Let vectors A = (2, 1, -4), B = (-3, 0, 1), and C = (-1, -1, 2). Calculate the following: A · B, What is the angle θAB between A and B? Express your answer using one significant fi... Let vectors A = (2, 1, -4), B = (-3, 0, 1), and C = (-1, -1, 2). Calculate the following: A · B, What is the angle θAB between A and B? Express your answer using one significant figure. 2B · 3C.
Understand the Problem
The question is asking to perform calculations involving vector operations, such as the dot product and finding the angle between vectors. The user is required to calculate the dot product of two vectors and the angle between them, based on the provided vector values.
Answer
$\mathbf{A} \cdot \mathbf{B} = -10$, $\theta_{AB} = \cos^{-1}\left(\frac{-10}{\sqrt{210}}\right)$, $2\mathbf{B} \cdot 3\mathbf{C} = 30$
Answer for screen readers
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The dot product $\mathbf{A} \cdot \mathbf{B} = -10$
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The angle $\theta_{AB} = \cos^{-1}\left(\frac{-10}{\sqrt{210}}\right)$
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The result of $2\mathbf{B} \cdot 3\mathbf{C} = 30$
Steps to Solve
- Calculate the dot product $\mathbf{A} \cdot \mathbf{B}$
Using the formula for the dot product of two vectors:
$$ \mathbf{A} \cdot \mathbf{B} = A_1 B_1 + A_2 B_2 + A_3 B_3 $$
Substituting the values of vectors:
$$ \mathbf{A} = (2, 1, -4) \quad \text{and} \quad \mathbf{B} = (-3, 0, 1) $$
We calculate:
$$ \mathbf{A} \cdot \mathbf{B} = (2)(-3) + (1)(0) + (-4)(1) $$
$$ = -6 + 0 - 4 = -10 $$
- Find the magnitude of each vector
The magnitude of vector $\mathbf{A}$:
$$ ||\mathbf{A}|| = \sqrt{A_1^2 + A_2^2 + A_3^2} = \sqrt{2^2 + 1^2 + (-4)^2} $$
$$ = \sqrt{4 + 1 + 16} = \sqrt{21} $$
The magnitude of vector $\mathbf{B}$:
$$ ||\mathbf{B}|| = \sqrt{(-3)^2 + 0^2 + 1^2} = \sqrt{9 + 0 + 1} = \sqrt{10} $$
- Calculate the angle $\theta_{AB}$ between vectors $\mathbf{A}$ and $\mathbf{B}$
Using the cosine formula:
$$ \cos(\theta_{AB}) = \frac{\mathbf{A} \cdot \mathbf{B}}{||\mathbf{A}|| \cdot ||\mathbf{B}||} $$
Substituting the known values:
$$ \cos(\theta_{AB}) = \frac{-10}{\sqrt{21} \cdot \sqrt{10}} $$
Calculate the angle:
$$ \theta_{AB} = \cos^{-1}\left(\frac{-10}{\sqrt{210}} \right) $$
- Calculate $2 \mathbf{B} \cdot 3 \mathbf{C}$
First, calculate $2\mathbf{B}$ and $3\mathbf{C}$:
$$ 2\mathbf{B} = 2(-3, 0, 1) = (-6, 0, 2) $$
$$ 3\mathbf{C} = 3(-1, -1, 2) = (-3, -3, 6) $$
Now find the dot product:
$$ 2\mathbf{B} \cdot 3\mathbf{C} = (-6)(-3) + (0)(-3) + (2)(6) $$
$$ = 18 + 0 + 12 = 30 $$
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The dot product $\mathbf{A} \cdot \mathbf{B} = -10$
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The angle $\theta_{AB} = \cos^{-1}\left(\frac{-10}{\sqrt{210}}\right)$
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The result of $2\mathbf{B} \cdot 3\mathbf{C} = 30$
More Information
The dot product gives a measure of how aligned the two vectors are. A negative dot product indicates they are pointing in somewhat opposite directions. The angle calculated shows the exact angle between the vectors in radians, which can yield useful geometric insights.
Tips
- Forgetting to square the components when calculating magnitudes.
- Using incorrect signs while multiplying vector components in the dot product.
- Overlooking to convert the angle from radians to degrees if needed, depending on context.
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