मराठी

Find the Equation of the Line Passing Through the Point (1, 2, −4) and Parallel to the Line X − 3 4 = Y − 5 2 = Z + 1 3 .

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प्रश्न

Find the equation of the line passing through the point (1, 2, −4) and parallel to the line \[\frac{x - 3}{4} = \frac{y - 5}{2} = \frac{z + 1}{3} .\] 

बेरीज
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उत्तर

The direction ratios of the line parallel to line  \[\frac{x - 3}{4} = \frac{y - 5}{2} = \frac{z + 1}{3}\] are proportional to 4, 2, 3.


Equation of the required line passing through the point (1, 2,-4) having direction ratios proportional to 4, 2, 3 is

\[\frac{x - 1}{4} = \frac{y - 2}{2} = \frac{z - \left( - 4 \right)}{3}\]

\[ = \frac{x - 1}{4} = \frac{y - 2}{2} = \frac{z + 4}{3}\]

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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 12 | पृष्ठ १६

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The Cartesian equation of a line is `(x-5)/3 = (y+4)/7 = (z-6)/2` Write its vector form.


Find the vector and Cartesian equations of a line passing through (1, 2, –4) and perpendicular to the two lines `(x - 8)/3 = (y + 19)/(-16) = (z - 10)/7` and `(x - 15)/3 = (y - 29)/8 = (z - 5)/(-5)`


A line passes through the point with position vector \[2 \hat{i} - 3 \hat{j} + 4 \hat{k} \] and is in the direction of  \[3 \hat{i} + 4 \hat{j} - 5 \hat{k} .\] Find equations of the line in vector and cartesian form. 


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).


Find the vector equation of a line passing through (2, −1, 1) and parallel to the line whose equations are \[\frac{x - 3}{2} = \frac{y + 1}{7} = \frac{z - 2}{- 3} .\]


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.


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