मराठी

Find the Equation of a Line Parallel to X-axis and Passing Through the Origin.

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प्रश्न

Find the equation of a line parallel to x-axis and passing through the origin.

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उत्तर

The direction ratios of the line parallel to x-axis are proportional to 1, 0, 0.

Equation of the line passing through the origin and parallel to x-axis is 

\[\frac{x - 0}{1} = \frac{y - 0}{0} = \frac{z - 0}{0}\]

\[ = \frac{x}{1} = \frac{y}{0} = \frac{z}{0}\]

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पाठ 27: Straight Line in Space - Exercise 28.2 [पृष्ठ १६]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 27 Straight Line in Space
Exercise 28.2 | Q 7 | पृष्ठ १६

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If the Cartesian equations of a line are ` (3-x)/5=(y+4)/7=(2z-6)/4` , write the vector equation for the line.


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ABCD is a parallelogram. The position vectors of the points AB and C are respectively, \[4 \hat{ i} + 5 \hat{j} -10 \hat{k} , 2 \hat{i} - 3 \hat{j} + 4 \hat{k}  \text{ and } - \hat{i} + 2 \hat{j} + \hat{k} .\]  Find the vector equation of the line BD. Also, reduce it to cartesian form.


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Find the cartesian equation of the line which passes through the point (−2, 4, −5) and parallel to the line given by  \[\frac{x + 3}{3} = \frac{y - 4}{5} = \frac{z + 8}{6} .\]


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Find the angle between the following pair of line:

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Determine the equations of the line passing through the point (1, 2, −4) and perpendicular to the two lines \[\frac{x - 8}{8} = \frac{y + 9}{- 16} = \frac{z - 10}{7} \text{    and    } \frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{- 5}\]


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Prove that the line \[\vec{r} = \left( \hat{i }+ \hat{j }- \hat{k} \right) + \lambda\left( 3 \hat{i} - \hat{j} \right) \text{ and } \vec{r} = \left( 4 \hat{i} - \hat{k} \right) + \mu\left( 2 \hat{i} + 3 \hat{k} \right)\] intersect and find their point of intersection.


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