Problem 12 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The angle between the vectors is approximately 1.83 radians.

Step 1: Determine the dot product of the vectors

Calculate the dot product of the vectors \( \mathbf{u} = \mathbf{i} + \sqrt{2} \mathbf{j} - \sqrt{2} \mathbf{k} \) and \( \mathbf{v} = -\mathbf{i} + \mathbf{j} + \mathbf{k} \).The dot product is calculated as follows:\[ \mathbf{u} \cdot \mathbf{v} = (1)(-1) + (\sqrt{2})(1) + (-\sqrt{2})(1) \]\[ \mathbf{u} \cdot \mathbf{v} = -1 + \sqrt{2} - \sqrt{2} = -1 \]

Step 2: Determine the magnitudes of the vectors

Calculate the magnitudes of both vectors.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 3: Use the dot product to find the cosine of the angle

Use the formula for the cosine of the angle between two vectors:\[ \cos(\theta) = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|} \]Substitute the completed calculations:\[ \cos(\theta) = \frac{-1}{\sqrt{5} \cdot \sqrt{3}} = \frac{-1}{\sqrt{15}} \]

Step 4: Calculate the angle using the inverse cosine

To find the angle \( \theta \), take the inverse cosine:\[ \theta = \cos^{-1}\left(\frac{-1}{\sqrt{15}}\right) \]Using a calculator, convert this into radians and round to the nearest hundredth:\[ \theta \approx 1.83 \text{ radians} \]

Step 5: Finalize and verify the angle

Verify that the angle found is reasonable, considering the components of the vectors. As both vectors are in 3-dimensional space and their dot product is negative, the angle should logically fall between \( \pi/2 \) and \( \pi \), which matches our calculation of \( \theta \approx 1.83 \text{ radians}.\)

Key Concepts

Understanding Dot Product

The dot product, also known as the scalar product, is a crucial operation when working with vectors. It's used to measure how much two vectors point in the same direction. The formula for the dot product of two vectors \( \mathbf{u} = a_1\mathbf{i} + b_1\mathbf{j} + c_1\mathbf{k} \) and \( \mathbf{v} = a_2\mathbf{i} + b_2\mathbf{j} + c_2\mathbf{k} \) is \( a_1 \cdot a_2 + b_1 \cdot b_2 + c_1 \cdot c_2 \). In our example, it simplifies to:

• Multiply the corresponding components of each vector: \( (1 \times -1), (\sqrt{2} \times 1), (-\sqrt{2} \times 1) \)
• Then, sum the results: \( -1 + \sqrt{2} - \sqrt{2} = -1 \)

You get a single number, which tells you how closely aligned the two vectors are. If the dot product is positive, the vectors point in a similar direction. If it's zero, they are perpendicular. If it's negative, like in our case, the vectors point in roughly opposite directions.

Vector Magnitudes Explained

Vector magnitude is the length or size of the vector, often seen as how far it reaches in space. To find a vector's magnitude, use the formula: \( \|\mathbf{u}\| = \sqrt{a^2 + b^2 + c^2} \). It's a simple way of using the Pythagorean theorem in three dimensions. Let's break it down:For the vector \( \mathbf{u} = \mathbf{i} + \sqrt{2} \mathbf{j} - \sqrt{2} \mathbf{k} \), the magnitude is calculated as follows:

• Take each component: \( 1, \sqrt{2}, -\sqrt{2} \)
• Square them: \( 1^2, (\sqrt{2})^2, (-\sqrt{2})^2 \)
• Add them together: \( 1 + 2 + 2 = 5 \)
• Finally, take the square root: \( \sqrt{5} \)

For vector \( \mathbf{v} \), follow similar steps: \( \|\mathbf{v}\| = \sqrt{(-1)^2 + 1^2 + 1^2} = \sqrt{3} \). The magnitudes tell us how long each vector is when drawn in space.

Discovering the Inverse Cosine

Once we have the dot product and the magnitudes of both vectors, the next step is to find the angle between them. This is where the inverse cosine, or arc cosine, comes into play. The formula to find the cosine of the angle \( \theta \) between two vectors \( \mathbf{u} \) and \( \mathbf{v} \) is:\[ \cos(\theta) = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|} \]Next, plug the values into the formula:

• Dot product: \(-1 \)
• Magnitude of \( \mathbf{u} \): \( \sqrt{5} \)
• Magnitude of \( \mathbf{v} \): \( \sqrt{3} \)
• Calculate: \( \cos(\theta) = \frac{-1}{\sqrt{5} \cdot \sqrt{3}} = \frac{-1}{\sqrt{15}} \)

The cosine value \( \frac{-1}{\sqrt{15}} \) tells us the ratio from cosine, which we need to convert back into an angle in radians. By using a calculator to apply the inverse cosine function, you'll find \( \theta \approx 1.83 \text{ radians}, \) giving you the angle between the vectors. Because our dot product is negative, it's likely the angle is between \( \pi/2 \) and \( \pi \), confirming our result.