Problem 18 - Essential Calculus Early Transcendentals

Question

Find, correct to the nearest degree, the three angles of the triangle with vertices \(A(1,0,-1), B(3,-2,0),\) and \(C(1,3,3) .\)

Step-by-Step Solution

Answer

The angles of the triangle are approximately 83°, 108°, and 69°.

Step 1: Find Vector AB, BC, and CA

First, calculate the vectors for each side of the triangle. The vectors can be found by subtracting the coordinates of the starting point from the ending point.\[\vec{AB} = B - A = (3-1, -2-0, 0-(-1)) = (2, -2, 1)\]\[\vec{BC} = C - B = (1-3, 3-(-2), 3-0) = (-2, 5, 3)\]\[\vec{CA} = A - C = (1-1, 0-3, -1-3) = (0, -3, -4)\]

Step 2: Compute Magnitudes of Vectors

Use the formula for the magnitude of a vector \(\sqrt{x^2 + y^2 + z^2}\) to find the magnitudes of vectors \(\vec{AB}, \vec{BC}, \text{ and } \vec{CA}\).\[|\vec{AB}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{9} = 3\]\[|\vec{BC}| = \sqrt{(-2)^2 + 5^2 + 3^2} = \sqrt{38}\]\[|\vec{CA}| = \sqrt{0^2 + (-3)^2 + (-4)^2} = \sqrt{25} = 5\]

Step 3: Use Dot Product to Find Cosines of Angles

The cosine of the angle between two vectors can be found using the dot product formula \(\vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos(\theta)\).Calculate the dot products:\[\vec{AB} \cdot \vec{BC} = 2(-2) + (-2)5 + 1(3) = -4 - 10 + 3 = -11\]\[\vec{BC} \cdot \vec{CA} = (-2)0 + 5(-3) + 3(-4) = -15 - 12 = -27\]\[\vec{CA} \cdot \vec{AB} = 0(2) + (-3)(-2) + (-4)(1) = 6 - 4 = 2\]

Step 4: Calculate Angles Using Inverse Cosine

Now find the angles by applying the inverse cosine function.For angle \(A\) at vertex \(A\):\[\cos(A) = \frac{\vec{CA} \cdot \vec{AB}}{|\vec{CA}| |\vec{AB}|} = \frac{2}{5 \times 3} = \frac{2}{15}\]\[A = \cos^{-1}\left(\frac{2}{15}\right) \approx 82.82^\circ\]For angle \(B\) at vertex \(B\):\[\cos(B) = \frac{\vec{AB} \cdot \vec{BC}}{|\vec{AB}| |\vec{BC}|} = \frac{-11}{3 \times \sqrt{38}}\]\[B = \cos^{-1}\left(\frac{-11}{3\sqrt{38}}\right) \approx 108.20^\circ\]For angle \(C\) at vertex \(C\):\[\cos(C) = \frac{\vec{BC} \cdot \vec{CA}}{|\vec{BC}| |\vec{CA}|} = \frac{-27}{\sqrt{38} \times 5}\]\[C = \cos^{-1}\left(\frac{-27}{5\sqrt{38}}\right) \approx 68.98^\circ\]

Step 5: Round the Angles to Nearest Degree

Round each angle calculated in the previous step to the nearest degree:- Angle \(A \approx 83^\circ\)- Angle \(B \approx 108^\circ\)- Angle \(C \approx 69^\circ\)

Key Concepts

Vector Operations in Geometry

In geometry, vectors are used to represent quantities that have both magnitude and direction. When working with triangles in three-dimensional space, vectors provide a convenient way to describe the positions and relationships of the triangle's vertices. To determine the sides of a triangle given its vertices, we perform simple vector operations.

Finding a vector between two points involves a straightforward subtraction of their coordinates. For example, if a triangle has vertices at coordinates \(A(1,0,-1)\), \(B(3,-2,0)\), and \(C(1,3,3)\), the vector \( \vec{AB} \) is calculated as \( (B - A) \):

• \(\vec{AB} = (3-1, -2-0, 0-(-1)) = (2, -2, 1)\)
• \(\vec{BC} = (1-3, 3-(-2), 3-0) = (-2, 5, 3)\)
• \(\vec{CA} = (1-1, 0-3, -1-3) = (0, -3, -4)\)

Vectors can be further used to explore geometric properties, such as the magnitude of each side of the triangle. The magnitude of a vector \( \vec{v} = (x, y, z) \) is given by: \( |\vec{v}| = \sqrt{x^2 + y^2 + z^2} \). This formula helps in calculating the sides' lengths required for further computations, like the dot product.

Dot Product for Angle Calculation

The dot product is a key operation in vector mathematics used to calculate the angles between two vectors. This is particularly useful when analyzing the angles within a triangle.

The dot product between two vectors \( \vec{u} = (u_1, u_2, u_3) \) and \( \vec{v} = (v_1, v_2, v_3) \) is calculated as \( \vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2 + u_3v_3 \). This value gives insight into the relationship between the vectors. To find the cosine of the angle \( \theta \) between vectors, the dot product formula is extended to include their magnitudes:

• \( \vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos(\theta) \)

By rearranging, the cosine of the angle is \( \cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|} \). This formula allows us to calculate the angle within the triangle. For example, to find the angle at vertex \( A \), you use vectors \( \vec{CA} \) and \( \vec{AB} \) and their magnitudes. With these, you can determine:

\( \cos(A) = \frac{\vec{CA} \cdot \vec{AB}}{|\vec{CA}| |\vec{AB}|} \), which is essential for the next calculation steps.

Inverse Cosine Function in Trigonometry

After determining the cosine of the angle using the dot product, the next step is to find the actual angle in degrees. This is achieved through the inverse cosine function, denoted as \( \cos^{-1} \). This function reverses the cosine operation, allowing us to retrieve the angle from its cosine value.

In trigonometry, the inverse cosine function is widely used to solve for angles in various applications, including triangles. Here's how you use it to find an angle \( A \) within the triangle:

• From the previous calculation, suppose \( \cos(A) = \frac{2}{15} \).
• To find \( A \), compute \( A = \cos^{-1} \left(\frac{2}{15}\right) \).

This computation yields the measure of angle \( A \) in degrees. For practical purposes, it is often necessary to round this result to the nearest degree. Similar computations can be carried out for other angles of the triangle, providing a comprehensive understanding of the triangle's geometric structure.