Problem 30 - Advanced Engineering Mathematics

Question

\( \mathbf{a}=\langle 2,8\rangle\) and \(\mathbf{b}=\langle 3,4\rangle .\) Find a unit vector in the same direction as the given vector. \(2 \mathbf{a}-3 \mathbf{b}\)

Step-by-Step Solution

Answer

The unit vector is \(\langle \frac{-5}{\sqrt{41}}, \frac{4}{\sqrt{41}} \rangle\).

Step 1: Calculate the Vector

First, calculate the vector resulting from the operation \(2 \mathbf{a} - 3 \mathbf{b}\). Given \(\mathbf{a} = \langle 2, 8 \rangle\) and \(\mathbf{b} = \langle 3, 4 \rangle\), the operation becomes:\[2 \cdot \langle 2, 8 \rangle - 3 \cdot \langle 3, 4 \rangle = \langle 4, 16 \rangle - \langle 9, 12 \rangle\]This yields:\[\langle 4 - 9, 16 - 12 \rangle = \langle -5, 4 \rangle\].So the resulting vector is \(\langle -5, 4 \rangle\).

Step 2: Calculate the Magnitude of the Vector

Next, find the magnitude (norm) of the vector \(\langle -5, 4 \rangle\). Use the formula for the magnitude of a vector \(v = \langle x, y \rangle\), which is \(\|v\| = \sqrt{x^2 + y^2}\).\[\|\langle -5, 4 \rangle\| = \sqrt{(-5)^2 + 4^2} = \sqrt{25 + 16} = \sqrt{41}\]

Step 3: Find the Unit Vector

To find the unit vector in the same direction as \(\langle -5, 4 \rangle\), divide the vector by its magnitude. The unit vector \(\mathbf{u}\) is given by:\[\mathbf{u} = \frac{1}{\sqrt{41}} \cdot \langle -5, 4 \rangle = \langle \frac{-5}{\sqrt{41}}, \frac{4}{\sqrt{41}} \rangle\]

Step 4: Simplify the Unit Vector

While the unit vector can remain in this form, sometimes it is rationalized for expression purposes:\[\mathbf{u} = \langle \frac{-5}{\sqrt{41}}, \frac{4}{\sqrt{41}} \rangle \approx \langle -0.780, 0.624 \rangle\]This is the unit vector in the same direction as \(2 \mathbf{a} - 3 \mathbf{b}\).

Key Concepts

Vector Operations

Vector operations form the building blocks of vector calculus. They allow us to manipulate vectors through simple algebraic steps.

One of the key operations is *scalar multiplication*. This multiplies each component of a vector by a scalar (number). For example, multiplying vector \( \mathbf{a} = \langle 2,8 \rangle\) by 2 gives \( 2 \cdot \langle 2, 8 \rangle = \langle 4, 16 \rangle\).

Next is *vector addition and subtraction*. This involves adding or subtracting corresponding components from two vectors. For instance, given vectors \(\mathbf{a} = \langle 4, 16 \rangle\) and \( \mathbf{b} = \langle 9, 12 \rangle\), vector subtraction yields \(\langle 4 - 9, 16 - 12 \rangle = \langle -5, 4 \rangle\).

Understanding these basic operations allows you to combine vectors and determine new directions and magnitudes.

Magnitude of a Vector

The magnitude of a vector, often represented by two vertical bars \(\|v\|\), measures its length or size. It's always a positive number.

For a vector \(\langle x, y \rangle\), you use the *Pythagorean theorem* to find its magnitude: \[\|v\| = \sqrt{x^2 + y^2}\]This formula comes from considering the vector as the hypotenuse of a right triangle formed by its components.

For example, the magnitude of the vector \(\langle -5, 4 \rangle\) is computed as:\[\sqrt{(-5)^2 + 4^2} = \sqrt{25 + 16} = \sqrt{41}\]The magnitude is useful to determine the overall 'size' of the vector, and it plays a crucial role in finding unit vectors.

Unit Vector

A unit vector is a vector with a magnitude of 1. It points in the same direction as the original vector but is scaled down to a length of one unit.

To find a unit vector \(\mathbf{u}\) from a given vector \(v\), divide each component of the vector by its magnitude: \[\mathbf{u} = \frac{1}{\|v\|} \cdot v\]This will not change the direction of the vector, only its length.

Take the vector \(\langle -5, 4 \rangle\) as an example. To find its unit vector, divide each component by \(\sqrt{41}\): \[\mathbf{u} = \langle \frac{-5}{\sqrt{41}}, \frac{4}{\sqrt{41}} \rangle\]This process ensures the new vector is still aligned along the same direction line as \(v\), but now precisely measures 1 unit in length.

Vector Subtraction

Vector subtraction effectively finds the difference between two vectors, showing how one vector changes into another.

This operation is quite straightforward: subtract the corresponding components of one vector from another. Using vectors \(\mathbf{a} = \langle 2, 8 \rangle\) and \(\mathbf{b} = \langle 3, 4 \rangle\), the subtraction \[2\mathbf{a} - 3\mathbf{b}\]becomes\[\langle 4, 16 \rangle - \langle 9, 12 \rangle = \langle 4-9, 16-12 \rangle = \langle -5, 4\rangle\]This new vector, \(\langle -5, 4 \rangle\), represents how far and in what direction you must move from the endpoint of \(3\mathbf{b}\) to the endpoint of \(2\mathbf{a}\).

Vector subtraction is crucial in various fields, from physics to computer graphics, as it provides insights into changes, trends, and relative positions.