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Find the Area of the Parallelogram Determined by the Vector 3 ^ I + ^ J − 2 ^ K and ^ I − 3 ^ J + 4 ^ K

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प्रश्न

Find the area of the parallelogram determined by the vector \[3 \hat{ i } + \hat{ j }  - 2 \hat{ k } \text{  and }  \hat{ i }  - 3 \hat{ j }  + 4 \hat{ k } \] .

 

योग
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उत्तर

\[ \text{ Let } : \]
\[ \vec{a} = 3 \hat{ i }  + \hat{ j }  - 2 \hat{ k }  \]
\[ \vec{b} = 1 \hat{ i } - 3 \hat{ j }  + 4 \hat{ k } \]
\[ \vec{a} \times \vec{b} = \begin{vmatrix}\hat{ i } & \hat{ j }  & \hat{ k }  \\ 3 & 1 & - 2 \\ 1 & - 3 & 4\end{vmatrix}\]
\[ = \hat{ i } \left( 4 - 6 \right) - \hat{ j }  \left( 12 + 2 \right) + \hat{ k } \left( - 9 - 1 \right)\]
\[ = - 2 \hat{ i } - 14 \hat{ j  }  - 10 \hat{ k }  \]
\[\text{ Area of the parallelogram } =\left| \vec{a} \times \vec{b} \right|\]
\[ = \sqrt{\left( - 2 \right)^2 + \left( - 14 \right)^2 + \left( - 10 \right)^2}\]
\[ = \sqrt{300}\]
\[ = 10\sqrt{3} \text{ sq. units } \]

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 24: Vector or Cross Product - Exercise 25.1 [पृष्ठ २९]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 24 Vector or Cross Product
Exercise 25.1 | Q 8.3 | पृष्ठ २९
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