Problem 54 - Algebra and Trigonometry with Analytic Geometry

Question

Find a vector that has the opposite direction of \(8 \mathbf{i}-5 \mathbf{j}\) and (a) three times the magnitude (b) one-third the magnitude

Step-by-Step Solution

Answer

The opposite vector with three times the magnitude is \(-24 \mathbf{i} + 15 \mathbf{j}\) and with one-third the magnitude is \(-\frac{8}{3} \mathbf{i} + \frac{5}{3} \mathbf{j}\).

Step 1: Find the magnitude of the original vector

The original vector is \(8 \mathbf{i} - 5 \mathbf{j}\). The magnitude of this vector is calculated using the formula for the magnitude of a vector: \(\sqrt{x^2 + y^2}\). So, the magnitude is \(\sqrt{8^2 + (-5)^2} = \sqrt{64 + 25} = \sqrt{89}\).

Step 2: Calculate the unit vector

To find a vector with the opposite direction, we first find the unit vector, which has a magnitude of 1. The unit vector can be found by dividing the vector by its magnitude. For the vector \(8 \mathbf{i} - 5 \mathbf{j}\), the unit vector will be \(\left( \frac{8}{\sqrt{89}} \right) \mathbf{i} - \left( \frac{5}{\sqrt{89}} \right) \mathbf{j}\). Reversing the direction means negating this unit vector, so we have: \[ -\left( \frac{8}{\sqrt{89}} \right) \mathbf{i} + \left( \frac{5}{\sqrt{89}} \right) \mathbf{j} \].

Step 3: Calculate vector for three times the magnitude

To find a vector three times the magnitude and in the opposite direction, multiply the negated unit vector by \(3 \times \sqrt{89}\) (which is three times the original magnitude). Hence, the new vector is: \[ -3 \cdot 8 \mathbf{i} + 3 \cdot 5 \mathbf{j} = -24 \mathbf{i} + 15 \mathbf{j} \].

Step 4: Calculate vector for one-third the magnitude

To find a vector one-third the magnitude and in the opposite direction, multiply the negated unit vector by \(\frac{1}{3}\) of the original magnitude, \(\frac{\sqrt{89}}{3}\). Hence, the calculated vector is: \[ -\frac{8}{3} \mathbf{i} + \frac{5}{3} \mathbf{j} \].

Key Concepts

Vector Magnitude

When working with vectors, one of the key components we need to address is their magnitude. The magnitude of a vector gives us an idea of its size or length. For a vector in two-dimensional space like \(8 \mathbf{i} - 5 \mathbf{j}\), the magnitude is found using the Pythagorean theorem formula:

• \( \sqrt{x^2 + y^2} \)

In this case, \(x = 8\) and \(y = -5\). Therefore, the magnitude becomes \( \sqrt{8^2 + (-5)^2} = \sqrt{64 + 25} = \sqrt{89} \).
This value of \(\sqrt{89}\) tells us how long the vector is. Understanding the magnitude is crucial when we need to scale a vector or find vectors of specific lengths, as illustrated in the exercise.

Unit Vector

A unit vector is a vector with a magnitude of 1. It's useful when we wish to retain the direction of a vector but not its magnitude. To find a unit vector, we divide each component of the original vector by its magnitude. For \(8 \mathbf{i} - 5 \mathbf{j}\), we get:

• \(\left(\frac{8}{\sqrt{89}}\right) \mathbf{i} - \left(\frac{5}{\sqrt{89}}\right) \mathbf{j}\)

Reversing the direction simply involves negating the unit vector components:

• \(-\left(\frac{8}{\sqrt{89}}\right) \mathbf{i} + \left(\frac{5}{\sqrt{89}}\right) \mathbf{j}\)

This new vector has the direction opposite to that of the original vector, while still maintaining a magnitude of 1.

Opposite Direction Vector

Finding a vector in the opposite direction involves more than just negating the components of the vector. Firstly, consider the original vector \(8 \mathbf{i} - 5 \mathbf{j}\). Its opposite direction vector is obtained by negating it completely, leading to

• \(-8 \mathbf{i} + 5 \mathbf{j}\)

If we also want to adjust the magnitude, such as tripling or taking one-third of it, use the negated unit vector and scale it. This ensures you maintain the opposite direction while adjusting magnitude according to the problem's requirements, as demonstrated in the next sections.

Vector Scaling

Vector scaling refers to changing the length, or magnitude, of a vector while keeping its direction constant (or reversed, if desired).
This can be achieved by multiplying the vector by a scalar.

Example 1: Three Times the Magnitude

To make a vector that is three times the magnitude in the opposite direction, use the negated unit vector and multiply by three times the original magnitude \(3 \times \sqrt{89}\).

• Resulting in \(-24 \mathbf{i} + 15 \mathbf{j}\)

Example 2: One-third the Magnitude

For one-third the magnitude in the opposite direction, multiply using \(\frac{1}{3}\) the magnitude:

• \(-\frac{8}{3} \mathbf{i} + \frac{5}{3} \mathbf{j}\)

In both examples, adjusting the magnitude properly ensures the vector retains its opposite direction while altering its length as needed.