Macdonald Gray
04/01/2023 · Junior High School
If \( \vec{A}=(2,-3) \) and \( \vec{B} / / \vec{A} \) and if \( \|\vec{B}\|=3 \sqrt{13} \) Find : \( \vec{B} \)
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Tutor-Verified Answer
Step-by-step Solution
Given that \( \vec{A} = (2, -3) \) and \( \vec{B} \) is parallel to \( \vec{A} \), we can express \( \vec{B} \) as a scalar multiple of \( \vec{A} \). Therefore, we can write:
\[
\vec{B} = k \vec{A} = k(2, -3) = (2k, -3k)
\]
where \( k \) is a scalar.
We are also given that the magnitude of \( \vec{B} \) is \( \|\vec{B}\| = 3\sqrt{13} \). The magnitude of \( \vec{B} \) can be calculated as follows:
\[
\|\vec{B}\| = \sqrt{(2k)^2 + (-3k)^2} = \sqrt{4k^2 + 9k^2} = \sqrt{13k^2} = |k| \sqrt{13}
\]
Setting this equal to the given magnitude:
\[
|k| \sqrt{13} = 3\sqrt{13}
\]
Dividing both sides by \( \sqrt{13} \) (assuming \( \sqrt{13} \neq 0 \)) gives:
\[
|k| = 3
\]
This means \( k \) can be either \( 3 \) or \( -3 \). Thus, we have two possible vectors for \( \vec{B} \):
1. If \( k = 3 \):
\[
\vec{B} = 3(2, -3) = (6, -9)
\]
2. If \( k = -3 \):
\[
\vec{B} = -3(2, -3) = (-6, 9)
\]
Therefore, the possible vectors \( \vec{B} \) are:
\[
\vec{B} = (6, -9) \quad \text{or} \quad \vec{B} = (-6, 9)
\]
Thus, the final answer is:
\[
\vec{B} = (6, -9) \quad \text{or} \quad \vec{B} = (-6, 9)
\]
Quick Answer
The possible vectors \( \vec{B} \) are \( (6, -9) \) or \( (-6, 9) \).
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