Problem 55 - Algebra and Trigonometry with Analytic Geometry

Question

Find a vector of magnitude 6 that has the opposite direction of \(\mathbf{a}=4 \mathbf{i}-7 \mathbf{j}\).

Step-by-Step Solution

Answer

The opposite vector with magnitude 6 is \( \mathbf{b} = -\frac{24}{\sqrt{65}}\mathbf{i} + \frac{42}{\sqrt{65}}\mathbf{j} \).

Step 1: Find the Magnitude of the Vector

Calculate the magnitude of the vector \( \mathbf{a} = 4\mathbf{i} - 7\mathbf{j} \). The formula for the magnitude (length) is \( |\mathbf{a}| = \sqrt{4^2 + (-7)^2} \). Simplifying this expression, we get \( |\mathbf{a}| = \sqrt{16 + 49} = \sqrt{65} \).

Step 2: Determine the Unit Vector in the Opposite Direction

To find the unit vector in the opposite direction of \( \mathbf{a} \), first find the unit vector of \( \mathbf{a} \) by dividing each component by its magnitude: \( \mathbf{u} = \frac{4}{\sqrt{65}}\mathbf{i} - \frac{7}{\sqrt{65}}\mathbf{j} \). To reverse the direction, multiply this unit vector by \(-1\), resulting in \( \mathbf{-u} = -\frac{4}{\sqrt{65}}\mathbf{i} + \frac{7}{\sqrt{65}}\mathbf{j} \).

Step 3: Scale the Unit Vector to the Desired Magnitude

Multiply the reversed unit vector \( \mathbf{-u} \) by the desired magnitude of 6: \( \mathbf{b} = 6 \left( -\frac{4}{\sqrt{65}}\mathbf{i} + \frac{7}{\sqrt{65}}\mathbf{j} \right) \). Performing the multiplication, we obtain \( \mathbf{b} = -\frac{24}{\sqrt{65}}\mathbf{i} + \frac{42}{\sqrt{65}}\mathbf{j} \). This vector is the desired vector with magnitude 6, in the opposite direction of \( \mathbf{a} \).

Key Concepts

Opposite Direction in Vectors

When we talk about vectors, direction is as important as magnitude. Imagine driving a car: moving speedily in the wrong direction won't get you where you need to go. Similarly, the opposite direction for vectors means pointing in a completely different way. If a vector originally points north, its opposite will point south.

To find a vector in the opposite direction, you simply change the signs of its components. For example, the vector \(\mathbf{a} = 4\mathbf{i} - 7\mathbf{j}\) has components \(4\) and \(-7\). Its opposite would be \(-4\mathbf{i} + 7\mathbf{j}\).

This change in sign reflects flipping the vector 180 degrees around the origin, meaning our new vector points exactly the other way from the original. This is particularly useful in navigation, physics, and any field where understanding the course or force needs adjustment.

Understanding Unit Vectors

A unit vector is a fundamental concept in vector mathematics. It is simply a vector with a magnitude of 1. Think of it as a direction indicator without any scale. Knowing the direction is often crucial, even if the actual size is not needed at the moment.

To calculate a unit vector \( \mathbf{u} \) for any given vector \( \mathbf{a} = 4\mathbf{i} - 7\mathbf{j} \), divide each component of the vector by the vector's magnitude. We've already seen how to compute the magnitude using \(|\mathbf{a}| = \sqrt{65}\). Thus, the unit vector in the same direction would be \( \frac{4}{\sqrt{65}}\mathbf{i} - \frac{7}{\sqrt{65}}\mathbf{j} \).

Unit vectors are instrumental because they provide a simple, standardized way to describe directions, making them a staple in physics and engineering applications. Once you have a unit vector, you can scale it to any desired magnitude.

Vector Scaling

Vector scaling involves adjusting the length of a vector without altering its direction, unless specifically directed. Take a unit vector: it shows direction without size. By multiplying it by a scalar, you can extend or shrink this vector to the precise magnitude required.

To see this in practice, consider the unit vector we derived, \(\mathbf{-u} = -\frac{4}{\sqrt{65}}\mathbf{i} + \frac{7}{\sqrt{65}}\mathbf{j}\). By scaling, which means multiplying each component by the desired magnitude, here 6, each component changes accordingly:

• \(-\frac{24}{\sqrt{65}}\mathbf{i}\)
• \(+\frac{42}{\sqrt{65}}\mathbf{j}\)

This results in a new vector, directing the same path but now with a specific magnitude of 6. Scaling is intensely useful in modeling real-world phenomena, such as calculating forces or distances, where both direction and specific magnitude are crucial.