Problem 7 - 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. $$\frac{3}{5}\mathbf{u}+\frac{4}{5} \mathbf{v}$$

Step-by-Step Solution

Answer

Component form: \( \langle \frac{1}{5}, \frac{14}{5} \rangle \). Magnitude: \( \frac{\sqrt{197}}{5} \).

Step 1: Scale Vector u

First, find the scaled version of vector \( \mathbf{u} = \langle 3, -2 \rangle \) by multiplying each component by \( \frac{3}{5} \). Follow the formula for scaling vectors: \( k\mathbf{a} = \langle kx, ky \rangle \). Thus, \( \frac{3}{5}\mathbf{u} = \left\langle \frac{3}{5} \times 3, \frac{3}{5} \times -2 \right\rangle = \langle \frac{9}{5}, -\frac{6}{5} \rangle \).

Step 2: Scale Vector v

Next, scale vector \( \mathbf{v} = \langle -2, 5 \rangle \) by multiplying each component by \( \frac{4}{5} \). So, \( \frac{4}{5}\mathbf{v} = \left\langle \frac{4}{5} \times -2, \frac{4}{5} \times 5 \right\rangle = \langle -\frac{8}{5}, \frac{20}{5} \rangle \).

Step 3: Add Scaled Vectors

Now add the scaled vectors from Step 1 and Step 2 to find the resulting vector. Use vector addition by adding the corresponding components: \( \left\langle \frac{9}{5}, -\frac{6}{5} \right\rangle + \left\langle -\frac{8}{5}, \frac{20}{5} \right\rangle = \left\langle \frac{9}{5} - \frac{8}{5}, -\frac{6}{5} + \frac{20}{5} \right\rangle = \langle \frac{1}{5}, \frac{14}{5} \rangle \).

Step 4: Find the Magnitude

To find the magnitude of a vector \( \langle x, y \rangle \), use the formula \( \sqrt{x^2 + y^2} \). For vector \( \langle \frac{1}{5}, \frac{14}{5} \rangle \), the magnitude is \( \sqrt{\left(\frac{1}{5}\right)^2 + \left(\frac{14}{5}\right)^2} \). Calculate this as: \( \sqrt{\frac{1}{25} + \frac{196}{25}} = \sqrt{\frac{197}{25}} = \frac{\sqrt{197}}{5} \).

Key Concepts

Vector Addition

Vector addition is a fundamental operation in vector mathematics that involves combining two or more vectors to produce a resultant vector. When we add vectors, we simply add their corresponding components. For example, if you have two vectors, say \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \), their sum is found as:

• \( \mathbf{a} + \mathbf{b} = \langle a_1 + b_1, a_2 + b_2 \rangle \)

In the context of our exercise, after scaling \( \mathbf{u} \) and \( \mathbf{v} \), we performed vector addition by adding the corresponding components of the scaled vectors:

• \( \langle \frac{9}{5}, -\frac{6}{5} \rangle + \langle -\frac{8}{5}, \frac{20}{5} \rangle \)
• \( = \langle \frac{9}{5} - \frac{8}{5}, -\frac{6}{5} + \frac{20}{5} \rangle \)
• \( = \langle \frac{1}{5}, \frac{14}{5} \rangle \)

Vector addition is graphical as well as algebraic, meaning it can be visualized as well by arranging vectors tip-to-tail to find the resultant vector. This operation is essential in physics and engineering, especially when dealing with forces.

Scalar Multiplication

Scalar multiplication involves multiplying a vector by a scalar (a real number). This operation scales a vector by stretching or compressing its magnitude, without changing its direction (unless the scalar is negative, which reverses it).

• Given a vector \( \mathbf{v} = \langle x, y \rangle \) and a scalar \( k \), the scalar multiplication is \( k \mathbf{v} = \langle kx, ky \rangle \).

Using the original exercise, to scale vector \( \mathbf{u} = \langle 3, -2 \rangle \) by \( \frac{3}{5} \), the operation was:

• \( \frac{3}{5} \mathbf{u} = \langle \frac{3}{5} \times 3, \frac{3}{5} \times -2 \rangle = \langle \frac{9}{5}, -\frac{6}{5} \rangle \)

Similarly, scaling vector \( \mathbf{v} = \langle -2, 5 \rangle \) by \( \frac{4}{5} \):

• \( \frac{4}{5} \mathbf{v} = \langle \frac{4}{5} \times -2, \frac{4}{5} \times 5 \rangle = \langle -\frac{8}{5}, \frac{20}{5} \rangle \)

This operation is commonly used to adjust vector magnitudes, which is useful in scenarios such as modifying velocities or forces in physics.

Magnitude of a Vector

The magnitude of a vector, also known as its length, is a measure of how long the vector is. It is calculated using the Pythagorean theorem. For a vector \( \mathbf{v} = \langle x, y \rangle \), the formula to find the magnitude \( \| \mathbf{v} \| \) is:

• \( \| \mathbf{v} \| = \sqrt{x^2 + y^2} \)

In the given problem, after performing vector addition and obtaining the resulting vector \( \langle \frac{1}{5}, \frac{14}{5} \rangle \), we used the magnitude formula to find its length:

• Calculate: \( \sqrt{\left(\frac{1}{5}\right)^2 + \left(\frac{14}{5}\right)^2} \)
• Result: \( = \sqrt{\frac{1}{25} + \frac{196}{25}} = \sqrt{\frac{197}{25}} \)
• Simplified: \( = \frac{\sqrt{197}}{5} \)

Understanding the magnitude is crucial for determining distances and lengths in physics and engineering applications.