Problem 8 - 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. $$-\frac{5}{13}\mathbf{u}+\frac{12}{13}\mathbf{v}$$

Step-by-Step Solution

Answer

The component form is \(\langle -3, \frac{70}{13} \rangle\); the magnitude is \(\frac{\sqrt{6421}}{13}\).

Step 1: Scale Vector \(\mathbf{u}\)

Multiply vector \(\mathbf{u}=\langle 3,-2\rangle\) by the scalar \(-\frac{5}{13}\). Compute: \(-\frac{5}{13} \times 3 = -\frac{15}{13}\) and \(-\frac{5}{13} \times (-2) = \frac{10}{13}\). The scaled vector is \(\langle -\frac{15}{13}, \frac{10}{13} \rangle\).

Step 2: Scale Vector \(\mathbf{v}\)

Multiply vector \(\mathbf{v}=\langle -2,5\rangle\) by the scalar \(\frac{12}{13}\). Compute: \(\frac{12}{13} \times (-2) = -\frac{24}{13}\) and \(\frac{12}{13} \times 5 = \frac{60}{13}\). The scaled vector is \(\langle -\frac{24}{13}, \frac{60}{13} \rangle\).

Step 3: Add the Scaled Vectors

Add the results from Step 1 and Step 2 to find the component form of the vector. \(\left\langle -\frac{15}{13}, \frac{10}{13} \right\rangle + \left\langle -\frac{24}{13}, \frac{60}{13} \right\rangle = \left\langle -\left(\frac{15}{13} + \frac{24}{13}\right), \left(\frac{10}{13} + \frac{60}{13}\right) \right\rangle \). This simplifies to \(\langle -\frac{39}{13}, \frac{70}{13} \rangle \), which further simplifies to \(\langle -3, \frac{70}{13} \rangle\).

Step 4: Calculate the Magnitude

Use the formula for the magnitude of a vector \(\langle a, b \rangle\), which is \(\sqrt{a^2 + b^2}\). Substitute \(a = -3\) and \(b = \frac{70}{13}\): \(\sqrt{(-3)^2 + \left(\frac{70}{13}\right)^2}\). Calculate \((-3)^2 = 9\) and \(\left(\frac{70}{13}\right)^2 = \frac{4900}{169}\). The magnitude is: \[\sqrt{9 + \frac{4900}{169}} = \sqrt{\frac{1521}{169} + \frac{4900}{169}} = \sqrt{\frac{6421}{169}} = \frac{\sqrt{6421}}{13}\].

Key Concepts

Vector Addition

Vector addition involves combining two or more vectors to produce a new single vector. This is done by adding the vectors' respective components. In the given exercise, we have the vectors \(\mathbf{u}=\langle 3,-2\rangle\) and \(\mathbf{v}=\langle -2,5\rangle\).
When these vectors are scaled by a scalar, as in the solution, they must be rewritten according to each new component's value. Once scaled, the corresponding components from each vector are added together:

• Add the first components: \(-\frac{15}{13} + -\frac{24}{13} = -\frac{39}{13}\) which simplifies to \(-3\).
• Add the second components: \(\frac{10}{13} + \frac{60}{13} = \frac{70}{13}\).

Thus, the resulting vector from this addition process is \(\langle -3, \frac{70}{13} \rangle\).

Scalar Multiplication

Scalar multiplication refers to multiplying a vector by a scalar, a real number, which changes the magnitude of the vector but not its direction unless the scalar is negative. In practical terms, each component of the vector is multiplied by the scalar.
In the problem, vector \(\mathbf{u}=\langle 3,-2\rangle\) is multiplied by \(-\frac{5}{13}\):

• First component: \(-\frac{5}{13} \times 3 = -\frac{15}{13}\)
• Second component: \(-\frac{5}{13} \times -2 = \frac{10}{13}\)

Similarly, vector \(\mathbf{v}=\langle -2,5\rangle\) is multiplied by \(\frac{12}{13}\):

• First component: \(\frac{12}{13} \times -2 = -\frac{24}{13}\)
• Second component: \(\frac{12}{13} \times 5 = \frac{60}{13}\)

This operation adjusts each vector to fit the calculated scale, ensuring accurate results in further calculations.

Magnitude of a Vector

The magnitude of a vector is a measure of its length, determined using the Pythagorean theorem in a geometric sense. It is calculated using the formula \(\sqrt{a^2 + b^2}\), where \(a\) and \(b\) are the components of the vector.
In our solution, we start with the vector \(\langle -3, \frac{70}{13} \rangle\) obtained from the previous steps.

• Square each component: \(a = -3 \, a^2 = 9\) and \(b = \frac{70}{13} \, b^2 = \frac{4900}{169}\).
• Add the squares: \(9 + \frac{4900}{169}\).
• Simplify and find the square root: \(\sqrt{\frac{6421}{169}}\).
• Finally, obtain the magnitude: \(\frac{\sqrt{6421}}{13}\).

This magnitude evaluation confirms the length of the resulting vector.

Component Form

The component form of a vector is a simple way of describing a vector in terms of its horizontal (x-axis) and vertical (y-axis) components. A vector is expressed in the form \(\langle a, b \rangle\), where \(a\) and \(b\) are its components along each axis.
When performing vector operations such as addition or scalar multiplication, the result must again be expressed in component form. In this exercise, after scaling and adding vectors \(\mathbf{u}\) and \(\mathbf{v}\), we arrived at the vector \(\langle -3, \frac{70}{13} \rangle\):

• The first component, -3, represents the horizontal change.
• The second component, \(\frac{70}{13}\), denotes the vertical change.

Expressing vectors in component form is vital for clear communication and further calculations in vector calculus.