The resultant of the two vectors having magnitude 2 and 3 is 1. What is their cross product
Text Solution
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Let the angle between the vectors be \(\theta\) .
Given:
|\vec{A}|=2, |\vec{B}|=3, |\vec{A}+\vec{B}|=1
Using
\(|\vec{A}+\vec{B}|^{2}=|\vec{A}|^{2}+|\vec{B}|^{2}+2|\vec{A}||\vec{B}|\cos\theta\)
\(1^{2} = 2^{2} + 3^{2} - 2(2)(3) \cos \theta\)
\(1 = 4 + 9 - 12 \cos \theta\)
\(12 \cos \theta = -12\)
\(\cos \theta = -1\)
Thus, \(\theta = 180^\circ\) .
Cross product:
\(|\vec{A} \times \vec{B}| = |\vec{A}| |\vec{B}| \sin \theta\)
\(= 2 \times 3 \times \sin 180^\circ = 0\)
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