We are tasked with proving that the vector $\mathbf{u} - \mathbf{u}_v$ is orthogonal (perpendicular) to the vector $\mathbf{v}$, where:

• $\mathbf{u}_v$ represents the vector projection of $\mathbf{u}$ in the direction of $\mathbf{v}$.

Step-by-step solution:

1. Formula for vector projection:

The vector projection of $\mathbf{u}$ onto $\mathbf{v}$ is given by: $\mathbf{u}_v = \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v}.$

Here:

• $\mathbf{u} \cdot \mathbf{v}$ is the dot product of $\mathbf{u}$ and $\mathbf{v}$,
• $\mathbf{v} \cdot \mathbf{v}$ is the squared magnitude of $\mathbf{v}$,
• $\frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}}$ is the scalar component of the projection.

2. Define the vector $\mathbf{u} - \mathbf{u}_v$:

Substitute $\mathbf{u}_v$ into the expression $\mathbf{u} - \mathbf{u}_v$: $\mathbf{u} - \mathbf{u}_v = \mathbf{u} - \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v}.$

Let us denote $\mathbf{w} = \mathbf{u} - \mathbf{u}_v$. So: $\mathbf{w} = \mathbf{u} - \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v}.$

3. Show that $\mathbf{w} \cdot \mathbf{v} = 0$:

The condition for $\mathbf{w}$ to be orthogonal to $\mathbf{v}$ is: $\mathbf{w} \cdot \mathbf{v} = 0.$

Compute $\mathbf{w} \cdot \mathbf{v}$: $\mathbf{w} \cdot \mathbf{v} = \left( \mathbf{u} - \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v} \right) \cdot \mathbf{v}.$

Using the distributive property of the dot product: $\mathbf{w} \cdot \mathbf{v} = \mathbf{u} \cdot \mathbf{v} - \left( \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v} \cdot \mathbf{v} \right).$

Simplify the second term:

• Since $\mathbf{v} \cdot \mathbf{v} = \|\mathbf{v}\|^2$, we have: $\mathbf{w} \cdot \mathbf{v} = \mathbf{u} \cdot \mathbf{v} - \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} (\mathbf{v} \cdot \mathbf{v}).$

The $\mathbf{v} \cdot \mathbf{v}$ terms cancel out: $\mathbf{w} \cdot \mathbf{v} = \mathbf{u} \cdot \mathbf{v} - \mathbf{u} \cdot \mathbf{v}.$

Thus: $\mathbf{w} \cdot \mathbf{v} = 0.$

4. Conclusion:

Since $\mathbf{w} \cdot \mathbf{v} = 0$, the vector $\mathbf{u} - \mathbf{u}_v$ is orthogonal to $\mathbf{v}$. Therefore, we have shown that: $\mathbf{u} - \mathbf{u}_v \text{ is orthogonal to } \mathbf{v}.$

Do you want further clarification or detailed examples?

5 Related Questions:

1. What is the geometric interpretation of a vector projection?
2. How do you compute the magnitude of a vector projection?
3. Can a projection vector ever be longer than the original vector $\mathbf{u}$?
4. What are some practical applications of vector projections in physics or engineering?
5. How would the result change if $\mathbf{u}$ and $\mathbf{v}$ were parallel?

Tip:

When solving vector projection problems, always verify the orthogonality condition by checking the dot product.