Chapter 1: Problem 45
Find the angle between each of these pairs of vectors: (a) \(\overrightarrow{A}\) = \(-\)2.00\(\hat{\imath}\) \(+\) 6.00\(\hat{\jmath}\) and \(\overrightarrow{B}\) = 2.00\(\hat{\imath}\) \(-\) 3.00\(\hat{\jmath}\) (b) \(\overrightarrow{A}\) = 3.00\(\hat{\imath}\) \(+\) 5.00\(\hat{\jmath}\) and \(\overrightarrow{B}\) = 10.00\(\hat{\imath}\) \(+\) 6.00\(\hat{\jmath}\) (c) \(\overrightarrow{A}\) = \(-\)4.00\(\hat{\imath}\) \(+\) 2.00\(\hat{\jmath}\) and \(\overrightarrow{B}\) = 7.00\(\hat{\imath}\) \(+\) 14.00\(\hat{\jmath}\)
Short Answer
Step by step solution
Use Dot Product Formula
Step 2(a): Calculate Dot Product for Part (a)
Step 3(a): Calculate Magnitudes for Part (a)
Step 4(a): Find Angle for Part (a)
Step 2(b): Calculate Dot Product for Part (b)
Step 3(b): Calculate Magnitudes for Part (b)
Step 4(b): Find Angle for Part (b)
Step 2(c): Calculate Dot Product for Part (c)
Step 3(c): Calculate Magnitudes for Part (c)
Step 4(c): Find Angle for Part (c)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Dot Product
- Multiply the corresponding components: \(a_1 \cdot b_1 + a_2 \cdot b_2\).
- Sum these products to obtain a scalar quantity.
Vector Magnitudes
- \(\sqrt{a_1^2 + a_2^2}\), where \(a_1\) and \(a_2\) are the vector components.
Angle Between Vectors
- \(\cos(\theta) = \frac{\overrightarrow{A} \cdot \overrightarrow{B}}{|\overrightarrow{A}| |\overrightarrow{B}|}\)
- If the dot product is zero, the angle is \(90^\circ\), indicating perpendicular vectors.
- A \(\cos(\theta)\) value of 1 means the vectors are parallel.
- If \(\cos(\theta)\) is \(-1\), vectors are in opposite directions.
Cosine Inverse Calculation
- Substitute the calculated \(\cos(\theta)\) value into the inverse cosine function: \(\theta = \cos^{-1}(\cos(\theta))\).