Question

Resultant of two vectors having magnitude 5 and 4 is 1 what is the magnitude of cross product.

Answer 1

GiveN :

• 1st vector Quantity (A) = 5
• 2nd vector quantity (B) = 5
• Resultant (|AB|) = 1

To FinD :

• Their Cross product

SolutioN :

Use formula for parallelogram law of vector addition :

$\implies \sf{|AB| \: = \: \sqrt{A^2 \: + \: B^2 \: + \: 2AB \cos \theta}} \\ \\ \implies \sf{1 \: = \: \sqrt{5^2 \: + \: 4^2 \: + \: 2(5 \: \times \: 4 \: \times \: \cos \theta) }} \\ \\ \implies \sf{1 \: = \: \sqrt{25 \: + \: 16 \: + \: 40 \cos \theta}} \\ \\ \implies \sf{1 \: = \: \sqrt{41 \: + \: 40 \cos \theta}} \\ \\ \implies \sf{1^2 \: = \: (\sqrt{41 \: + \: 40 \cos \theta})^2} \\ \\ \implies \sf{1 \: = \: 41 \: + \: 40 \cos \theta} \\ \\ \implies \sf{40 \cos \theta \: = \: 1 \: - \: 41} \\ \\ \implies \sf{40 \cos \theta \: = \: - \: 40} \\ \\ \implies \sf{\cos \theta \: = \: \dfrac{-40}{40}} \\ \\ \implies \sf{\cos \theta \: = \: -1} \\ \\ \implies \sf{\sin \theta \: = \: 0}$

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Now, we will find value of cosØ

$\implies \sf{A \: \times \: B} \\ \\ \implies \sf{A \: \times \: B \: = \: AB \sin \theta} \\ \\ \implies \sf{A \: \times \: B \: = \: 4 \: \times \: 5 \: \times \: \sin \theta} \\ \\ \implies \sf{A \: \times \: B \: = \: 20 \sin \theta} \\ \\ \implies \sf{A \: \times \: B \: = \: 20(0)} \\ \\ \implies \sf{A \: \times \: B \: = \: 0} \\ \\ \underline{\sf{\therefore \: Cross \: product \: is \: 0}}$