Question
TASK 1: Finding the Angle Between Two Vectors (6 pts)
Given the vectors:
a=<3,4>, b=<6,8>
Compute the dot product ab.
Find the magnitude of each vector |a| and |b|.
Use the formula to determine the angle between the vectors.
cos = a b||a||||b||
NOTE: Numerator is DOT PRODUCT. Denominator is MULTIPLICATION
Interpret the result: What does this angle tell us about the relationship between the vectors?
Given the vectors:
a=<3,4>, b=<6,8>
Compute the dot product ab.
Find the magnitude of each vector |a| and |b|.
Use the formula to determine the angle between the vectors.
cos = a b||a||||b||
NOTE: Numerator is DOT PRODUCT. Denominator is MULTIPLICATION
Interpret the result: What does this angle tell us about the relationship between the vectors?
Answers
Answered by
GPT-4o mini
To find the angle between the vectors \( a \) and \( b \), we will follow the steps provided.
1. **Calculate the dot product \( a \cdot b \)**:
The dot product of two vectors \( a = \langle a_1, a_2 \rangle \) and \( b = \langle b_1, b_2 \rangle \) is given by:
\[
a \cdot b = a_1b_1 + a_2b_2
\]
For our vectors \( a = \langle 3, 4 \rangle \) and \( b = \langle 6, 8 \rangle \):
\[
a \cdot b = 3 \cdot 6 + 4 \cdot 8 = 18 + 32 = 50
\]
2. **Calculate the magnitudes of each vector**:
The magnitude of a vector \( a = \langle a_1, a_2 \rangle \) is given by:
\[
|a| = \sqrt{a_1^2 + a_2^2}
\]
For vector \( a = \langle 3, 4 \rangle \):
\[
|a| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5
\]
For vector \( b = \langle 6, 8 \rangle \):
\[
|b| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10
\]
3. **Use the formula to determine the cosine of the angle \( \theta \)**:
The formula for the cosine of the angle between the vectors is:
\[
\cos(\theta) = \frac{a \cdot b}{|a| |b|}
\]
Substituting in the values we calculated:
\[
\cos(\theta) = \frac{50}{5 \cdot 10} = \frac{50}{50} = 1
\]
4. **Determine the angle \( \theta \)**:
Since \( \cos(\theta) = 1 \), we can find the angle \( \theta \):
\[
\theta = \cos^{-1}(1) = 0^\circ
\]
5. **Interpret the result**:
The angle of \( 0^\circ \) between the vectors indicates that the vectors \( a \) and \( b \) are parallel and point in the same direction. In this case, the vector \( b \) is a scalar multiple of vector \( a \) (specifically, \( b = 2a \)). This means that the two vectors share the same direction, and there is a direct proportionality between their magnitudes.
1. **Calculate the dot product \( a \cdot b \)**:
The dot product of two vectors \( a = \langle a_1, a_2 \rangle \) and \( b = \langle b_1, b_2 \rangle \) is given by:
\[
a \cdot b = a_1b_1 + a_2b_2
\]
For our vectors \( a = \langle 3, 4 \rangle \) and \( b = \langle 6, 8 \rangle \):
\[
a \cdot b = 3 \cdot 6 + 4 \cdot 8 = 18 + 32 = 50
\]
2. **Calculate the magnitudes of each vector**:
The magnitude of a vector \( a = \langle a_1, a_2 \rangle \) is given by:
\[
|a| = \sqrt{a_1^2 + a_2^2}
\]
For vector \( a = \langle 3, 4 \rangle \):
\[
|a| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5
\]
For vector \( b = \langle 6, 8 \rangle \):
\[
|b| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10
\]
3. **Use the formula to determine the cosine of the angle \( \theta \)**:
The formula for the cosine of the angle between the vectors is:
\[
\cos(\theta) = \frac{a \cdot b}{|a| |b|}
\]
Substituting in the values we calculated:
\[
\cos(\theta) = \frac{50}{5 \cdot 10} = \frac{50}{50} = 1
\]
4. **Determine the angle \( \theta \)**:
Since \( \cos(\theta) = 1 \), we can find the angle \( \theta \):
\[
\theta = \cos^{-1}(1) = 0^\circ
\]
5. **Interpret the result**:
The angle of \( 0^\circ \) between the vectors indicates that the vectors \( a \) and \( b \) are parallel and point in the same direction. In this case, the vector \( b \) is a scalar multiple of vector \( a \) (specifically, \( b = 2a \)). This means that the two vectors share the same direction, and there is a direct proportionality between their magnitudes.