Question

For the following vectors u and \(\mathbf{v}\) express u as the sum \(\mathbf{u}=\mathbf{p}+\mathbf{n},\) where \(\mathbf{p}\) is parallel to \(\mathbf{v}\) and \(\mathbf{n}\) is orthogonal to \(\mathbf{v}\). \(\mathbf{u}=\langle 4,3\rangle, \mathbf{v}=\langle 1,1\rangle\)

Short answer

Question: Express the vector \(\mathbf{u} = \langle 4,3\rangle\) as the sum of a vector \(\mathbf{p}\) parallel to \(\mathbf{v} = \langle 1,1\rangle\) and a vector \(\mathbf{n}\) orthogonal to \(\mathbf{v}\). Answer: \(\mathbf{u} = \langle 4,3\rangle = \langle \frac{7}{2},\frac{7}{2} \rangle + \langle \frac{1}{2}, -\frac{1}{2}\rangle\), where \(\mathbf{p} = \langle \frac{7}{2},\frac{7}{2} \rangle\) and \(\mathbf{n} = \langle \frac{1}{2}, -\frac{1}{2}\rangle\).

Step-by-step solution

Find the scalar projection of \(\mathbf{u}\) onto \(\mathbf{v}\)

To find the scalar projection of \(\mathbf{u}\) onto \(\mathbf{v}\), we use the formula: $$c = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|}.$$ First, calculate the dot product \(\mathbf{u} \cdot \mathbf{v}\): $$\mathbf{u} \cdot \mathbf{v}=\langle 4,3\rangle \cdot \langle 1,1\rangle = 4\times1 + 3\times1 = 7.$$ Next, calculate the magnitude of vector \(\mathbf{v}\): $$\|\mathbf{v}\| = \sqrt{1^2 + 1^2} = \sqrt2.$$ Now find the scalar projection \(c\): $$c = \frac{7}{\sqrt2}.$$

Find the vector projection \(\mathbf{p}\) of \(\mathbf{u}\) onto \(\mathbf{v}\)

Using the scalar projection \(c\) and the direction of \(\mathbf{v}\), we can calculate the vector projection \(\mathbf{p}\): $$\mathbf{p} = c \frac{\mathbf{v}}{\|\mathbf{v}\|} = \frac{7}{\sqrt2} \frac{\langle 1,1\rangle}{\sqrt2} = \frac{7}{2} \langle 1,1\rangle = \langle \frac{7}{2},\frac{7}{2} \rangle.$$

Find the orthogonal component \(\mathbf{n}\)

To find the orthogonal component \(\mathbf{n}\), subtract the vector projection \(\mathbf{p}\) from vector \(\mathbf{u}\): $$\mathbf{n} = \mathbf{u} - \mathbf{p} = \langle 4,3\rangle - \langle\frac{7}{2},\frac{7}{2}\rangle= \langle \frac{1}{2}, -\frac{1}{2}\rangle.$$

Express \(\mathbf{u}\) as the sum of \(\mathbf{p}\) and \(\mathbf{n}\)

We have calculated the vector projection \(\mathbf{p}\) and the orthogonal component \(\mathbf{n}\). To express \(\mathbf{u}\) as the sum of these two components, we simply add them together: $$\mathbf{u} = \mathbf{p} + \mathbf{n} = \langle \frac{7}{2},\frac{7}{2} \rangle + \langle \frac{1}{2}, -\frac{1}{2}\rangle = \langle 4,3 \rangle.$$ Thus, we have expressed the given vector \(\mathbf{u}\) as the sum of the parallel and orthogonal components to \(\mathbf{v}\): $$\mathbf{u} = \langle 4,3\rangle = \langle \frac{7}{2},\frac{7}{2} \rangle + \langle \frac{1}{2}, -\frac{1}{2}\rangle.$$

Key concepts

Scalar Projection

The scalar projection is a simple yet powerful idea that helps us measure how one vector "falls" onto another. Imagine two vectors, \(\mathbf{u}\) and \(\mathbf{v}\), on a flat plane. The scalar projection of \(\mathbf{u}\) onto \(\mathbf{v}\) tells us how much of \(\mathbf{u}\) is pointing in the direction of \(\mathbf{v}\). This is calculated using the dot product and the magnitude (or length) of \(\mathbf{v}\). To find it, we use the formula:

• \(c = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|}\)

The dot product, \(\mathbf{u} \cdot \mathbf{v}\), gives us a measure of the vectors’ alignment by multiplying their corresponding components:
\(\langle 4,3\rangle \cdot \langle 1,1\rangle = 4 \times 1 + 3 \times 1 = 7\). Then, we find the magnitude of \(\mathbf{v}\), which is \(\sqrt{1^2 + 1^2} = \sqrt{2}\). Dividing these gives us the scalar projection \(\frac{7}{\sqrt{2}}\), which represents how much \(\mathbf{u}\) extends along \(\mathbf{v}\).
This aspect is crucial for breaking down the vector into components.

Vector Projection

Once we have the scalar projection, we can find the vector projection. Think of this as translating the scalar information into a full vector that's parallel to \(\mathbf{v}\). To achieve this, we scale \(\mathbf{v}\) by the scalar projection. The vector projection, \(\mathbf{p}\), is calculated with:

• \(\mathbf{p} = c \frac{\mathbf{v}}{\|\mathbf{v}\|}\)

Essentially, we're multiplying the vector's direction by the scalar \(c\) to ensure it has the correct magnitude. For our vectors, after normalizing \(\langle 1,1\rangle\) to its magnitude \(\sqrt{2}\), we find \(\mathbf{p} = \frac{7}{\sqrt{2}} \cdot \frac{\langle 1,1\rangle}{\sqrt{2}} = \langle \frac{7}{2}, \frac{7}{2} \rangle\).
By constructing \(\mathbf{p}\), we identify the part of \(\mathbf{u}\) that lies exactly along the direction of \(\mathbf{v}\). It shows us how much of \(\mathbf{u}\) is "in line" with \(\mathbf{v}\).

Orthogonal Component

The orthogonal component is the part of \(\mathbf{u}\) that is entirely perpendicular to \(\mathbf{v}\). It represents any portion of \(\mathbf{u}\) that doesn't align with \(\mathbf{v}\). After finding the vector projection \(\mathbf{p}\), we can determine this perpendicular portion by subtracting \(\mathbf{p}\) from \(\mathbf{u}\):

• \(\mathbf{n} = \mathbf{u} - \mathbf{p}\)

For our exercise, this calculation becomes \(\langle 4,3\rangle - \langle \frac{7}{2}, \frac{7}{2} \rangle = \langle \frac{1}{2}, -\frac{1}{2} \rangle\).
Thus, \(\mathbf{n}\) stands as the portion of \(\mathbf{u}\) that is orthogonal (or at a 90-degree angle) to \(\mathbf{v}\). Adding \(\mathbf{p}\) to \(\mathbf{n}\) will reproduce the original vector \(\mathbf{u}\), verifying our decomposition.

Dot Product

The dot product is a foundational operation in vector mathematics that allows us to determine the angle or alignment between two vectors. It combines corresponding components of vectors, \(\mathbf{u}\) and \(\mathbf{v}\), through multiplication and then sum these products:

• \(\mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2\)

This formula provides a single scalar value. If the dot product is positive, the vectors are pointing in roughly the same direction; if negative, they point in opposite directions.
In our exercise, this dot product calculation \(4 \times 1 + 3 \times 1 = 7\) tells us that \(\mathbf{u}\) has a strong alignment with \(\mathbf{v}\). Moreover, this operation is crucial for determining the scalar projection, guiding us on how much of \(\mathbf{u}\) lies along \(\mathbf{v}\). Thus, it acts as a bridge from vector alignment to component breakdown.