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The Value of ( → a × → B ) 2 is - Mathematics

MCQ

The value of \[\left( \vec{a} \times \vec{b} \right)^2\] is 

 

Options

  • \[\left| \vec{a} \right|^2 + \left| \vec{b} \right|^2 - \left( \vec{a} \cdot \vec{b} \right)^2\]

     

  • \[\left| \vec{a} \right|^2 \left| \vec{b} \right|^2 - \left( \vec{a} \cdot \vec{b} \right)^2\]

     

  • \[\left| \vec{a} \right|^2 + \left| \vec{b} \right|^2 - 2\left( \vec{a} \cdot \vec{b} \right)\]

     

  • \[\left| \vec{a} \right|^2 + \left| \vec{b} \right|^2 - \vec{a} \cdot \vec{b}\]

     

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Solution

\[\left( \vec{a} . \vec{b} \right)^2 + \left| \vec{a} \times \vec{b} \right|^2 \]
\[ = \left( \left| \vec{a} \right| \left| \vec{b} \right| \cos \theta \right)^2 + \left( \left| \vec{a} \right| \left| \vec{b} \right| \sin \theta \right)^2 \]
\[ = \left| \vec{a} \right|^2 \left| \vec{b} \right|^2 \left( \cos^2 \theta + \sin^2 \theta \right)\]
\[ = \left| \vec{a} \right|^2 \left| \vec{b} \right|^2 \left( 1 \right)\]
\[ = \left| \vec{a} \right|^2 \left| \vec{b} \right|^2 \]
\[ \therefore \left| \vec{a} \times \vec{b} \right|^2 = \left| \vec{a} \right|^2 \left| \vec{b} \right|^2 - \left( \vec{a} . \vec{b} \right)^2 \]
\[\text{ Thus, the value of }  \left( \vec{a} \times \vec{b} \right)^2 \text{ is } \left| \vec{a} \right|^2 \left| \vec{b} \right|^2 - \left( \vec{a} . \vec{b} \right)^2 .\]

  Is there an error in this question or solution?
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APPEARS IN

RD Sharma Class 12 Maths
Chapter 25 Vector or Cross Product
MCQ | Q 13 | Page 36
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