Problem 26 - University Calculus: Early Transcendentals

Question

Express each vector as a product of its length and direction. $$9 \mathbf{i}-2 \mathbf{j}+6 \mathbf{k}$$

Step-by-Step Solution

Answer

The vector is expressed as \( 11 \left( \frac{9}{11} \mathbf{i} - \frac{2}{11} \mathbf{j} + \frac{6}{11} \mathbf{k} \right) \).

Step 1: Identify the Vector

The vector is \( \mathbf{v} = 9 \mathbf{i} - 2 \mathbf{j} + 6 \mathbf{k} \). This vector is given in terms of the standard unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \).

Step 2: Calculate the Length of the Vector

The length of a vector \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \) is given by the formula \( \| \mathbf{v} \| = \sqrt{a^2 + b^2 + c^2} \). For our vector, \( a = 9 \), \( b = -2 \), and \( c = 6 \). Therefore, \( \| \mathbf{v} \| = \sqrt{9^2 + (-2)^2 + 6^2} = \sqrt{81 + 4 + 36} = \sqrt{121} = 11 \).

Step 3: Determine the Direction of the Vector

The direction of a vector \( \mathbf{v} \) is given by its unit vector, which is \( \frac{\mathbf{v}}{\| \mathbf{v} \|} \). Thus, the unit vector for our vector is \( \frac{1}{11} (9 \mathbf{i} - 2 \mathbf{j} + 6 \mathbf{k}) = \left( \frac{9}{11} \right) \mathbf{i} + \left( -\frac{2}{11} \right) \mathbf{j} + \left( \frac{6}{11} \right) \mathbf{k} \).

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

We express the original vector as a product of its length (\(11\)) and its direction, which is the unit vector. So, \( \mathbf{v} = 11 \times \left( \frac{9}{11} \mathbf{i} - \frac{2}{11} \mathbf{j} + \frac{6}{11} \mathbf{k} \right) \). Therefore, the vector is expressed as a product of its length and direction.

Key Concepts

Vector Length

The length of a vector, often referred to as its magnitude, can be thought of as the distance from the vector's starting point to its endpoint in space. This measurement is crucial because it gives us a sense of how "long" the vector is, regardless of its direction.

To find the length of a vector with components like \( a \), \( b \), and \( c \), you use the formula:

\[ \| \mathbf{v} \| = \sqrt{a^2 + b^2 + c^2} \]

This formula originates from the Pythagorean theorem. In the case of the vector \( 9 \mathbf{i} - 2 \mathbf{j} + 6 \mathbf{k} \), the length is:

\[ \| \mathbf{v} \| = \sqrt{9^2 + (-2)^2 + 6^2} = \sqrt{81 + 4 + 36} = \sqrt{121} = 11 \]

Knowing the vector's length helps in various applications, such as determining forces, velocity, and other physical quantities.

Unit Vector

A unit vector is a vector that has a length of exactly 1 unit. It is primarily used to describe the direction of a vector, without concern for its magnitude.

To convert any vector into a unit vector, you divide each component of the vector by its length. This process scales the vector to have a magnitude of 1 while maintaining its original direction.

For instance, using the vector \( 9 \mathbf{i} - 2 \mathbf{j} + 6 \mathbf{k} \) with a length of 11, the unit vector is:

\[ \frac{1}{11} (9 \mathbf{i} - 2 \mathbf{j} + 6 \mathbf{k}) = \left( \frac{9}{11} \right) \mathbf{i} + \left( -\frac{2}{11} \right) \mathbf{j} + \left( \frac{6}{11} \right) \mathbf{k} \]

Unit vectors are helpful when you need a standardized measure of direction. They are often used in physics and engineering to indicate directions in vector compositions.

Standard Unit Vectors

Standard unit vectors are the building blocks of vector space, especially in three-dimensional space.

They are defined as:

• \( \mathbf{i} \), which is the unit vector in the direction of the x-axis: \( (1,0,0) \).
• \( \mathbf{j} \), which represents the direction of the y-axis: \( (0,1,0) \).
• \( \mathbf{k} \), which indicates direction along the z-axis: \( (0,0,1) \).

These vectors are essential because they form the basis for expressing any vector in three-dimensional space.

For the vector \( 9 \mathbf{i} - 2 \mathbf{j} + 6 \mathbf{k} \), each component is a scalar multiple of these standard unit vectors. By understanding standard unit vectors, you can decompose and simplify vector operations effectively.