Problem 25 - University Calculus: Early Transcendentals

Question

Express each vector as a product of its length and direction. $$2 \mathbf{i}+\mathbf{j}-2 \mathbf{k}$$

Step-by-Step Solution

Answer

The vector can be expressed as \( 3 \left(\frac{2}{3} \mathbf{i} + \frac{1}{3} \mathbf{j} - \frac{2}{3} \mathbf{k}\right) \).

Step 1: Identify the Vector Components

The given vector is expressed as \( \mathbf{v} = 2 \mathbf{i} + \mathbf{j} - 2 \mathbf{k} \). The components are: \( x = 2 \), \( y = 1 \), and \( z = -2 \).

Step 2: Calculate the Length of the Vector

To find the length (magnitude) of the vector \( \mathbf{v} \), use the formula: \( \|\mathbf{v}\| = \sqrt{x^2 + y^2 + z^2} \). Plug in the components: \[ \|\mathbf{v}\| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3. \]

Step 3: Determine the Direction of the Vector

The direction of the vector is given by the unit vector, which is \( \frac{\mathbf{v}}{\|\mathbf{v}\|} \). Divide each component by the magnitude: \[ \text{Unit vector} = \left(\frac{2}{3}\right) \mathbf{i} + \left(\frac{1}{3}\right) \mathbf{j} + \left(\frac{-2}{3}\right) \mathbf{k}. \]

Step 4: Express the Vector as a Product of Length and Direction

Combine the magnitude and the unit vector to write the original vector as a product of its length and direction: \[ \mathbf{v} = 3 \left(\frac{2}{3} \mathbf{i} + \frac{1}{3} \mathbf{j} - \frac{2}{3} \mathbf{k}\right). \]

Key Concepts

Magnitude of a Vector

The magnitude of a vector is like its length in space. Imagine it as the "distance" the vector covers from its starting point to its endpoint. To find this length, we use a special formula. It's just an extension of the Pythagorean theorem to three dimensions. For any vector \( \mathbf{v} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k} \),we calculate the magnitude, noted as \( \|\mathbf{v}\| \),by using:

• 
  \( \|\mathbf{v}\| = \sqrt{x^2 + y^2 + z^2} \).

Plugging in values is all that's needed. In our example, for \( \mathbf{v} = 2 \mathbf{i} + \mathbf{j} - 2 \mathbf{k} \),we get:

• 
  \( \|\mathbf{v}\| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \).

This tells us the vector is 3 units long. Simple, right? It’s a matter of plugging into the formula and solving.

Unit Vector

A unit vector points in the same direction as the original vector, but has a length of exactly one. Think of it as a "direction indicator" without worrying about the length. Finding the unit vector is straightforward. You take your original vector \( \mathbf{v} \)and divide it by its magnitude \( \|\mathbf{v}\| \).This way, you shrink (or stretch) it to length one.For a vector \( \mathbf{v} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k} \),its unit vector is:

• 
  \( \text{Unit vector} = \frac{x}{\|\mathbf{v}\|} \mathbf{i} + \frac{y}{\|\mathbf{v}\|} \mathbf{j} + \frac{z}{\|\mathbf{v}\|} \mathbf{k} \).

In our case, for \( \mathbf{v} = 2 \mathbf{i} + \mathbf{j} - 2 \mathbf{k} \),we calculated a unit vector as:

• 
  \( \left(\frac{2}{3}\right) \mathbf{i} + \left(\frac{1}{3}\right) \mathbf{j} + \left(\frac{-2}{3}\right) \mathbf{k} \).

This is just a tiny version of the original vector, showing where it points.

Vector Components

Vector components tell you how much the vector moves in the directions of the coordinate axes. Think of it as breaking a vector down into parts that explain how far it stretches along the x, y, and z axes. It's like saying: "how much do I move east, north, and up to make this trip?".For the vector \( \mathbf{v} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k} \),the components, \( x \), \( y \),and \( z \),represent its influence in the x, y, and z directions, respectively. In our vector example \( 2 \mathbf{i} + \mathbf{j} - 2 \mathbf{k} \):

• The x component is 2 (meaning 2 steps in the x direction).
• The y component is 1 (meaning 1 step in the y direction).
• The z component is -2 (meaning going backwards 2 steps in the z direction).

These components are the building blocks of the vector, showing the total effect along each axis.