Problem 33 - Thomas Calculus

Question

Find a vector of magnitude 7 in the direction of \(\mathbf{v}=12 \mathbf{i}-5 \mathbf{k}\)

Step-by-Step Solution

Answer

The vector is \( \frac{84}{13}\mathbf{i} - \frac{35}{13}\mathbf{k} \).

Step 1: Find the unit vector

First, we need to find the unit vector in the direction of the given vector \( \mathbf{v} = 12\mathbf{i} - 5\mathbf{k} \). The unit vector \( \mathbf{u} \) in the direction of \( \mathbf{v} \) is given by \( \mathbf{u} = \frac{\mathbf{v}}{||\mathbf{v}||} \).

Step 2: Calculate the magnitude of \( \mathbf{v} \)

The magnitude of the vector \( \mathbf{v} = 12\mathbf{i} - 5\mathbf{k} \) is calculated using the formula \( ||\mathbf{v}|| = \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \).

Step 3: Express the unit vector

Now that we know the magnitude of \( \mathbf{v} \) is 13, the unit vector \( \mathbf{u} \) is \( \mathbf{u} = \frac{\mathbf{v}}{13} = \frac{12\mathbf{i} - 5\mathbf{k}}{13} = \frac{12}{13}\mathbf{i} - \frac{5}{13}\mathbf{k} \).

Step 4: Scale the unit vector

To find a vector of magnitude 7 in the same direction, we multiply the unit vector by 7. Thus, the vector is \( 7 \mathbf{u} = 7 \left( \frac{12}{13}\mathbf{i} - \frac{5}{13}\mathbf{k} \right) = \frac{84}{13}\mathbf{i} - \frac{35}{13}\mathbf{k} \).

Step 5: Verify the magnitude

Check the magnitude of the resulting vector \( 7 \mathbf{u} = \frac{84}{13}\mathbf{i} - \frac{35}{13}\mathbf{k} \): \( \sqrt{\left(\frac{84}{13}\right)^2 + \left(\frac{-35}{13}\right)^2} = \sqrt{\frac{7056}{169} + \frac{1225}{169}} = \sqrt{\frac{8281}{169}} = \sqrt{49} = 7 \). The calculation confirms the correct magnitude.

Key Concepts

Unit Vector

A unit vector is a vector that has a magnitude of exactly 1. It points in the same direction as the original vector, but is rescaled to have this unit length. To find the unit vector, we take the original vector and divide it by its magnitude.

This allows us to strip the vector down to its essence: its direction. Consider a vector \( \mathbf{v} = 12\mathbf{i} - 5\mathbf{k} \). First, you need to calculate its magnitude, which in this case is 13 (as shown in the solution). The unit vector \( \mathbf{u} \) is then given by:

• \( \mathbf{u} = \frac{1}{13}(12\mathbf{i} - 5\mathbf{k}) \)
• This ensures our vector is of unit length, pointing in the same direction as \( \mathbf{v} \).

Grasping unit vectors is crucial when scaling vectors to different magnitudes without altering their original direction.

Vector Calculation

Vector calculation involves operations such as addition, subtraction, and multiplication of vectors to obtain another vector. In this exercise, vector multiplication is crucial. Specifically, scalar multiplication where the vector is multiplied by a real number.

Consider the unit vector \( \mathbf{u} = \frac{12}{13} \mathbf{i} - \frac{5}{13} \mathbf{k} \). If we want a vector in the same direction but with a magnitude of 7, we multiply each component of \( \mathbf{u} \) by 7, giving:

• \( 7 \mathbf{u} = 7 \left( \frac{12}{13} \mathbf{i} - \frac{5}{13} \mathbf{k} \right) = \frac{84}{13} \mathbf{i} - \frac{35}{13} \mathbf{k} \)
• This results in a vector that is exactly 7 times longer than the unit vector in the same direction.

In vector calculations, always ensure the direction remains consistent when altering the magnitude.

Magnitude of Vector

The magnitude of a vector is its length. It is a measure of how long the vector is in the space it's placed. For a vector expressed in rectangular coordinates, like \( \mathbf{v} = 12\mathbf{i} - 5\mathbf{k} \), we find its magnitude using the Pythagorean theorem.

The formula for the magnitude of vector \( \mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \) (considering \( b = 0 \) in this case) is:

• \( ||\mathbf{v}|| = \sqrt{a^2 + b^2 + c^2} \)
• For our vector, this becomes \( ||\mathbf{v}|| = \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \).

Understanding how to compute vector magnitudes is fundamental in fields ranging from physics to computer graphics, where exact dimensions matter.