cosine of the angle between them. Mathematically, it is represented as: \[ \text{u} \cdot \text{v} = \|\text{u}\|\text{v}\text{cos} \theta \] Here, \(\text{cos} \theta\) is the cosine of the angle between the two vectors. A fundamental property to remember is that the dot product of two orthogonal vectors is zero because the cosine of 90 degrees (or \(\frac{\pi}{2}\)) is zero. Hence, \[ \text{u} \cdot \text{v} = 0 \] when \(\text{u}\) and \(\text{v}\) are orthogonal. This property is crucial for proving orthogonality in vector problems.

Orthogonality

Orthogonality in vector algebra means that two vectors are perpendicular to each other. As mentioned earlier, the dot product of orthogonal vectors is zero. For example, if vectors \(\text{u}\) and \(\text{v}\) are orthogonal, then: \[ \text{u} \cdot \text{v} = 0 \] To prove that a vector \(\text{v} - \alpha \text{w}\) is orthogonal to \(\text{w}\), we utilize the condition of orthogonality. From the given exercise, we substitute \(\text{alpha}\) with \(\frac{\text{v} \cdot \text{w}}{\|\text{w}\text{\text{2}}}\), which simplifies our expression. The orthogonality condition can then be expanded using the dot product properties to show that the result is indeed zero. Hence, \(\text{v} - \left( \frac{\text{v} \cdot \text{w}}{\|\text{w}\text{\text{2}}}\right) \text{w}\) is orthogonal to \(\text{w}\).

Vector Algebra

In vector algebra, vectors are mathematical objects that have both magnitude and direction. Vectors can be added or scaled by real numbers (scalars). Key operations include:

• Vector addition: Summing vectors component-wise.
• Scalar multiplication: Scaling a vector by multiplying each component by a scalar.
• Dot product: Combining magnitudes and the cosine of the angle for two vectors.

For the given exercise, combining vector algebra concepts is essential. First, we understand the vector \(\text{v} - \alpha \text{w}\) needs to be orthogonal to \(\text{w}\). By leveraging the dot product properties and the provided scalar value of \(\text{alpha}\), we transform the vector and simplify the expression, proving orthogonality. The problem also reinforces understanding of how vector algebra operations can be applied to geometric properties like angles and perpendicularity.