Problem 25 - Thomas Calculus

Question

In Exercises 25–30, 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 is \( 3 \left( \frac{2}{3} \mathbf{i} + \frac{1}{3} \mathbf{j} - \frac{2}{3} \mathbf{k} \right) \).

Step 1: Find the Magnitude of the Vector

To express the vector as a product of its length and direction, first calculate the magnitude of the vector \( \mathbf{v} = 2 \mathbf{i} + \mathbf{j} - 2 \mathbf{k} \). Use the formula for the magnitude of a vector: \[\|\mathbf{v}\| = \sqrt{a^2 + b^2 + c^2}\]where \( a = 2 \), \( b = 1 \), and \( c = -2 \).Substitute these values into the formula:\[\|\mathbf{v}\| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3\]

Step 2: Determine the Direction of the Vector

The direction of the vector is given by the unit vector, which is the original vector divided by its magnitude. The unit vector \( \mathbf{u} \) is calculated as:\[\mathbf{u} = \frac{1}{\|\mathbf{v}\|}(2 \mathbf{i} + \mathbf{j} - 2 \mathbf{k})\]Substitute the magnitude we found:\[\mathbf{u} = \frac{1}{3}(2 \mathbf{i} + \mathbf{j} - 2 \mathbf{k}) = \frac{2}{3} \mathbf{i} + \frac{1}{3} \mathbf{j} - \frac{2}{3} \mathbf{k}\]

Step 3: Express Vector as a Product of Length and Direction

Finally, express the vector \( \mathbf{v} = 2 \mathbf{i} + \mathbf{j} - 2 \mathbf{k} \) as a product of its length and direction. Using the magnitude (length) 3 and the unit vector \( \mathbf{u} = \frac{2}{3} \mathbf{i} + \frac{1}{3} \mathbf{j} - \frac{2}{3} \mathbf{k} \), the expression is:\[\mathbf{v} = 3 \left( \frac{2}{3} \mathbf{i} + \frac{1}{3} \mathbf{j} - \frac{2}{3} \mathbf{k} \right)\]

Key Concepts

Vector Magnitude

Understanding the concept of vector magnitude is crucial in vector algebra. The magnitude of a vector can be perceived as the vector's "length" or "size" in the mathematical space. It's a numerical representation of how long the vector is.To calculate the magnitude of a vector, use the formula:

• For a vector \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \), the magnitude \( \|\mathbf{v}\| \) is calculated as \( \sqrt{a^2 + b^2 + c^2} \).

This showcases the Pythagorean theorem extension into three dimensions, where each component is squared, summed, and then square-rooted to find the total length.For example in the solution above, with components \( a = 2, b = 1, \) and \( c = -2 \), the magnitude is calculated as \( \|\mathbf{v}\| = \sqrt{4 + 1 + 4} = 3 \). This means the vector's length in space is 3 units.

Unit Vector

A unit vector is a key component in vector algebra, representing the direction of a vector by reducing it to a standard length of 1. Essentially, a unit vector retains the direction of the original vector but standardizes its length.To find a unit vector \( \mathbf{u} \) from a given vector \( \mathbf{v} \), divide the vector by its magnitude:

• \( \mathbf{u} = \frac{1}{\|\mathbf{v}\|} \mathbf{v} \)

In our exercise, after finding the vector magnitude as 3, the unit vector is calculated as \( \mathbf{u} = \frac{1}{3}(2 \mathbf{i} + \mathbf{j} - 2 \mathbf{k}) \).This results in a unit vector expressed as \( \frac{2}{3} \mathbf{i} + \frac{1}{3} \mathbf{j} - \frac{2}{3} \mathbf{k} \). By doing this, we essentially normalize the vector, maintaining its direction but simplifying calculations involving direction-only.

Direction of a Vector

The direction of a vector indicates where the vector is pointing in space. It is independent of the vector's magnitude and can be described by a unit vector.When you express a vector in terms of its magnitude and direction, you simplify it into a form that multiplies its magnitude by the unit vector. This decomposition is incredibly useful in many scientific and engineering applications where direction and magnitude are considered separately.For the vector \( \mathbf{v} = 2 \mathbf{i} + \mathbf{j} - 2 \mathbf{k} \), already scaled down by its magnitude, becomes:

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

This expression clearly shows that the vector's spatial orientation doesn't change with size, yet retains a distinct path regardless of context or scale.