Problem 4 - University Calculus: Early Transcendentals

Question

Let \(\mathbf{u}=\langle 3,-2\rangle\) and \(\mathbf{v}=\langle-2,5\rangle .\) Find the (a) component form and (b) magnitude (length) of the vector. $$\mathbf{u}-\mathbf{v}$$

Step-by-Step Solution

Answer

(a) \( \langle 5, -7 \rangle \), (b) \( \sqrt{74} \).

Step 1: Understanding the Problem

We are given two vectors, \(\mathbf{u}=\langle 3,-2\rangle\) and \(\mathbf{v}=\langle-2,5\rangle\). The task is to find the component form and magnitude of the vector \(\mathbf{u} - \mathbf{v}\).

Step 1: Subtract Vectors to Find Component Form

To find the component form of \(\mathbf{u}-\mathbf{v}\), subtract each corresponding component of the vectors \(\mathbf{u}\) and \(\mathbf{v}\). The subtraction is calculated as follows:\[ \mathbf{u} - \mathbf{v} = \langle 3, -2 \rangle - \langle -2, 5 \rangle = \langle 3 - (-2), -2 - 5 \rangle. \]This simplifies to: \[ \mathbf{u} - \mathbf{v} = \langle 3+2, -2-5 \rangle = \langle 5, -7 \rangle. \]

Step 2: Compute the Magnitude of the Vector

Now that we have the component form of \(\mathbf{u} - \mathbf{v}\) as \(\langle 5, -7 \rangle\), we calculate its magnitude. The magnitude of a vector \(\langle a, b \rangle\) is found using the formula:\[ \|\mathbf{a}\| = \sqrt{a^2 + b^2}. \]Substitute \(a = 5\) and \(b = -7\):\[ \|\mathbf{u} - \mathbf{v}\| = \sqrt{5^2 + (-7)^2} = \sqrt{25 + 49} = \sqrt{74}. \]

Key Concepts

Component Form

Understanding the component form of a vector is essential for performing operations like vector subtraction. The component form of a vector tells you about its direction and magnitude in a geometric space, typically a plane. When we have two vectors, such as \( \mathbf{u} = \langle 3, -2 \rangle \) and \( \mathbf{v} = \langle -2, 5 \rangle \), their component form is expressed in terms of their coordinates in a Cartesian plane. To perform vector subtraction and get the component form of \( \mathbf{u} - \mathbf{v} \), we subtract the corresponding components of each vector:

• Subtract the x-components: \( 3 - (-2) = 3 + 2 = 5 \)
• Subtract the y-components: \( -2 - 5 = -7 \)

Thus, the component form of \( \mathbf{u} - \mathbf{v} \) is \( \langle 5, -7 \rangle \). This vector takes into account both the changes in the x-direction and the y-direction of the original vectors.

Magnitude of a Vector

The magnitude of a vector is a measure of its length and represents its size irrespective of direction. Once you have the component form of a vector like \( \langle 5, -7 \rangle \), calculating its magnitude is straightforward using the Pythagorean theorem. The formula for magnitude \( \|\mathbf{a}\| \) of a vector \( \langle a, b \rangle \) is:\[ \|\mathbf{a}\| = \sqrt{a^2 + b^2} \]This equation essentially gives the distance from the origin to the point \( (a, b) \) in the plane, visualizing it as the hypotenuse of a right triangle. To find the magnitude of \( \mathbf{u} - \mathbf{v} = \langle 5, -7 \rangle \):

• Square each component of the vector: \( 5^2 = 25 \) and \( (-7)^2 = 49 \)
• Add the squared components: \( 25 + 49 = 74 \)
• Take the square root to find the magnitude: \( \sqrt{74} \)

Therefore, the magnitude of \( \mathbf{u} - \mathbf{v} \) is \( \sqrt{74} \). This value represents the overall length of the resulting vector in space.

Vector Operations

Vector operations allow us to perform various arithmetic involving vectors, which include addition, subtraction, and scalar multiplication. These operations enable us to simplify complex vector problems, such as changes in displacement or force in physics. In our situation, we are concentrating on vector subtraction.
To perform vector subtraction, express each vector in component form and subtract the corresponding components. For two vectors, the general form is:

• \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \)
• Result of subtraction: \( \mathbf{a} - \mathbf{b} = \langle a_1 - b_1, a_2 - b_2 \rangle \)

For \( \mathbf{u} - \mathbf{v} \), subtraction can represent the difference in position or change in direction between two points, making these operations very practical in fields like physics and engineering.
Moreover, understanding vector operations opens a pathway to more advanced topics like dot product and cross product, which involve multiplying vectors in ways that reveal information about their direction and orientation.
Overall, mastering vector operations gives you a strong foundation for solving more complex problems in multidimensional spaces.