Chapter 2: Problem 279
If \(\mathrm{A}^{\rightarrow}=2 \mathrm{i} \wedge+5 \mathrm{j} \wedge-\mathrm{k} \wedge\) and \(\mathrm{B}^{\rightarrow}=3 \mathrm{i} \wedge-2 \mathrm{j} \wedge-4 \mathrm{k} \wedge\) the angle between \(\mathrm{A}^{\rightarrow}\) and \(\mathrm{B}^{\rightarrow}\) is (A) 0 (B) \((\pi / 2)\) (C) \((\pi / 4)\) (D) \((\pi / 6)\)
Short Answer
Step by step solution
Find the dot product of A and B
Calculate the dot product
Calculate the magnitude of A and B
Use the dot product formula to find the angle
Solve for θ
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Magnitude
Here's how:
- Square each component: \(2^2 = 4\), \(5^2 = 25\), \((-1)^2 = 1\)
- Add the squared values: \(4 + 25 + 1 = 30\)
- Take the square root of the sum: \(|A| = \sqrt{30}\)
The concept of magnitude is fundamental in linear algebra as it helps in determining the vector's size and comparing it with other vectors. Similarly, for vector \( \mathbf{B} = (3i, -2j, -4k) \), the magnitude is calculated as \( \sqrt{29} \) following the same steps.
Cosine of Angle Between Vectors
The formula is: \( A \cdot B = |A| |B| \cos(\theta) \).
- The dot product \(A \cdot B\) gives us a scalar value which, in this case, was found to be \(0\).
- The magnitudes \(|A|\) and \(|B|\) were calculated as \(\sqrt{30}\) and \(\sqrt{29}\), respectively.
Angle Between Vectors
For vectors \( \mathbf{A} = 2i + 5j - k \) and \( \mathbf{B} = 3i - 2j - 4k \), after finding their dot product and magnitudes, we observe
- If the dot product is zero, the vectors are perpendicular, as cosine \(\theta = 0\)
- If vectors are parallel, cosine \(\theta\) would be \(\pm 1\)
Linear Algebra Concepts
- Vectors are fundamental building blocks used to express direction and magnitude.
- The dot product assesses how much of one vector goes in the direction of another.
- Magnitude offers a measure of the length or size of a vector.