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For Any Two Vectors → a and → B , Prove that ∣ ∣ → a × → B ∣ ∣ 2 = ∣ ∣ ∣ → a . → a → a . → B → B . → a → B . → B ∣ ∣ ∣ - Mathematics

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प्रश्न

For any two vectors \[\vec{a} \text{ and }  \vec{b}\] , prove that \[\left| \vec{a} \times \vec{b} \right|^2 = \begin{vmatrix}\vec{a} . \vec{a} & & \vec{a} . \vec{b} \\ \vec{b} . \vec{a} & & \vec{b} . \vec{b}\end{vmatrix}\]

 
 
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उत्तर

\[RHS=\begin{vmatrix}\vec{a} . \vec{a} & \vec{a} . \vec{b} \\ \vec{b} . \vec{a} & \vec{b} . \vec{b}\end{vmatrix}\]

\[ =\begin{vmatrix}\left| \vec{a} \right|^2 & \left| \vec{a} \right| \left| \vec{b} \right| \cos \theta \\ \left| \vec{a} \right| \left| \vec{b} \right| \cos \theta & \left| \vec{b} \right|^2\end{vmatrix}\]

\[ = \left| \vec{a} \right|^2 \left| \vec{b} \right|^2 - \left| \vec{a} \right|^2 \left| \vec{b} \right|^2 \cos^2 \theta\]

\[ = \left| \vec{a} \right|^2 \left| \vec{b} \right|^2 \left( 1 - \cos^2 \theta \right)\]

\[ = \left| \vec{a} \right|^2 \left| \vec{b} \right|^2 \sin^2 \theta\]

\[ = \left( \left| \vec{a} \right|\left| \vec{b} \right| \sin \theta \right)^2 \]

\[ = \left| \vec{a} \times \vec{b} \right|^2 \]

\[ = LHS\]

\[ \text{ Hence proved } . \]

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अध्याय 25: Vector or Cross Product - Exercise 25.1 [पृष्ठ ३०]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 25 Vector or Cross Product
Exercise 25.1 | Q 23 | पृष्ठ ३०

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