Problem 78 - Precalculus with Calculus Previews

Question

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

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: Calculate the Magnitude of Vector

To find the magnitude of vector \( \mathbf{v} \), we use the formula \( \|\mathbf{v}\| = \sqrt{a^2 + b^2 + c^2} \) for a vector \( \mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \). Here, \( a = 1 \), \( b = -3 \), and \( c = 2 \). So the magnitude is \( \|\mathbf{v}\| = \sqrt{1^2 + (-3)^2 + 2^2} = \sqrt{1 + 9 + 4} = \sqrt{14} \).

Step 2: Express the Unit Vector

The unit vector \( \mathbf{u} \) in the direction of \( \mathbf{v} \) is given by \( \mathbf{u} = \frac{1}{\|\mathbf{v}\|} \mathbf{v} \). Substitute the magnitude from Step 1: \( \mathbf{u} = \frac{1}{\sqrt{14}}(\mathbf{i} - 3\mathbf{j} + 2\mathbf{k}) \).

Step 3: Simplify the Unit Vector Expression

Simplify the expression for the unit vector \( \mathbf{u} \) by distributing \( \frac{1}{\sqrt{14}} \): \( \mathbf{u} = \frac{1}{\sqrt{14}}\mathbf{i} - \frac{3}{\sqrt{14}}\mathbf{j} + \frac{2}{\sqrt{14}}\mathbf{k} \).

Key Concepts

Vector Magnitude

The magnitude of a vector is a crucial concept in understanding its length or size without regard to direction. Calculating the magnitude involves finding the "length" of the vector using a formula derived from the Pythagorean theorem. For a 3D vector like \( \mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \), the magnitude, denoted as \( \|\mathbf{v}\| \), is given by:\[\|\mathbf{v}\| = \sqrt{a^2 + b^2 + c^2}\]This formula breaks down as follows:- **Square** each component of the vector: \( a^2 \), \( b^2 \), and \( c^2 \).- **Sum** the squares: \( a^2 + b^2 + c^2 \).- Take the **square root** of this sum to find the magnitude.In our exercise, with the vector \( \mathbf{v} = \mathbf{i} - 3\mathbf{j} + 2\mathbf{k} \), when we plug in \( a = 1 \), \( b = -3 \), and \( c = 2 \), the magnitude comes out to be \( \sqrt{14} \). Understanding the magnitude helps us use the vector effectively in further calculations, such as determining a unit vector.

Unit Vector Calculation

A unit vector is a vector that has a magnitude of 1. It is used to describe the direction of a vector without considering its length. The unit vector in the direction of \( \mathbf{v} \) is calculated by dividing each component of \( \mathbf{v} \) by its magnitude:\[\mathbf{u} = \frac{1}{\|\mathbf{v}\|} \mathbf{v}\]Here's how to do it:- **Find** the magnitude of \( \mathbf{v} \). In our exercise, this was \( \sqrt{14} \).- **Divide** each component of \( \mathbf{v} \) by this magnitude.So for \( \mathbf{v} = \mathbf{i} - 3\mathbf{j} + 2\mathbf{k} \), it leads to:\[ \mathbf{u} = \frac{1}{\sqrt{14}}\mathbf{i} - \frac{3}{\sqrt{14}}\mathbf{j} + \frac{2}{\sqrt{14}}\mathbf{k} \]Through this process, the unit vector \( \mathbf{u} \) maintains the direction of \( \mathbf{v} \) but is scaled down to a length of 1. This is a practical method to find direction by stripping away the magnitude.

Vector Direction

Grasping the vector direction is essential for applications where orientation is more important than length. Direction is represented by a unit vector which retains the vector's path while neutralizing the magnitude to 1. It boils down to understanding which way the vector points in space.The exercise provides \( \mathbf{v} = \mathbf{i} - 3\mathbf{j} + 2\mathbf{k} \), which establishes a specific direction by indicating how far the vector moves along the x, y, and z axes. By converting \( \mathbf{v} \) into a unit vector, we solely focus on this direction aspect. Unit vectors like \( \mathbf{u} = \frac{1}{\sqrt{14}}\mathbf{i} - \frac{3}{\sqrt{14}}\mathbf{j} + \frac{2}{\sqrt{14}}\mathbf{k} \) enable us to maintain directionality while reducing complexity by removing magnitude.These unit vectors are beneficial in:- **Graphical computations**, helping decide positions relative to each other.- **Physics**, where direction, not magnitude, dictates certain laws of motion and force.Understanding vector direction helps model realistic behaviors and accurately predicts movements and interactions in a 3D space.