Problem 17 - Essential Calculus Early Transcendentals

Question

Find a unit vector with the same direction as \(8 \mathbf{i}-\mathbf{j}+4 \mathbf{k}\)

Step-by-Step Solution

Answer

The unit vector is \( \frac{8}{9} \mathbf{i} - \frac{1}{9} \mathbf{j} + \frac{4}{9} \mathbf{k} \).

Step 1: Understand the Given Vector

We are given a vector \( \mathbf{v} = 8 \mathbf{i} - \mathbf{j} + 4 \mathbf{k} \), where \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) are the unit vectors in the x, y, and z directions, respectively. We need to find a unit vector in the same direction as this vector.

Step 2: Calculate the Magnitude of the Vector

Calculate the magnitude \( ||\mathbf{v}|| \) of the vector \( \mathbf{v} \) using the formula: \[ ||\mathbf{v}|| = \sqrt{8^2 + (-1)^2 + 4^2} \]which simplifies to \[ ||\mathbf{v}|| = \sqrt{64 + 1 + 16} = \sqrt{81} = 9 \].

Step 3: Determine the Unit Vector

A unit vector \( \mathbf{u} \) in the direction of \( \mathbf{v} \) is given by:\[ \mathbf{u} = \frac{\mathbf{v}}{||\mathbf{v}||} = \frac{8 \mathbf{i} - \mathbf{j} + 4 \mathbf{k}}{9} \]Thus, \[ \mathbf{u} = \frac{8}{9} \mathbf{i} - \frac{1}{9} \mathbf{j} + \frac{4}{9} \mathbf{k} \].

Key Concepts

Vector Magnitude

The magnitude of a vector, often referred to as its length, is a measure of the vector's extent regardless of its direction. It is calculated using an extension of the Pythagorean theorem, adding up the squares of its components. For a vector \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \), the magnitude is calculated as:\[ ||\mathbf{v}|| = \sqrt{a^2 + b^2 + c^2} \]This formula essentially finds the distance from the vector's tail to its tip in 3D space.
Understanding vector magnitude is crucial because it serves as a foundation for determining the unit vector.

• It tells us about the size of the vector without considering its direction.
• This is helpful in normalizing vectors, allowing us to create unit vectors.

In our example, \( \mathbf{v} = 8 \mathbf{i} - \mathbf{j} + 4 \mathbf{k} \), the magnitude calculated was 9.

Directional Vector

A directional vector maintains the direction of the original vector but scales it to a length of one, becoming what we refer to as a unit vector. Directional vectors are found by dividing each component of the original vector by its magnitude.
This process is known as normalization. For the vector \( \mathbf{v} = 8 \mathbf{i} - \mathbf{j} + 4 \mathbf{k} \) with magnitude 9, the unit or directional vector will be:

• Divide the \( x \)-component by the magnitude: \( \frac{8}{9} \).
• Divide the \( y \)-component by the magnitude: \( \frac{-1}{9} \).
• Divide the \( z \)-component by the magnitude: \( \frac{4}{9} \).

This gives us the updated vector:\[ \mathbf{u} = \frac{8}{9} \mathbf{i} - \frac{1}{9} \mathbf{j} + \frac{4}{9} \mathbf{k} \] Thus, the vector retains its direction but with a length of one unit.

3D Vector Operations

3D vectors like those involving\( \mathbf{i}, \mathbf{j}, \mathbf{k} \) allow us to perform various operations that are essential in physics and engineering. One major operation is the calculation of vector magnitude, which we have shown previously.
Besides finding magnitudes, there are other important operations:

• Addition and Subtraction: Vectors can be added or subtracted component-wise to find resultant vectors.
• Dot Product: This operation calculates the scalar product, indicating how much one vector extends in the direction of another.
• Cross Product: The cross product of two vectors provides a third vector that is perpendicular to both, determining areas and volumes.

Understanding these operations helps in effectively manipulating and interpreting vectors in 3D space, allowing for better analytical and practical applications in real-life scenarios.