Problem 30 - Thomas Calculus

Question

In Exercises 25–30, express each vector as a product of its length and direction. $$ \frac{\mathbf{i}}{\sqrt{3}}+\frac{\mathbf{j}}{\sqrt{3}}+\frac{\mathbf{k}}{\sqrt{3}} $$

Step-by-Step Solution

Answer

The vector is \( 1 \times \left( \frac{\mathbf{i}}{\sqrt{3}} + \frac{\mathbf{j}}{\sqrt{3}} + \frac{\mathbf{k}}{\sqrt{3}} \right) \).

Step 1: Identify the Given Vector

The given vector is \( \frac{\mathbf{i}}{\sqrt{3}} + \frac{\mathbf{j}}{\sqrt{3}} + \frac{\mathbf{k}}{\sqrt{3}} \). The components of this vector are \( \frac{1}{\sqrt{3}} \) for each of \( \mathbf{i}, \mathbf{j}, \mathbf{k} \).

Step 2: Calculate the Magnitude of the Vector

The magnitude \( \| \mathbf{v} \| \) of a vector \( \mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \) is calculated using the formula \( \| \mathbf{v} \| = \sqrt{a^2 + b^2 + c^2} \). Substitute the components into the formula: \[ \| \mathbf{v} \| = \sqrt{ \left(\frac{1}{\sqrt{3}}\right)^2 + \left(\frac{1}{\sqrt{3}}\right)^2 + \left(\frac{1}{\sqrt{3}}\right)^2 } = \sqrt{ \frac{1}{3} + \frac{1}{3} + \frac{1}{3} } = \sqrt{1} = 1 \].

Step 3: Express Vector as Product of Length and Direction

The direction vector is obtained by dividing each component of the original vector by its magnitude. Since the magnitude is 1, the direction vector is the same as the original vector, \( \mathbf{v} = \frac{\mathbf{i}}{\sqrt{3}} + \frac{\mathbf{j}}{\sqrt{3}} + \frac{\mathbf{k}}{\sqrt{3}} \). Thus, the vector expressed as a product of its length and direction is \( 1 \times \left( \frac{\mathbf{i}}{\sqrt{3}} + \frac{\mathbf{j}}{\sqrt{3}} + \frac{\mathbf{k}}{\sqrt{3}} \right) \).

Key Concepts

Vector Magnitude

When we talk about vector magnitude, we are essentially discussing the length of a vector. Imagine a vector as an arrow pointing in space - the magnitude would be how long that arrow is. To calculate this, we use a special formula that takes into account each of the vector's components. These components are the pieces of the vector in the x, y, and z directions.

• For a vector represented as \( \mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \), the magnitude is given by \( \| \mathbf{v} \| = \sqrt{a^2 + b^2 + c^2} \).
• This calculates the square root of the sum of the squares of each component.

For the vector \( \frac{\mathbf{i}}{\sqrt{3}} + \frac{\mathbf{j}}{\sqrt{3}} + \frac{\mathbf{k}}{\sqrt{3}} \), each component is \( \frac{1}{\sqrt{3}} \). When we plug these into our magnitude formula, we get: \[\| \mathbf{v} \| = \sqrt{ \left(\frac{1}{\sqrt{3}}\right)^2 + \left(\frac{1}{\sqrt{3}}\right)^2 + \left(\frac{1}{\sqrt{3}}\right)^2 } = \sqrt{1} = 1.\] This means our vector is actually equal in length to 1 unit, making it very straightforward to work with!

Direction Vectors

Direction vectors are important because they describe the orientation or the direction of a vector in space, without worrying about its length. It's like knowing which way your arrow points, without caring how long it is.
To find a direction vector, you divide the vector’s components by its magnitude.

• This is because direction vectors are essentially unit vectors, meaning they have a magnitude of 1.
• Unit vectors simplify the comparison of vector directions because they are reduced to just showing direction.

Our original vector \( \frac{\mathbf{i}}{\sqrt{3}} + \frac{\mathbf{j}}{\sqrt{3}} + \frac{\mathbf{k}}{\sqrt{3}} \) has a magnitude of 1, which means it's already a unit vector! This implies it's already in its simplest direction form.You'll often use direction vectors to determine angles between vectors, assist in navigating paths, or describe orientations.

Vector Components

Understanding vector components is like breaking down a vector into its essential parts, showing how much of the vector is in each direction - think of this as the x, y, and z directions in 3D space. These directions are often represented by the unit vectors \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \).

• A component is a scalar multiplier of these unit vectors.
• In formulas, a vector \( \mathbf{v} \) is expressed as \( a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \), where \( a \), \( b \), and \( c \) are the components.

For our specific problem, the vector \( \mathbf{v} = \frac{\mathbf{i}}{\sqrt{3}} + \frac{\mathbf{j}}{\sqrt{3}} + \frac{\mathbf{k}}{\sqrt{3}} \) showcases that each component in the i, j, and k directions is \( \frac{1}{\sqrt{3}} \).
This tells us that the vector is equally distributed along each direction, forming an equal angle with the x, y, and z axes. Understanding this distribution helps compute various phenomena in physics and engineering like equilibrium positions, forces, and velocities.