Problem 57 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the projection of \(\mathbf{u}\) onto \(\mathbf{v}\). Then write \(\mathbf{u}\) as the sum of two orthogonal vectors, one of which is \({\mathrm{proj}_v}\mathrm{u}.\) $$\begin{aligned} &\mathbf{u}=\langle 2,2\rangle\\\ &\mathbf{v}=\langle 6,1\rangle \end{aligned}$$

Step-by-Step Solution

Answer

The projection of \(\mathbf{u}\) onto \(\mathbf{v}\), denoted \({\mathrm{proj}_v}\mathrm{u}\), is \(\langle 1.846,0.308\rangle\). The orthogonal component is \(\langle 0.154,1.692\rangle\). Therefore, \(\mathbf{u}\) is the sum of these two orthogonal vectors: \(\mathbf{u} = {\mathrm{proj}_v}\mathrm{u} + \mathbf{u}_{\perp} = \langle 1.846,0.308\rangle + \langle 0.154,1.692\rangle\).

Step 1: Finding the Projection

The projection of \(\mathbf{u}\) onto \(\mathbf{v}\) can be found using the formula \({\mathrm{proj}_v}\mathrm{u} = \frac{\mathbf{u} . \mathbf{v}}{\|\mathbf{v}\|^2} \mathbf{v}\) . Here, \(\mathbf{u} . \mathbf{v}\) represents the dot product of \(\mathbf{u}\) and \(\mathbf{v}\), and \(\|\mathbf{v}\|^2\) represents the magnitude of \(\mathbf{v}\) squared. Compute the dot product, calculate the magnitude of vector \(\mathbf{v}\), then substitute these values into the formula to find \({\mathrm{proj}_v}\mathrm{u}.\)

Step 2: Orthogonal Decomposition

Once the projection \({\mathrm{proj}_v}\mathrm{u}\) is found, we can get the orthogonal vector by subtracting \({\mathrm{proj}_v}\mathrm{u}\) from \(\mathbf{u}\), denoted as \(\mathbf{u}_{\perp}\). Finally, we express \(\mathbf{u}\) as the sum of \({\mathrm{proj}_v}\mathrm{u}\) and \(\mathbf{u}_{\perp}\), which are orthogonal to each other because they compose the original vector \(\mathbf{u}\).

Key Concepts

Dot Product

The dot product is a fundamental operation that combines two vectors and results in a scalar value. For two vectors, \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \), the dot product is calculated as:

• \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \)

This calculation shows how parallel two vectors are to each other. If the dot product is positive, the vectors tend to point in the same direction, whereas a negative result indicates opposite tendencies.
In the original exercise, the dot product of \( \mathbf{u} \) and \( \mathbf{v} \) is calculated as follows:

• \( \mathbf{u} \cdot \mathbf{v} = 2 \cdot 6 + 2 \cdot 1 = 12 + 2 = 14 \)

This result is integral in determining how much of vector \( \mathbf{u} \) lies in the direction of vector \( \mathbf{v} \). The dot product is vital to finding the projection of one vector onto another.

Orthogonal Vectors

Orthogonal vectors are vectors that are at right angles to each other.
This means that when two vectors are orthogonal, their dot product is zero. This property of vectors allows us to decompose any vector into two orthogonal components.
In the context of the original exercise, we decompose vector \( \mathbf{u} \) into:

• The projection of \( \mathbf{u} \) onto \( \mathbf{v} \)
• The orthogonal part of \( \mathbf{u} \) relative to \( \mathbf{v} \), denoted as \( \mathbf{u}_{\perp} \)

These components are orthogonal to one another and their dot product equals zero:

• \( \mathbf{u}_{\perp} \cdot {\mathrm{proj}_v} \mathbf{u} = 0 \)

Therefore, vector \( \mathbf{u} \) can be written as \( \mathbf{u} = {\mathrm{proj}_v} \mathbf{u} + \mathbf{u}_{\perp} \). This form is useful for applications in physics, computer graphics, and engineering where resolving forces or decomposing vectors into independent directions is necessary.

Magnitude of a Vector

The magnitude of a vector, also known as its length, is a measure of the distance from the vector's origin to its endpoint in the coordinate space.
The magnitude of a vector \( \mathbf{v} = \langle v_1, v_2 \rangle \) is calculated using the formula:

• \( \|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2} \)

This measurement is crucial when determining how much of one vector extends in the direction of another, as seen in vector projection computations.
In the original problem, the magnitude of vector \( \mathbf{v} \) is calculated as:

• \( \|\mathbf{v}\| = \sqrt{6^2 + 1^2} = \sqrt{36 + 1} = \sqrt{37} \)

The magnitude squared, \( \|\mathbf{v}\|^2 = 37 \), is then used to compute the projection of \( \mathbf{u} \) onto \( \mathbf{v} \), ensuring we scale the direction of \( \mathbf{v} \) properly for the correct decomposition.