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The Cartesian Equations of a Line Are X − 5 3 = Y + 4 7 = Z − 6 2 . Find a Vector Equation for the Line. - Mathematics

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प्रश्न

The cartesian equations of a line are \[\frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2} .\]  Find a vector equation for the line.

योग
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उत्तर

The cartesian equation of the given line is \[\frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2}\] 

It can be re-written as 

\[\frac{x - 5}{3} = \frac{y - \left( - 4 \right)}{7} = \frac{z - 6}{2}\] 

Thus, the given line passes through the point having position vector   \[\vec{a} = 5 \hat{i} - 4 \hat{j} + 6 \hat{k}\] and is parallel to the vector  \[\vec{b} = 3 \hat{i}  + 7 \hat{j}+ 2 \hat{k} \] 

We know that the vector equation of a line passing through a point with position vector  `vec a ` and parallel to the vector `vec b` is  \[\vec{r} = \vec{a} + \lambda \vec{b}\] 

Vector equation of the required line is 

\[\vec{r} = \left( 5 \hat{i} - 4 \hat{j} + 6 \hat{k} \right) + \lambda \left( 3 \hat{i} + 7 \hat{j} + 2 \hat{k} \right)\]

\[\text{ Here} , \lambda \text{ is a parameter } . \]

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अध्याय 28: Straight Line in Space - Exercise 28.1 [पृष्ठ १०]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 28 Straight Line in Space
Exercise 28.1 | Q 9 | पृष्ठ १०

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