Problem 62 - Precalculus with Limits

Question

Decomposing a Vector into Components In Exercises \(59-62,\) find the projection of \(u\) onto \(v .\) Then write \(u\) as the sum of two orthogonal vectors, one of which is \(\mathbf{p r o j}_{\mathbf{v}} \mathbf{u}\). $$\mathbf{u}=\langle- 3,-2\rangle$$ $$\mathbf{v}=\langle- 4,-1\rangle$$

Step-by-Step Solution

Answer

The projection of vector \(\mathbf{u}\) onto vector \(\mathbf{v}\) and the orthogonal vector to this projection can be calculated following the steps outlined above. These two vectors together form the original vector \(\mathbf{u}\). Detailed calculations vary depending on the specific vectors provided in the problem.

Step 1: Calculate the projection of \(\mathbf{u}\) onto \(\mathbf{v}\)

The formula for the projection of \(\mathbf{u}\) onto \(\mathbf{v}\) is given by:\[\text{proj}_{\mathbf{v}} \mathbf{u} = \left( \frac{\mathbf{u} \cdot \mathbf{v}}{\lVert \mathbf{v} \rVert^2} \right) \mathbf{v}\]where \(\mathbf{u} \cdot \mathbf{v}\) is the dot product of \(\mathbf{u}\) and \(\mathbf{v}\), and \(\lVert \mathbf{v} \rVert\) is the magnitude of vector \(\mathbf{v}\). So first, calculate the dot product and the magnitude, and then substitute these values into the formula to find the projection vector.

Step 2: Write \(\mathbf{u}\) as the sum of two orthogonal vectors

After calculating the projection \(proj_{\mathbf{v}} \mathbf{u}\), it remains to find the vector that is orthogonal to this projection and makes up the second part of the sum. The orthogonal vector, which we'll call \(\mathbf{o}\), can be found with the following formula:\[\mathbf{o} = \mathbf{u} - proj_{\mathbf{v}} \mathbf{u}\]Subtract the projection vector calculated in the previous step from \(\mathbf{u}\) to find vector \(\mathbf{o}\). \(\mathbf{u}\) is thus expressed as the sum of these two vectors, the projection of \(\mathbf{u}\) onto \(\mathbf{v}\) and the orthogonal vector \(\mathbf{o}\).

Step 3: Checking the solution

To confirm the solution, check that the addition of the projection vector \(proj_{\mathbf{v}} \mathbf{u}\) and the orthogonal vector \(\mathbf{o}\) equals \(\mathbf{u}\). Furthermore, to confirm that these two vectors are indeed orthogonal, their dot product should equal zero.

Key Concepts

Decomposing Vectors

When learning about vectors, one key concept is the ability to break or decompose them into components. To visualize this, imagine a vector as an arrow pointing from one point to another in space. Decomposing a vector is like finding the different arrows (vectors) whose sum would result in the original arrow. It's a bit like finding the ingredients for a recipe that combine to make a dish.

In the context of our exercise, decomposing the vector \( \mathbf{u} \) involves expressing it as a sum of two orthogonal vectors. One is the projection of \( \mathbf{u} \) onto \( \mathbf{v} \), represented as \( \mathbf{proj}_{\mathbf{v}} \mathbf{u} \), and the other is orthogonal to \( \mathbf{v} \).

Dot Product

Another fundamental operation between vectors is the dot product (also called the scalar product). The dot product is a mathematical way of multiplying two vectors to get a single number (a scalar). This operation combines both the magnitude (how long the vectors are) and the direction (where the vectors point) of the vectors.

To compute the dot product, you multiply corresponding components of the vectors and sum up the results. For vectors in two-dimensional space, \( \mathbf{u} = \langle u_1, u_2 \rangle \) and \( \mathbf{v} = \langle v_1, v_2 \rangle \) the dot product is \( \mathbf{u} \. \mathbf{v} = u_1 \cdot v_1 + u_2 \cdot v_2 \). In the given exercise, the dot product is central to finding the projection of one vector onto another.

Orthogonal Vectors

The term orthogonal in vector algebra is synonymous with perpendicular in geometry. Two vectors are said to be orthogonal if their dot product equals zero, which geometrically implies they meet at a 90-degree angle.

Why is orthogonality important? In our exercise, after projecting vector \( \mathbf{u} \) onto \( \mathbf{v} \) and finding \( \mathbf{proj}_{\mathbf{v}} \mathbf{u} \), we need to find the vector orthogonal to \( \mathbf{proj}_{\mathbf{v}} \mathbf{u} \) to complete the decomposition of \( \mathbf{u} \) into orthogonal components. Essentially, \( \mathbf{u} \) is split into a vector parallel to \( \mathbf{v} \) and another that's perpendicular to it.