Question

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

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

Step-by-step solution

Understand the Vectors

Given vectors are $\mathbf{\text{A}}=-2\mathbf{\text{i}}+\mathbf{\text{j}}-2\mathbf{\text{k}}$ and $\mathbf{\text{B}}=2\mathbf{\text{i}}-2\mathbf{\text{j}}$.

Use the Dot Product Formula

The dot product of two vectors $\mathbf{\text{A}}$ and $\mathbf{\text{B}}$ is given by $\mathbf{\text{A}}\bullet \mathbf{\text{B}}=A_x{B}_x+A_y{B}_y+A_z{B}_z$.

Calculate the Dot Product

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

Find the Magnitude of Each Vector

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

Use the Dot Product to Find the Angle

The angle between two vectors is found using the formula: $\mathbf{\text{A}}\bullet \mathbf{\text{B}}=|\mathbf{\text{A}}||\mathbf{\text{B}}|cos(\theta )$. \ \ 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.}}

Key concepts

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, $\mathbf{\text{A}}=A_x\mathbf{\text{i}}+A_y\mathbf{\text{j}}+A_z\mathbf{\text{k}}$ and $\mathbf{\text{B}}=B_x\mathbf{\text{i}}+B_y\mathbf{\text{j}}+B_z\mathbf{\text{k}}$, you use the formula: $\mathbf{\text{A}}\bullet \mathbf{\text{B}}=A_x{B}_x+A_y{B}_y+A_z{B}_z$.

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

For our example vectors $\mathbf{\text{A}}=-2\mathbf{\text{i}}+\mathbf{\text{j}}-2\mathbf{\text{k}}$ and $\mathbf{\text{B}}=2\mathbf{\text{i}}-2\mathbf{\text{j}}$, 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: $\mathbf{\text{A}}\bullet \mathbf{\text{B}}=-6$. 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 $\mathbf{\text{A}}=A_x\mathbf{\text{i}}+A_y\mathbf{\text{j}}+A_z\mathbf{\text{k}}$ is given by the formula: $|\mathbf{\text{A}}|=\sqrt{A_x^2+A_y^2+A_z^2}$. 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 $\mathbf{\text{A}}=-2\mathbf{\text{i}}+\mathbf{\text{j}}-2\mathbf{\text{k}}$: $|\mathbf{\text{A}}|=\sqrt{(-2{)}^2+1^2+(-2{)}^2}=\sqrt{4+1+4}=\sqrt{9}=3$.
• Vector $\mathbf{\text{B}}=2\mathbf{\text{i}}-2\mathbf{\text{j}}$: $|\mathbf{\text{B}}|=\sqrt{2^2+(-2{)}^2+0^2}=\sqrt{4+4+0}=\sqrt{8}=2\text{√2}$.

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: $\mathbf{\text{A}}\bullet \mathbf{\text{B}}=|\mathbf{\text{A}}||\mathbf{\text{B}}|\text{cos}(\theta )$.

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

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.