Question

Find the angle between the vectors \(12 \boldsymbol{i}-\boldsymbol{j}\), and \(2 \boldsymbol{i}+\boldsymbol{j}+\boldsymbol{k}\).

Short answer

Question: Find the angle between the vectors \(\boldsymbol{u} = 12 \boldsymbol{i}-\boldsymbol{j}\) and \(\boldsymbol{v} = 2 \boldsymbol{i}+\boldsymbol{j}+\boldsymbol{k}\). Answer: \(\theta=\arccos(\frac{23}{\sqrt{145}\sqrt{6}})\)

Step-by-step solution

1. Define the vectors

Define vector \(\boldsymbol{u} = 12 \boldsymbol{i}-\boldsymbol{j}\) and vector \(\boldsymbol{v} = 2 \boldsymbol{i}+\boldsymbol{j}+\boldsymbol{k}\).

2. Calculate the dot product

To calculate the dot product \(\boldsymbol{u} \cdot \boldsymbol{v}\), we have: \(\boldsymbol{u} \cdot \boldsymbol{v} = (12 \boldsymbol{i}-\boldsymbol{j}) \cdot (2 \boldsymbol{i}+\boldsymbol{j}+\boldsymbol{k})\) \(= 12(2) + (-1)(1) + 0(1) = 24 - 1 = 23\)

3. Calculate the magnitudes of the vectors

To calculate the magnitude of \(\boldsymbol{u}\), we have: \(\lVert \boldsymbol{u} \rVert = \sqrt{(12)^2 + (-1)^2 + 0^2} = \sqrt{144 + 1} = \sqrt{145}\) To calculate the magnitude of \(\boldsymbol{v}\), we have: \(\lVert \boldsymbol{v} \rVert = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{4+1+1} = \sqrt{6}\)

4. Substitute into the formula

Now, we substitute the values into the formula: \(\cos{\theta} = \frac{\boldsymbol{u} \cdot \boldsymbol{v}}{\lVert \boldsymbol{u} \rVert \lVert \boldsymbol{v} \rVert} = \frac{23}{\sqrt{145}\sqrt{6}}\)

5. Find the angle

Finally, we solve for the angle \(\theta\): \(\theta=\arccos(\frac{23}{\sqrt{145}\sqrt{6}})\) This will give us the angle between the two vectors.

Key concepts

Dot Product

The dot product, also known as scalar product or inner product, is a fundamental operation in vector algebra. It's a way to multiply two vectors that results in a scalar (a single number), rather than another vector. For two vectors \textbf{u} and \textbf{v}, the dot product is calculated by multiplying corresponding components and then summing up those products.

For example, if \textbf{u} has components (u1, u2, u3) and \textbf{v} has components (v1, v2, v3), the dot product is given by: \[ \textbf{u} \bullet \textbf{v} = u1 \times v1 + u2 \times v2 + u3 \times v3 \].
The dot product of vectors can tell us about the angle between them—if the dot product is zero, the vectors are orthogonal, meaning they are at a 90-degree angle to each other.

Vector Magnitude

In physics and mathematics, the magnitude of a vector represents its length and is denoted by \(||\textbf{v}||\). To calculate the magnitude of a vector \textbf{v} with components (v1, v2, v3), we use the Pythagorean theorem extended into three dimensions: \[ ||\textbf{v}|| = \textbf{sqrt}{v1^2 + v2^2 + v3^2} \].
The magnitude is always a non-negative number and provides insight into the vector's size without regard to its direction. Determining the magnitude is essential when calculating the angle between two vectors as it forms part of the denominator in the cosine formula.

Cosine of Angle

The cosine of an angle is a trigonometric function that measures the ratio of the adjacent side to the hypotenuse of a right triangle. In the context of vectors, the cosine of the angle between two vectors is crucially used to determine their directional relationship.

To find the cosine of the angle between two non-zero vectors \textbf{u} and \textbf{v}, we use the formula: \[ \textbf{cos}\theta = \frac{\textbf{u} \bullet \textbf{v}}{||\textbf{u}|| ||\textbf{v}||} \], where \(\textbf{u} \bullet \textbf{v}\) is the dot product, and \(||\textbf{u}||\) and \(||\textbf{v}||\) are their magnitudes. If the vectors point in the same direction, the cosine of the angle is 1; if they point in the exact opposite directions, the cosine is -1.

Arccos Function

The arccos function, also known as the inverse cosine function, is used to determine the angle whose cosine is a given number. If you know the cosine of an angle, you can use the arccos function to find the actual angle in radians (or degrees after conversion).

For example, if the cosine of an angle is 0.5, then the angle is \[ \textbf{arccos}(0.5) \]. This function is important in vector calculus to determine the angle between vectors if we have calculated the cosine of the angle using the dot product and vector magnitudes. The output of arccos falls in the range from 0 to \(\textbf{pi}\) radians (0 to 180 degrees).