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Find a vector whose length is 3 and which is perpendicular to the vector → a = 3 ^ i + ^ j − 4 ^ k and → b = 6 ^ i + 5 ^ j − 2 ^ k . - Mathematics

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प्रश्न

Find a vector whose length is 3 and which is perpendicular to the vector \[\vec{a} = 3 \hat{ i }  + \hat{ j  } - 4 \hat{ k }  \text{ and }  \vec{b} = 6 \hat{ i }  + 5 \hat{ j }  - 2 \hat{ k } .\]

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उत्तर

\[\text{ Given } : \]

\[ \vec{a} = 3 \hat{ i }  + \hat{ j }  - 4 \hat{ k }  \]

\[ \vec{b} = 6 \hat{ i }  + 5 \hat{ j }  - 2 \hat{ k }  \]

\[ \therefore a^\to \times \vec{b} = \begin{vmatrix}\hat{ i }  & \hat{ j }  & \hat { k }  \\ 3 & 1 & - 4 \\ 6 & 5 & - 2\end{vmatrix}\]

\[ = \left( - 2 + 20 \right) \hat{ i } - \left( - 6 + 24 \right) \hat{ j }  + \left( 15 - 6 \right) \hat{ k }  \]

\[ = 18 \hat{ i }  - 18 \hat{ j }  + 9 \hat{ k }  \]

\[ \Rightarrow \left| \vec{a} \times \vec{b} \right| = \sqrt{{18}^2 + \left( - {18}^2 \right) + 9^2}\]

\[ = \sqrt{729}\]

\[ = 27\]

\[\text{ Required vector } = 3 \times \left\{ \frac{\vec{a} \times \vec{b}}{\left| \vec{a} \times \vec{b} \right|} \right\}\]

\[ = 3 \times \frac{18 \hat{ i }  - 18 \hat{ j }  + 9 \hat{ k } }{27}\] 

\[ = \frac{3\left( 2 \hat{ i } - 2 \hat{ j }  + \hat{ k }  \right)}{3}\]

\[ = 2 \hat{ i }  - 2 \hat{ j }  + \hat{ k }  \]

 

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अध्याय 25: Vector or Cross Product - Exercise 25.1 [पृष्ठ २९]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 25 Vector or Cross Product
Exercise 25.1 | Q 7.2 | पृष्ठ २९

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