Problem 5 - Active Calculus Multivariable

Question

What is the angle in radians between the vectors $$ \begin{array}{l} \mathbf{a}=(6,-4,9) \text { and } \\ \mathbf{b}=(5,-1,6) ? \end{array} $$ Angle:_______(radians)

Step-by-Step Solution

Answer

Angle: \(≈ 0.592\) radians

Step 1: Calculate the dot product of \(\mathbf{a}\) and \(\mathbf{b}\)

To calculate the dot product of \(\mathbf{a} = (6, -4, 9)\) and \(\mathbf{b} = (5, -1, 6)\), multiply corresponding components and sum them up: \[\mathbf{a} · \mathbf{b} = 6*5 + (-4)*(-1) + 9*6 = 30 + 4 + 54 = 88\]

Step 2: Calculate the magnitudes of \(\mathbf{a}\) and \(\mathbf{b}\)

Find the magnitudes of \(\mathbf{a}\) and \(\mathbf{b}\) using the square root of the sum of the squares of their components: \(||\mathbf{a}|| = \sqrt{6^2 + (-4)^2 + 9^2} = \sqrt{36 + 16 + 81} = \sqrt{133}\) \(||\mathbf{b}|| = \sqrt{5^2 + (-1)^2 + 6^2} = \sqrt{25 + 1 + 36} = \sqrt{62}\)

Step 3: Calculate the angle θ in radians between \(\mathbf{a}\) and \(\mathbf{b}\)

Now, we will use the formula we mentioned in the analysis: \[cos(θ) = \frac{\mathbf{a} · \mathbf{b}}{||\mathbf{a}|| ||\mathbf{b}||} = \frac{88}{\sqrt{133} \sqrt{62}}\] To find the angle θ, we will take the inverse cosine (arccos) of the expression above: \[θ = arccos\left( \frac{88}{\sqrt{133} \sqrt{62}} \right)\] Using a calculator, we get: \[θ ≈ 0.592 \, radians\] So, the angle between the vectors \(\mathbf{a}=(6,-4,9)\) and \(\mathbf{b}=(5,-1,6)\) is approximately 0.592 radians.

Key Concepts

Dot Product

The dot product, also known as the scalar product, is a fundamental operation in vector calculus. It combines two vectors to produce a single number, or scalar. This is particularly useful when you want to determine how much one vector extends in the direction of another.
To compute the dot product of two vectors, each component of the first vector is multiplied by the corresponding component of the second vector. The resulting products are then added together.
For example, if we have vectors \( \mathbf{a} = (a_1, a_2, a_3) \) and \( \mathbf{b} = (b_1, b_2, b_3) \), the dot product \( \mathbf{a} \cdot \mathbf{b} \) can be found using the formula:

• \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \)

For vectors \( \mathbf{a} = (6, -4, 9) \) and \( \mathbf{b} = (5, -1, 6) \), this yields:

• \( \mathbf{a} \cdot \mathbf{b} = 6 \times 5 + (-4) \times (-1) + 9 \times 6 = 88 \)

The result, 88, reflects how these vectors align with each other: a positive result indicates they point in generally the same direction.

Magnitude of a Vector

The magnitude of a vector, often referred to as its length or norm, is a measure of how long or large a vector is. It is computed using the Euclidean distance formula, which is the square root of the sum of the squares of the vector's components.
For any vector \( \mathbf{a} = (a_1, a_2, a_3) \), the magnitude \( ||\mathbf{a}|| \) can be calculated by:

• \( ||\mathbf{a}|| = \sqrt{a_1^2 + a_2^2 + a_3^2} \)

Applying this to vector \( \mathbf{a} = (6, -4, 9) \), we find:

• \( ||\mathbf{a}|| = \sqrt{6^2 + (-4)^2 + 9^2} = \sqrt{133} \)

Similarly, for vector \( \mathbf{b} = (5, -1, 6) \), we calculate:

• \( ||\mathbf{b}|| = \sqrt{5^2 + (-1)^2 + 6^2} = \sqrt{62} \)

The magnitudes reflect the lengths of the vectors in three-dimensional space, essential for comparing and analyzing their scale and directions.

Angle Between Vectors

Understanding the angle between two vectors helps in determining their relative orientation. In vector calculus, this angle can be found using the dot product alongside the magnitudes of the vectors.
The formula to find the cosine of the angle \( \theta \) between two vectors \( \mathbf{a} \) and \( \mathbf{b} \) is:

• \( \cos(\theta) = \frac{\mathbf{a} \cdot \mathbf{b}}{||\mathbf{a}|| \cdot ||\mathbf{b}||} \)

For our specific vectors \( \mathbf{a} \) and \( \mathbf{b} \), this becomes:

• \( \cos(\theta) = \frac{88}{\sqrt{133} \times \sqrt{62}} \)

To find the angle itself, we use the inverse cosine function:

• \( \theta = \arccos\left( \frac{88}{\sqrt{133} \times \sqrt{62}} \right) \)

Using a calculator, this gives \( \theta \approx 0.592 \) radians. This indicates that the vectors are not aligned or orthogonal, but have a specific angle between them that measures approximately 0.592 radians. This measure is crucial in understanding how vectors relate spatially.