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Magazine Chapter 8: Problem 10
Find the angle between the vectors. $$ \langle-4,1\rangle ;\langle 8,-2\rangle $$
Short Answer
Expert verified
The angle between the vectors is \( 180^{\circ} \) or \( \pi \) radians.
Step by step solution
01
Write Down the Formula
To find the angle \( \theta \) between two vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \), we use the formula: \[ \cos(\theta) = \frac{\mathbf{a} \cdot \mathbf{b}}{\| \mathbf{a} \| \| \mathbf{b} \|} \] where \( \mathbf{a} \cdot \mathbf{b} \) is the dot product and \( \| \mathbf{a} \| \) and \( \| \mathbf{b} \| \) are the magnitudes of the vectors.
02
Calculate the Dot Product
Calculate the dot product of \( \mathbf{a} = \langle -4, 1 \rangle \) and \( \mathbf{b} = \langle 8, -2 \rangle \): \[ \mathbf{a} \cdot \mathbf{b} = (-4 \times 8) + (1 \times -2) = -32 - 2 = -34 \] So, the dot product \( \mathbf{a} \cdot \mathbf{b} = -34 \).
03
Calculate the Magnitudes
Calculate the magnitude of both vectors. For \( \mathbf{a} \): \[ \| \mathbf{a} \| = \sqrt{(-4)^2 + 1^2} = \sqrt{16 + 1} = \sqrt{17} \] For \( \mathbf{b} \): \[ \| \mathbf{b} \| = \sqrt{8^2 + (-2)^2} = \sqrt{64 + 4} = \sqrt{68} \] So, \( \| \mathbf{a} \| = \sqrt{17} \) and \( \| \mathbf{b} \| = \sqrt{68} \).
04
Solve for Cosine of the Angle
Substitute the dot product and magnitudes into the formula: \[ \cos(\theta) = \frac{-34}{\sqrt{17} \times \sqrt{68}} \] Simplify: \[ \sqrt{68} = \sqrt{4 \times 17} = 2\sqrt{17} \] Therefore: \[ \cos(\theta) = \frac{-34}{\sqrt{17} \times 2\sqrt{17}} = \frac{-34}{2 \times 17} = \frac{-34}{34} = -1 \]
05
Find the Angle
Since \( \cos(\theta) = -1 \), the angle \( \theta = \cos^{-1}(-1) \) is \( 180^{\circ} \) or \( \pi \) radians.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Dot Product
The dot product is a fundamental operation in vector algebra that allows us to measure how two vectors "point" in relation to each other. To compute the dot product of two vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \), we multiply corresponding components and then sum them up:
- \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \)
Vector Magnitudes
The magnitude of a vector is like its "length" or how much space it occupies. For a vector \( \mathbf{a} = \langle a_1, a_2 \rangle \), the magnitude is given by the square root of the sum of the squares of its components:
- \( \| \mathbf{a} \| = \sqrt{a_1^2 + a_2^2} \)
- For \( \mathbf{a} = \langle -4, 1 \rangle \), \( \| \mathbf{a} \| = \sqrt{(-4)^2 + 1^2} = \sqrt{17} \)
- For \( \mathbf{b} = \langle 8, -2 \rangle \), \( \| \mathbf{b} \| = \sqrt{8^2 + (-2)^2} = \sqrt{68} \)
Cosine Formula
The cosine formula is the hero of this calculation, as it connects the dot product and magnitudes to the angle between two vectors. The formula is:
- \( \cos(\theta) = \frac{\mathbf{a} \cdot \mathbf{b}}{\| \mathbf{a} \| \| \mathbf{b} \|} \)
- \( \cos(\theta) = \frac{-34}{\sqrt{17} \times \sqrt{68}} \)
- Simplified further because \( \sqrt{68} = 2\sqrt{17} \)
- Finally, \( \cos(\theta) = \frac{-34}{34} = -1 \)
Trigonometry
Trigonometry is the mathematical study of triangles and the relationships between their sides and angles. In this context, it helps us determine angles using cosine, a fundamental trigonometric function. The inverse cosine, \(\cos^{-1}\), allows us to find the angle when we know its cosine value.
Here, once we find \( \cos(\theta) = -1 \), we determine that \( \theta = \cos^{-1}(-1) \) leads us to an angle of \(180^\circ\). This tells us that if the two vectors lie on a coordinate plane, they form a straight line extending in opposite directions, highlighting trigonometry's power to reveal geometric insights from algebraic expressions.
Here, once we find \( \cos(\theta) = -1 \), we determine that \( \theta = \cos^{-1}(-1) \) leads us to an angle of \(180^\circ\). This tells us that if the two vectors lie on a coordinate plane, they form a straight line extending in opposite directions, highlighting trigonometry's power to reveal geometric insights from algebraic expressions.
