If \(\left| {\vec a\;} \right| = 13,\;\left| {\vec b} \right| = 5\) and \(\vec a\;.\;\vec b = 60\) , then the value of \(\left| {\vec a \times \vec b} \right|\) is

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TNUSRB SI 2020 Official Paper (Held on 12 January 2020)
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  1. 45
  2. 35
  3. 25
  4. 15

Answer (Detailed Solution Below)

Option 3 : 25
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Detailed Solution

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Explanation:
The dot product and the cross product of two vectors have a particular relationship. For two vectors, a and b, their dot product is given by:
 
\( {\vec a . \vec b} \)= \(\left| {\vec a } \right|\) * \(\left| {\vec b } \right|\) * cos(θ)
 
and the magnitude of their cross product is given by:
 
\(\left| {\vec a \times \vec b} \right|\) =  \(\left| {\vec a } \right|\) * \(\left| {\vec b } \right|\) * sin(θ)
 
where θ is the angle between the vectors.
 
Given that \(\left| {\vec a . \vec b} \right|\)=  \(\left| {\vec a } \right|\) * \(\left| {\vec b } \right|\) * cos(θ) = 60, we can solve for cos(θ):
 
cos(θ) = \( {\vec a . \vec b} \) /  \(\left| {\vec a } \right|\) * \(\left| {\vec b } \right|\) ) = 60 / (13 * 5) = 60 / 65 = 12/13.
 
Then, we can find sin(θ) using the Pythagorean identity sin²(θ) + cos²(θ) = 1:
 
sin(θ) = sqrt(1 - cos²(θ)) = sqrt(1 - (12/13)²) = sqrt(1 - 144/169) = sqrt(25/169) = 5/13.
 
Finally, we can substitute\(\left| {\vec a } \right|\) , \(\left| {\vec b } \right|\) and sin(θ) into the equation for \(\left| {\vec a \times \vec b} \right|\) to find:
 
\(\left| {\vec a \times \vec b} \right|\) =  \(\left| {\vec a } \right|\) * \(\left| {\vec b } \right|\)  * sin(θ) = 13 * 5 * (5/13) = 25.
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