Problem 59 - 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 \({\mathrm{proj}_v}\mathrm{u}.\) $$\begin{aligned} &\mathbf{u}=\langle 0,3\rangle\\\ &\mathbf{v}=\langle 2,15\rangle \end{aligned}$$

Step-by-Step Solution

Answer

\({\mathrm{proj}_v}\mathbf{u} = \langle \frac{90}{229}, \frac{675}{229} \rangle\), the orthogonal vector to \(\mathbf{v}\) is \(\langle -\frac{90}{229}, \frac{12}{229} \rangle\), and \(\mathbf{u} = \langle 0,3\rangle = \langle \frac{90}{229}, \frac{675}{229} \rangle + \langle -\frac{90}{229}, \frac{12}{229} \rangle\).

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

This is done by computing the dot product of \(\mathbf{u}\) and \(\mathbf{v}\), then dividing this value by the length of \(\mathbf{v}\) squared. The resulting scalar is then multiplied with vector \(\mathbf{v}\). The formula is \({\mathrm{proj}_v}(\mathbf{u}) = \frac{\mathbf{u}\cdot\mathbf{v}}{\|\mathbf{v}\|^2}\mathbf{v}\).Substitute \(\mathbf{u} = \langle 0,3 \rangle\) and \(\mathbf{v} = \langle 2,15 \rangle\) into the formula. \({\mathrm{proj}_v}\mathbf{u} = \frac{(0)(2) + (3)(15)}{2^2 + 15^2}\langle 2,15 \rangle = \frac{45}{229}\langle 2,15 \rangle = \langle \frac{90}{229}, \frac{675}{229} \rangle\)

Step 2: Calculate the Vector Orthogonal to \(\mathbf{v}\)

To get the vector orthogonal to \(\mathbf{v}\), subtract the projection of \(\mathbf{u}\) onto \(\mathbf{v}\) from \(\mathbf{u}\). In other words, \(\mathbf{u}_{ortho} = \mathbf{u} - {\mathrm{proj}_v}\mathbf{u}\). Substitute \(\mathbf{u} = \langle 0,3 \rangle\) and \({\mathrm{proj}_v}\mathbf{u} = \langle \frac{90}{229}, \frac{675}{229} \rangle\) into the formula. \(\mathbf{u}_{ortho} = \langle 0,3 \rangle - \langle \frac{90}{229}, \frac{675}{229} \rangle = \langle -\frac{90}{229}, \frac{12}{229} \rangle\)

Step 3: Write \(\mathbf{u}\) as the Sum of Two Orthogonal Vectors

Finally, write \(\mathbf{u}\) as the sum of its projection onto \(\mathbf{v}\) and the orthogonal vector to \(\mathbf{v}\). So \(\mathbf{u} = {\mathrm{proj}_v}\mathbf{u} + \mathbf{u}_{ortho}\). Substitute \({\mathrm{proj}_v}\mathbf{u} = \langle \frac{90}{229}, \frac{675}{229} \rangle\) and \(\mathbf{u}_{ortho} = \langle -\frac{90}{229}, \frac{12}{229} \rangle\) to get: \(\mathbf{u} = \langle \frac{90}{229}, \frac{675}{229} \rangle + \langle -\frac{90}{229}, \frac{12}{229} \rangle = \langle 0,3\rangle\).

Key Concepts

Orthogonal Vectors

Orthogonal vectors are vectors that are perpendicular to each other in space. When two vectors are orthogonal, their dot product equals zero. This means they have no components in the same direction, and they meet at a right angle (90 degrees). In a geometric sense, if you were to place them on a plane, they would intersect at right angles.
In mathematical exercises, identifying orthogonal vectors is essential for various calculations including finding vector components or simplifying vector problems. When a vector \( \mathbf{u} \) is expressed as the sum of two orthogonal vectors, one might be in the direction of another vector \( \mathbf{v} \) and the other orthogonal to \( \mathbf{v} \). Understanding the orthogonality helps to decompose a vector into projections easily.

Dot Product

The dot product, also known as the scalar product, is an operation that takes two vectors and returns a scalar. It is calculated by multiplying corresponding entries of the vectors and summing those products. The formula for the dot product of two vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \) is \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \).
The dot product is very useful because it helps in determining angles between vectors, checking orthogonality, and finding vector projections. If the dot product of two vectors is zero, it implies the vectors are orthogonal. For projections, the dot product is used in computing how much of one vector goes in the direction of another, making it crucial in vector decomposition tasks.

Vector Subtraction

Vector subtraction involves taking one vector and subtracting another from it. This operation is used to find the difference between two vectors. Given vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \), their difference is calculated by \( \mathbf{a} - \mathbf{b} = \langle a_1 - b_1, a_2 - b_2 \rangle \).
Vector subtraction is particularly helpful when you need to find a vector orthogonal to another, as in the exercise's step 2. By subtracting the projection vector from the original vector, we effectively find the part of the original vector that points in the orthogonal direction. This concept simplifies breaking down vectors into components along and perpendicular to another vector.