Chapter 2: Problem 152
Consider points \(A(1,1), \quad B(2,-7), \quad\) and \(C(6,3)\) a. Determine vectors \(\overrightarrow{B A}\) and \(\overrightarrow{B C}\) . Express the answer in component form. b. Determine the measure of angle \(B\) in triangle \(A B C\) . Express the answer in degrees rounded to two decimal places.
Short Answer
Step by step solution
Find Vector BA
Find Vector BC
Find Dot Product of BA and BC
Calculate Magnitudes of BA and BC
Use the Dot Product to Find Angle B
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Dot Product
The formula for the dot product of two vectors \(\overrightarrow{u} = (u_1, u_2)\) and \(\overrightarrow{v} = (v_1, v_2)\) is:
- \(\overrightarrow{u} \cdot \overrightarrow{v} = u_1 \cdot v_1 + u_2 \cdot v_2\)
In our example, vectors \(\overrightarrow{BA} = (-1, 8)\) and \(\overrightarrow{BC} = (4, 10)\), the dot product is calculated as:
- \((-1 \times 4) + (8 \times 10) = -4 + 80 = 76\)
Magnitude of Vectors
To find the magnitude of a vector \(\overrightarrow{V} = (a, b)\), use the formula:
- \(||\overrightarrow{V}|| = \sqrt{a^2 + b^2}\)
- \(||\overrightarrow{BA}|| = \sqrt{(-1)^2 + 8^2} = \sqrt{1 + 64} = \sqrt{65}\)
- \(||\overrightarrow{BC}|| = \sqrt{4^2 + 10^2} = \sqrt{16 + 100} = \sqrt{116}\)
Angle Between Vectors
The formula used is:
- \(\cos(\theta) = \frac{\overrightarrow{u} \cdot \overrightarrow{v}}{||\overrightarrow{u}|| \times ||\overrightarrow{v}||}\)
- \(\overrightarrow{BA} \cdot \overrightarrow{BC} = 76\)
- \(||\overrightarrow{BA}|| = \sqrt{65}\)
- \(||\overrightarrow{BC}|| = \sqrt{116}\)
- \(\cos(B) = \frac{76}{\sqrt{65} \times \sqrt{116}} \approx 0.8742\)
- \(B = \cos^{-1}(0.8742) \approx 29.48^\circ\)
Component Form of Vectors
The component form of a vector with coordinates \((x_1, y_1)\) to \((x_2, y_2)\) is specified as:
- \((x_2 - x_1, y_2 - y_1)\)
- \(\overrightarrow{BA} = (1 - 2, 1 + 7) = (-1, 8)\)
- \(\overrightarrow{BC} = (6 - 2, 3 + 7) = (4, 10)\)