Problem 12 - University Calculus: Early Transcendentals

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 is approximately 1.84 radians.

Step 1: Understand Vector Components

First, break down the given vectors into their components. For the vector \( \mathbf{u} = \mathbf{i} + \sqrt{2} \mathbf{j} - \sqrt{2} \mathbf{k} \), the components are \( u_1 = 1 \), \( u_2 = \sqrt{2} \), and \( u_3 = -\sqrt{2} \). Similarly, for the vector \( \mathbf{v} = -\mathbf{i} + \mathbf{j} + \mathbf{k} \), the components are \( v_1 = -1 \), \( v_2 = 1 \), and \( v_3 = 1 \).

Step 2: Calculate Dot Product

The dot product of two vectors \( \mathbf{u} \) and \( \mathbf{v} \) is given by the formula: \( \mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2 + u_3 v_3 \). Plug in the values: \( \mathbf{u} \cdot \mathbf{v} = (1)(-1) + (\sqrt{2})(1) + (-\sqrt{2})(1) = -1 + \sqrt{2} - \sqrt{2} = -1 \).

Step 3: Calculate Magnitudes of Vectors

The magnitude of a vector \( \mathbf{u} = (u_1, u_2, u_3) \) is \( \|\mathbf{u}\| = \sqrt{u_1^2 + u_2^2 + u_3^2} \). So 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: Use the Dot Product to Find the Cosine of the Angle

Use the formula \( \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|} \). Substitute the values: \( \cos \theta = \frac{-1}{\sqrt{5} \times \sqrt{3}} = \frac{-1}{\sqrt{15}} \).

Step 5: Calculate the Angle \( \theta \)

Find \( \theta \) by applying the inverse cosine: \( \theta = \cos^{-1}\left(\frac{-1}{\sqrt{15}}\right) \). Compute this using a calculator to get \( \theta \approx 1.84 \) radians to the nearest hundredth.

Key Concepts

Dot Product

Understanding the dot product is crucial when dealing with angles between vectors. The dot product, or scalar product, is a way to multiply two vectors and get a scalar (a single number) as a result. For two vectors \( \mathbf{u} = (u_1, u_2, u_3) \) and \( \mathbf{v} = (v_1, v_2, v_3) \), the dot product \( \mathbf{u} \cdot \mathbf{v} \) is computed by multiplying corresponding components of the vectors and then summing up those products:
\[ \mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2 + u_3 v_3 \]

• If the dot product is positive, the angle between the vectors is acute.
• If it is zero, the vectors are perpendicular to each other.
• If the dot product is negative, the angle is obtuse.

In our exercise, we find the dot product of \( \mathbf{u} \) and \( \mathbf{v} \) is \(-1\), indicating an obtuse angle. The dot product is fundamental because it connects vector calculations with cosine of the angle, allowing us to find the angle between the vectors.

Vector Magnitude

The magnitude of a vector, sometimes referred to as the vector's "length" or "norm," helps us understand the size or extent of the vector. Calculating the magnitude of a vector is similar to applying the Pythagorean theorem in three dimensions. For a vector \( \mathbf{u} = (u_1, u_2, u_3) \), the magnitude is given by:
\[ \|\mathbf{u}\| = \sqrt{u_1^2 + u_2^2 + u_3^2} \]

• Magnitude is always a non-negative number.
• A zero magnitude implies the vector is a zero vector - essentially a point.

In this exercise, the magnitudes were calculated as \( \sqrt{5} \) for \( \mathbf{u} \) and \( \sqrt{3} \) for \( \mathbf{v} \). These magnitudes are necessary for determining the angle between the vectors by placing them into the formula that connects dot product, magnitudes, and the cosine of the angle.

Inverse Cosine

The inverse cosine function is the bridge that helps us find the angle between two vectors once we have their dot product and magnitudes. By applying the inverse cosine, we can determine the exact angle in radians. The relevant formula is:
\[ \theta = \cos^{-1}\left(\frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|}\right) \]

• This formula returns an angle \( \theta \) in radians.
• Radian is a unit of angular measure used in many areas of mathematics.

For the vectors in our exercise, after calculating the cosine of the angle, using inverse cosine resulted in \( \theta \approx 1.84 \) radians. The application of inverse cosine is what translates numerical data from vectors into a geometric angle.