Problem 27 - Algebra and Trigonometry

Question

Find a vector of magnitude 15 that is parallel to \(4 \mathbf{i}-3 \mathbf{j}\)

Step-by-Step Solution

Answer

The vector is \ \(12 \mathbf{i} - 9 \mathbf{j}\).

Step 1: Find the magnitude of the given vector

The given vector is \ \(4 \mathbf{i} - 3 \mathbf{j}\). Calculate its magnitude using the formula \ \(\|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2}\), where \ \(a_1\) and \ \(a_2\) are the components of the vector. Here, \ \(a_1 = 4\) and \ \(a_2 = -3\). \(\|4 \mathbf{i} - 3 \mathbf{j}\| = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5\).

Step 2: Determine the unit vector in the direction of the given vector

A unit vector in the direction of a vector \ \(\mathbf{a}\) is given by \ \(\frac{\mathbf{a}}{\|\mathbf{a}\|}\). Use the components of the original vector and its magnitude obtained in Step 1. The unit vector is: \ \(\frac{4 \mathbf{i} - 3 \mathbf{j}}{5} = \frac{4}{5} \mathbf{i} - \frac{3}{5} \mathbf{j}\).

Step 3: Scale the unit vector to the desired magnitude

To get a vector of magnitude 15 in the same direction, multiply the unit vector by 15: \ \(15 \left(\frac{4}{5} \mathbf{i} - \frac{3}{5} \mathbf{j}\right) = 15 \times \frac{4}{5} \mathbf{i} - 15 \times \frac{3}{5} \mathbf{j} = 12 \mathbf{i} - 9 \mathbf{j}\).

Key Concepts

Vector Calculation

Vectors are mathematical objects with both magnitude and direction. They help us represent quantities like displacement, velocity, and force.
When dealing with vectors, we often need to perform calculations. These include addition, subtraction, dot product, and cross product.

• Addition: Combine two vectors by adding their corresponding components.
• Subtraction: Subtract one vector from another by subtracting their components.
• Dot Product: Multiply corresponding components of two vectors and add the results.
• Cross Product: A vector perpendicular to two given vectors, found using the determinant of a matrix.

In this exercise, we specifically focus on transforming an existing vector by adjusting its magnitude.

Vector Magnitude

The magnitude of a vector is its length and is found using the Pythagorean theorem in a 2-dimensional space.
For a vector \(\boldsymbol{a} = a_1 \boldsymbol{i} + a_2 \boldsymbol{j}\), you calculate its magnitude using:
\[\|\boldsymbol{a}\| = \sqrt{a_1^2 + a_2^2}\]
In our example, the vector is \(4 \boldsymbol{i} - 3 \boldsymbol{j}\).
Calculate its magnitude as follows: \(\|\boldsymbol{a}\| = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5\)
A vector magnitude is crucial for determining other properties like direction and unit vectors.

Unit Vector

A unit vector has a magnitude of 1 and points in the same direction as the original vector.
It is found by dividing the vector by its magnitude.
For our vector \(4 \boldsymbol{i} - 3 \boldsymbol{j}\) with magnitude 5, the unit vector is given by:
\[ \frac{4 \boldsymbol{i} - 3 \boldsymbol{j}}{5}\] which simplifies to:
\[ \frac{4}{5} \boldsymbol{i} - \frac{3}{5} \boldsymbol{j}\]
Unit vectors are useful because they allow us to scale vectors easily.
To get a vector of magnitude 15 in the same direction, multiply this unit vector by 15:
\[ 15 \left( \frac{4}{5} \boldsymbol{i} - \frac{3}{5} \boldsymbol{j} \right) = 12 \boldsymbol{i} - 9 \boldsymbol{j}\]