Problem 50 - Advanced Engineering Mathematics

Question

Find a unit vector in the same direction as \(\mathbf{a}=\mathbf{i}-3 \mathbf{j}+2 \mathbf{k}\).

Step-by-Step Solution

Answer

The unit vector is \(\frac{1}{\sqrt{14}} \mathbf{i} - \frac{3}{\sqrt{14}} \mathbf{j} + \frac{2}{\sqrt{14}} \mathbf{k}.\)

Step 1: Find the Magnitude of \(\mathbf{a}\)

The magnitude of a vector \(\mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k}\) is given by \(\|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2 + a_3^2}\). For the vector \(\mathbf{a} = \mathbf{i} - 3 \mathbf{j} + 2 \mathbf{k}\), the components are \(a_1=1\), \(a_2=-3\), and \(a_3=2\). Thus, the magnitude is \(\|\mathbf{a}\| = \sqrt{1^2 + (-3)^2 + 2^2} = \sqrt{1 + 9 + 4} = \sqrt{14}\).

Step 2: Normalize the Vector \(\mathbf{a}\)

To find a unit vector in the same direction as \(\mathbf{a}\), divide each component of \(\mathbf{a}\) by the magnitude calculated in Step 1. The unit vector \(\mathbf{a}^{\ast}\) is given by: \[\mathbf{a}^{\ast} = \frac{\mathbf{a}}{\|\mathbf{a}\|} = \frac{1}{\sqrt{14}} \mathbf{i} - \frac{3}{\sqrt{14}} \mathbf{j} + \frac{2}{\sqrt{14}} \mathbf{k}.\]

Key Concepts

Unit Vector

A unit vector is a vector that has a magnitude of exactly 1. It retains the same direction as the original vector, but its length is adjusted. This is achieved by dividing the vector by its magnitude. Unit vectors are essential in vector mathematics as they are often used to define directions in space without concern for magnitude.

Think of a unit vector as a way to normalize a vector. Normalization scales the vector to a length of 1 while preserving its direction. When working with vectors, it is sometimes useful to separate the direction from the magnitude, especially in applications like computer graphics and physics simulations.

To find a unit vector in the same direction as a vector \(\mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k}\) you would calculate:

• Calculate the magnitude of vector \(\mathbf{a}\).
• Divide each component of \(\mathbf{a}\) by its magnitude.

This results in a vector that is parallel to the original, but with a total length of precisely 1.

Magnitude of a Vector

The magnitude of a vector is a measure of its length. It quantifies how long a vector is and is always a non-negative number. Understanding the magnitude is crucial because it tells us the scale of the vector, which is often related to the magnitude of a quantity in real-world applications, such as speed or force.

For a vector \(\mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k}\), its magnitude \(\|\mathbf{a}\|\)is calculated using the formula:\(\|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2 + a_3^2}\).

This is essentially the application of the Pythagorean theorem extended to three dimensions.

Here's how it works step by step:

• Square each component of the vector.
• Add the results of these squares together.
• Take the square root of the sum.

Calculating the magnitude is the first step in vector normalization, providing the scale needed to adjust the vector to become a unit vector.

Direction of a Vector

The direction of a vector is described by its orientation in space. When we talk about the direction, we're focusing on where the vector points, regardless of how far it stretches.

A vector's direction can be visualized as an arrow pointing from the origin to a point in a coordinate system. Each vector can point in one unique direction, given by its components. These components \(a_1, a_2, a_3\) define how far the vector moves along each of the coordinate planes.

In some cases, angles relative to coordinate axes are used to describe direction more precisely.

Understanding a vector’s direction is essential in navigation, physics calculations, and any application where orientation is as important as magnitude. When normalizing a vector, we preserve this direction but reduce the vector to a unit vector. This retention makes it possible to work with direction as an abstract component without the influence of vector length.