Problem 61 - 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 \(\mathbf{p r o j}_{\mathbf{v}} \mathbf{u}\). $$\mathbf{u}=\langle 0,3\rangle$$ $$\mathbf{v}=\langle 2,15\rangle$$

Step-by-Step Solution

Answer

The projection of vector u onto vector v is <0.39214, 2.96052> and vector u can be written as the sum of two orthogonal vectors: <0.39214, 2.96052> and <-0.39214, 0.03948>.

Step 1: Find the projection of u onto v

Firstly, the formula to find the projection of u onto v:Proj_{v}u = ((u . v )/(||v||^2)) * vWe start by finding the dot product of vectors u and v:u . v = (0*2 + 3*15) = 45Then, calculate the magnitude of vector v squared:||v||^2 = (2^2 + 15^2) = 229.Using these values, the projection vector of u onto v would be Proj_{v}u = ((u.v)/(||v||^2)) * v= (45/229) * <2,15> = <0.39214, 2.96052>.

Step 2: Write u as the sum of two orthogonal vectors

Secondly, we need to express vector u as the sum of two orthogonal vectors. The projection vector Proj_{v}u from the previous step will be one of these vectors. By the properties of projection, the second vector should be u - Proj_{v}u.Subtracting the projection vector from vector u to get the second vector:u - Proj_{v}u = <0,3> - <0.39214, 2.96052> = <-0.39214, 0.03948>. These two vectors are orthogonal to each other.

Key Concepts

Projection of Vectors

The projection of one vector onto another is a fundamental concept in vector calculus. It essentially measures how much of one vector lies in the direction of another. To find the projection of vector \( \mathbf{u} \) onto vector \( \mathbf{v} \), we use the following formula:
\[\text{Proj}_{\mathbf{v}} \mathbf{u} = \left( \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{v}||^2} \right) \mathbf{v}\]
This formula involves the dot product of the vectors \( \mathbf{u} \) and \( \mathbf{v} \), as well as the square of the magnitude of vector \( \mathbf{v} \).

Once calculated, the projection gives us a new vector that indicates the influence of \( \mathbf{u} \) in the direction of \( \mathbf{v} \). It's important to note that this projection will be collinear with vector \( \mathbf{v} \).

In our specific example with \( \mathbf{u} = \langle 0, 3 \rangle \) and \( \mathbf{v} = \langle 2, 15 \rangle \), the projection was calculated to be approximately \( \langle 0.39214, 2.96052 \rangle \), which lies along \( \mathbf{v} \). This vector provides a component of \( \mathbf{u} \) aligned with \( \mathbf{v} \).

Dot Product

The dot product is a crucial operation when working with vectors. It results in a scalar value that tells us about the relationship between two vectors. To compute the dot product of two vectors \( \mathbf{u} = \langle u_1, u_2 \rangle \) and \( \mathbf{v} = \langle v_1, v_2 \rangle \), we use the formula:
\[\mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2\]

This operation gives us a single number, which can indicate:

• If the dot product is positive, the vectors have an acute angle between them.
• If it is zero, the vectors are orthogonal, meaning they are at a right angle.
• If it is negative, the vectors have an obtuse angle between them.

In our example, the dot product of \( \mathbf{u} = \langle 0, 3 \rangle \) and \( \mathbf{v} = \langle 2, 15 \rangle \) is calculated as \( 0 \cdot 2 + 3 \cdot 15 = 45 \). This value is crucial for determining the projection of \( \mathbf{u} \) onto \( \mathbf{v} \).

Magnitude of a Vector

The magnitude of a vector, also known as the vector's length or norm, quantifies how long or large the vector is. For a vector \( \mathbf{v} = \langle v_1, v_2 \rangle \), the formula to calculate its magnitude is:
\[||\mathbf{v}|| = \sqrt{v_1^2 + v_2^2}\]

This equation is derived from the Pythagorean theorem. The magnitude gives us the distance from the origin to the point defined by the vector in space.

In the exercise, we found the magnitude of vector \( \mathbf{v} = \langle 2, 15 \rangle \) to be \( \sqrt{2^2 + 15^2} = \sqrt{229} \). The square of this magnitude, \( 229 \), is used in the projection formula to scale the influence of the vector correctly.
Understanding the magnitude is essential when working with both projections and normalization of vectors.

Orthogonal Vectors

Vectors are orthogonal if they meet at a right angle, meaning their dot product is zero. This property is pivotal when decomposing a vector into components, such as when expressing a vector as the sum of two orthogonal vectors.

In this context, when we decompose vector \( \mathbf{u} \), we obtain two vectors: the projection \( \text{Proj}_{\mathbf{v}} \mathbf{u} \) and the vector \( \mathbf{u} - \text{Proj}_{\mathbf{v}} \mathbf{u} \). These vectors are orthogonal, ensuring that one component lies entirely in the direction of \( \mathbf{v} \) while the other lies perpendicularly.

To determine orthogonality in our given exercise, calculate the vector resulting from the subtraction:
\[\mathbf{u} - \text{Proj}_{\mathbf{v}} \mathbf{u} = \langle 0, 3 \rangle - \langle 0.39214, 2.96052 \rangle = \langle -0.39214, 0.03948 \rangle\]
These vectors are orthogonal if they give a dot product of zero when calculated.

This decomposition is helpful because it provides a clear geometric interpretation of vector components, where each part serves a distinct role in relation to vector \( \mathbf{v} \).