Chapter 3: Problem 12

Find the angle between the vectors \(\mathbf{A}=-2 \mathbf{i}+\mathbf{j}-2 \mathbf{k}\) and \(\mathbf{B}=2 \mathbf{i}-2 \mathbf{j}\).

Short Answer

Expert verified

The angle between the vectors is \( 135 \text{ degrees} \).

Step by step solution

Understand the Vectors

Given vectors are \(\textbf{A}=-2 \textbf{i} + \textbf{j} - 2 \textbf{k} \) and \(\textbf{B}=2 \textbf{i} - 2 \textbf{j}\).

Use the Dot Product Formula

The dot product of two vectors \( \textbf{A} \) and \( \textbf{B} \) is given by \(\textbf{A} \bullet \textbf{B} = A_x B_x + A_y B_y + A_z B_z\).

Calculate the Dot Product

\(\textbf{A} \bullet \textbf{B} = (-2)(2) + (1)(-2) + (-2)(0) \) \ \(-4 - 2 + 0 = -6\)

Find the Magnitude of Each Vector

The magnitude of vector \( \textbf{A} \) is \(\textbf{|A|} = \sqrt{(-2)^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3\). \ \ \ \ The magnitude of vector \( \textbf{B} \) is \(\textbf{|B|} = \sqrt{2^2 + (-2)^2 + 0^2} = \sqrt{4 + 4 + 0} = \sqrt{8} = 2\text{√2}\).

Use the Dot Product to Find the Angle

The angle between two vectors is found using the formula: \(\textbf{A} \bullet \textbf{B} = |\textbf{A}| \ |\textbf{B}| \ cos(\theta)\). \ \ Solving for \ \theta,\: \ \ \cos(\theta) = \frac{\textbf{A} \bullet \textbf{B}} {\textbf{|A| |\textbf{B}|}} = \frac{-6}{3*2\text{√2}} = \frac{-6}{6\text{√2}} = \frac{-1}{\text{√2}} = -\text{√2}/2\. \ \ \theta = \arccos\bigg(\-\frac{\text{√2}}{2} \bigg) \ = \ \frac{3\pi}{4} \text{or} \ 135 \ \text{{degrees.}}

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dot Product

The dot product is a fundamental operation for vectors in mathematics. It provides a scalar (a single number), rather than a vector itself. To calculate the dot product of two vectors, \(\textbf{A} = A_x \textbf{i} + A_y \textbf{j} + A_z \textbf{k}\) and \(\textbf{B} = B_x \textbf{i} + B_y \textbf{j} + B_z \textbf{k}\), you use the formula: \(\textbf{A} \bullet \textbf{B} = A_x B_x + A_y B_y + A_z B_z\).

Pairs of corresponding components are multiplied.
The products are then summed.

For our example vectors \(\textbf{A} = -2 \textbf{i} + \textbf{j} - 2 \textbf{k}\) and \(\textbf{B} = 2 \textbf{i} - 2 \textbf{j}\), the calculation steps are:

Multiply the \(i\) components: \(-2 \times 2 = -4 \).
Multiply the \(j\) components: \(1 \times -2 = -2 \).
Multiply the \(k\) components: \(-2 \times 0 = 0 \).

Summing these results, the dot product is: \(\textbf{A} \bullet \textbf{B} = -6 \). This step is critical for finding the angle between two vectors.

Vector Magnitude

Understanding the magnitude of a vector is crucial when working with vectors. The magnitude (or length) of a vector \(\textbf{A} = A_x \textbf{i} + A_y \textbf{j} + A_z \textbf{k}\) is given by the formula: \(|\textbf{A}| = \sqrt{A_x^2 + A_y^2 + A_z^2} \). This is essentially the Euclidean distance from the origin to the point defined by the vector. Let's calculate the magnitude for our vectors:

Vector \(\textbf{A} = -2 \textbf{i} + \textbf{j} - 2 \textbf{k}\): \(|\textbf{A}| = \sqrt{(-2)^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \).
Vector \(\textbf{B} = 2 \textbf{i} - 2 \textbf{j}\): \(|\textbf{B}| = \sqrt{2^2 + (-2)^2 + 0^2} = \sqrt{4 + 4 + 0} = \sqrt{8} = 2\text{√2} \).

The magnitude helps us normalize vectors and is often used in the formula to find the angle between two vectors.

Arccosine

The arccosine function, denoted as \(\text{arccos}\), is the inverse of the cosine function. It helps us find the angle from a given cosine value. When we know the result of a cosine function, \(\text{arccos}\) tells us the corresponding angle. To find the angle \(\theta\) between two vectors, we use the dot product and the magnitudes of the vectors in the formula: \(\textbf{A} \bullet \textbf{B} = |\textbf{A}| |\textbf{B}| \text{cos}(\theta)\).

First, solve for \(\text{cos}(\theta)\): \(\text{cos}(\theta) = \frac{\textbf{A} \bullet \textbf{B}}{|\textbf{A}| |\textbf{B}|} \).
Then, find \(\theta\) using \(\text{arccos}\bigg(\frac{\textbf{A} \bullet \textbf{B}}{|\textbf{A}| |\textbf{B}|}\bigg) \).

For the vectors given, we calculated \(\text{cos}(\theta) = \frac{-6}{6\text{√2}} = -\frac{\text{√2}}{2} \). Applying arccosine: \(\theta = \text{arccos}(-\text{√2}/2) \). This yields \( \theta = 135 \text{degrees} \). Thus, arccosine converts our cosine value back into the angle, completing our process.