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Magazine Chapter 8: Problem 12
Find the angle between the vectors. $$ \langle 5,3\rangle ;\langle-6,10\rangle $$
Short Answer
Expert verified
The angle is 90 degrees.
Step by step solution
01
Understand the Problem
We need to find the angle between two vectors: \( \mathbf{a} = \langle 5,3 \rangle \) and \( \mathbf{b} = \langle -6,10 \rangle \). This can be done using the dot product formula.
02
Recall the Formula
The formula to find the angle \( \theta \) between two vectors \( \mathbf{a} \) and \( \mathbf{b} \) is given by:\[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} \]Where \( \mathbf{a} \cdot \mathbf{b} \) is the dot product and \( \|\mathbf{a}\| \) and \( \|\mathbf{b}\| \) are the magnitudes of the vectors.
03
Calculate the Dot Product
The dot product is calculated as follows:\[ \mathbf{a} \cdot \mathbf{b} = 5 \times (-6) + 3 \times 10 = -30 + 30 = 0 \]
04
Calculate the Magnitudes
The magnitude of vector \( \mathbf{a} \) is:\[ \|\mathbf{a}\| = \sqrt{5^2 + 3^2} = \sqrt{25 + 9} = \sqrt{34} \]The magnitude of vector \( \mathbf{b} \) is:\[ \|\mathbf{b}\| = \sqrt{(-6)^2 + 10^2} = \sqrt{36 + 100} = \sqrt{136} \]
05
Solve for Cosine of the Angle
Using the dot product and magnitudes, substitute into the formula:\[ \cos \theta = \frac{0}{\sqrt{34} \times \sqrt{136}} = 0 \]
06
Find the Angle
Since \( \cos \theta = 0 \), the angle \( \theta = \cos^{-1}(0) = \frac{\pi}{2} \) radians or 90 degrees.
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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 method that allows us to multiply two vectors, giving us a scalar value. This scalar is crucial when finding the angle between two vectors.
To calculate the dot product of vectors \(\mathbf{a} = \langle 5, 3 \rangle\) and \(\mathbf{b} = \langle -6, 10 \rangle\), we use the formula:
To calculate the dot product of vectors \(\mathbf{a} = \langle 5, 3 \rangle\) and \(\mathbf{b} = \langle -6, 10 \rangle\), we use the formula:
- Multiply the corresponding components: \(5 \times (-6)\) and \(3 \times 10\).
- Add the results: \((-30) + 30\).
Magnitude of Vectors
The magnitude of a vector gives us its length. It鈥檚 like finding the diagonal of a box using the Pythagorean theorem.
To find the magnitude of \(\mathbf{a} = \langle 5, 3 \rangle\), apply the formula \(\|\mathbf{a}\| = \sqrt{5^2 + 3^2}\). This simplifies to \(\sqrt{25 + 9} = \sqrt{34}\).
For \(\mathbf{b} = \langle -6, 10 \rangle\), we calculate \(\|\mathbf{b}\| = \sqrt{(-6)^2 + 10^2}\), resulting in \(\sqrt{36 + 100} = \sqrt{136}\).
To find the magnitude of \(\mathbf{a} = \langle 5, 3 \rangle\), apply the formula \(\|\mathbf{a}\| = \sqrt{5^2 + 3^2}\). This simplifies to \(\sqrt{25 + 9} = \sqrt{34}\).
For \(\mathbf{b} = \langle -6, 10 \rangle\), we calculate \(\|\mathbf{b}\| = \sqrt{(-6)^2 + 10^2}\), resulting in \(\sqrt{36 + 100} = \sqrt{136}\).
- These magnitudes tell us how "long" each vector is.
- Altogether, they're essential to determine the angle between vectors with the cosine formula.
Cosine Formula
The cosine formula is a powerful tool for calculating the angle between two vectors. It relates the dot product of the vectors to their magnitudes. The formula used is: \[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} \] In our problem, since the dot product \(\mathbf{a} \cdot \mathbf{b} = 0\), we have: \[ \cos \theta = \frac{0}{\sqrt{34} \times \sqrt{136}} = 0 \]
- When the cosine of an angle is 0, it indicates the vectors are perpendicular.
- This means \(\theta = \cos^{-1}(0)\), which gives us \(\theta = \frac{\pi}{2}\) radians or 90 degrees.
Vector Components
Vector components break down a vector into its individual parts, which are easier to manage when doing calculations like dot products or magnitudes.
Each vector is typically represented as \(\langle x, y \rangle\), which captures its direction and magnitude in two-dimensional space. For our vectors:
Each vector is typically represented as \(\langle x, y \rangle\), which captures its direction and magnitude in two-dimensional space. For our vectors:
- \(\mathbf{a} = \langle 5, 3 \rangle\) has components 5 and 3.
- \(\mathbf{b} = \langle -6, 10 \rangle\) has components -6 and 10.
- We can easily apply mathematical operations, like addition or scalar multiplication.
- Components also simplify applying geometric interpretations, such as finding angles or determining alignments.
