मराठी

Find the Equation of the Line Passing Through the Point (2, −1, 3) and Parallel to the Line → R = ( ^ I − 2 ^ J + ^ K ) + λ ( 2 ^ I + 3 ^ J − 5 ^ K ) . - Mathematics

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प्रश्न

Find the equation of the line passing through the point (2, −1, 3) and parallel to the line  \[\overrightarrow{r} = \left( \hat{i} - 2 \hat{j} + \hat{k} \right) + \lambda\left( 2 \hat{i} + 3 \hat{j} - 5 \hat{k} \right) .\]

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उत्तर

The given line is parallel to the vector  \[2 \hat{i} + 3 \hat{j} - 5 \hat{k} \] and the required line is parallel to the given line. So, the required line is parallel to the vector \[2 \hat{i} + 3 \hat{j} - 5 \hat{k} \]  Hence, the equation of the required line passing through the point (2,-1, 3) and parallel to the vector  \[2 \hat{i} + 3 \hat{j} - 5 \hat{k} \]  is  \[\overrightarrow{r} = \left( 2 \hat{i} - \hat{j} + 3 \hat{k} \right) + \lambda\left( 2 \hat{i} + 3 \hat{j} - 5 \hat{k} \right)\] 

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पाठ 28: Straight Line in Space - Exercise 28.2 [पृष्ठ १६]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 28 Straight Line in Space
Exercise 28.2 | Q 14 | पृष्ठ १६

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संबंधित प्रश्‍न

 

A line passes through (2, −1, 3) and is perpendicular to the lines `vecr=(hati+hatj-hatk)+lambda(2hati-2hatj+hatk) and vecr=(2hati-hatj-3hatk)+mu(hati+2hatj+2hatk)` . Obtain its equation in vector and Cartesian from. 

 

 

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Find in vector form as well as in cartesian form, the equation of the line passing through the points A (1, 2, −1) and B (2, 1, 1).


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