Problem 12 - Thomas Calculus

Question

Find the angles between the vectors in Exercises \(9-12\) to the nearest hundredth of a radian. $$ \mathbf{u}=\mathbf{i}+\sqrt{2} \mathbf{j}-\sqrt{2} \mathbf{k}, \quad \mathbf{v}=-\mathbf{i}+\mathbf{j}+\mathbf{k} $$

Step-by-Step Solution

Answer

The angle between the vectors is approximately 1.85 radians.

Step 1: Understand the Dot Product

The angle between two vectors \( \mathbf{u} \) and \( \mathbf{v} \) can be found using the dot product formula: \[ \mathbf{u} \cdot \mathbf{v} = \|\mathbf{u}\| \|\mathbf{v}\| \cos(\theta) \] where \( \theta \) is the angle between the vectors.

Step 2: Compute the Dot Product

Calculate the dot product \( \mathbf{u} \cdot \mathbf{v} \). The dot product for two vectors \( \mathbf{u} = \langle a_1, b_1, c_1 \rangle \) and \( \mathbf{v} = \langle a_2, b_2, c_2 \rangle \) is:\[ \mathbf{u} \cdot \mathbf{v} = a_1a_2 + b_1b_2 + c_1c_2 \].For \( \mathbf{u} = \mathbf{i}+\sqrt{2} \mathbf{j}-\sqrt{2} \mathbf{k} \) and \( \mathbf{v} = -\mathbf{i}+\mathbf{j}+\mathbf{k} \):- \( a_1 = 1, \; b_1 = \sqrt{2}, \; c_1 = -\sqrt{2} \)- \( a_2 = -1, \; b_2 = 1, \; c_2 = 1 \)Substituting these into the dot product formula gives:\[ \mathbf{u} \cdot \mathbf{v} = (1)(-1) + (\sqrt{2})(1) + (-\sqrt{2})(1) = -1 + \sqrt{2} - \sqrt{2} = -1 \].

Step 3: Compute the Magnitude of Each Vector

Compute the magnitudes \( \|\mathbf{u}\| \) and \( \|\mathbf{v}\| \). The magnitude of a vector \( \mathbf{a} = \langle a_1, b_1, c_1 \rangle \) is calculated as:\[ \|\mathbf{a}\| = \sqrt{a_1^2 + b_1^2 + c_1^2} \].For \( \mathbf{u} \):\[ \|\mathbf{u}\| = \sqrt{1^2 + (\sqrt{2})^2 + (-\sqrt{2})^2} = \sqrt{1+2+2} = \sqrt{5} \].For \( \mathbf{v} \):\[ \|\mathbf{v}\| = \sqrt{(-1)^2 + 1^2 + 1^2} = \sqrt{1+1+1} = \sqrt{3} \].

Step 4: Apply the Cosine Formula

We know the dot product \( \mathbf{u} \cdot \mathbf{v} = -1 \), and the magnitudes \( \|\mathbf{u}\| = \sqrt{5} \) and \( \|\mathbf{v}\| = \sqrt{3} \) from previous steps. Substitute these into the formula:\[ \cos(\theta) = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|} = \frac{-1}{\sqrt{5} \times \sqrt{3}} \].This simplifies to:\[ \cos(\theta) = \frac{-1}{\sqrt{15}} \].

Step 5: Solve for Theta

To find \( \theta \), take the inverse cosine:\[ \theta = \cos^{-1}\left(\frac{-1}{\sqrt{15}}\right) \].Calculate \( \theta \) using a calculator:\[ \theta \approx 1.85 \text{ radians (to the nearest hundredth)} \].

Key Concepts

Vector Magnitude

The vector magnitude is a measure of how long or how big a vector is. You can think of it as the vector's length in space. To calculate the magnitude of a vector \( \mathbf{a} = \langle a_1, b_1, c_1 \rangle \), you use the formula: \[ \|\mathbf{a}\| = \sqrt{a_1^2 + b_1^2 + c_1^2} \]This formula follows directly from the Pythagorean theorem, extending it to three dimensions.
Consider the vector \( \mathbf{u} = \mathbf{i} + \sqrt{2} \mathbf{j} - \sqrt{2} \mathbf{k} \). The magnitude \( \|\mathbf{u}\| \) is \[ \sqrt{1^2 + (\sqrt{2})^2 + (-\sqrt{2})^2} = \sqrt{5} \].
It's always helpful to remember that magnitudes are non-negative, which reflects the vector’s absolute length. Just like measuring a step, the direction doesn’t affect the length itself, only the path.

Cosine Formula

When finding the angle between two vectors, the cosine formula comes into play. It bridges the dot product with the cosine of the angle \( \theta \) between the vectors. The formula is:\[ \mathbf{u} \cdot \mathbf{v} = \|\mathbf{u}\| \|\mathbf{v}\| \cos(\theta) \]Here, \( \mathbf{u} \cdot \mathbf{v} \) represents the dot product, and \( \|\mathbf{u}\| \) and \( \|\mathbf{v}\| \) are the magnitudes of the vectors.
Think about this formula as a way to relate directional alignment between vectors. Perfect alignment (or opposite alignment) leads to a maximum absolute value of \( \cos(\theta) \), whereas perpendicular vectors yield a cosine of zero, meaning no directional overlap or alignment at all.
This formula is not just a numerical pathway but a conceptual doorway to understanding how two vectors interact directionally in space.

Angle Between Vectors

The angle between vectors provides insight into their directional relationship. To find this angle, \( \theta \), the inverse cosine is applied to the ratio of their dot product and the product of their magnitudes: \[ \cos(\theta) = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|} \]
By rearranging for \( \theta \), we have:\[ \theta = \cos^{-1}\left(\frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|}\right) \]For example, in solving the problem, we found \( \cos(\theta) = \frac{-1}{\sqrt{15}} \), andafter computing the inverse cosine, the angle \( \theta \) was approximately 1.85 radians.
Understanding this angle tells us about proximity in direction. If the angle is small, vectors are closely aligned. At right angles, they are orthogonal, while large angles point to opposite directions. This knowledge not only aids in spatial reasoning but is foundational in physics and engineering applications.