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महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता १२ वी

Find the Cartesian equation of the line passing through the origin which is perpendicular to x – 1 = y – 2 = z – 1 and intersect the line x-12=y+13=z-14.

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प्रश्न

Find the Cartesian equation of the line passing through the origin which is perpendicular to x – 1 = y – 2 = z – 1 and intersect the line `(x - 1)/(2) = (y + 1)/(3) = (z - 1)/(4)`.

बेरीज
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उत्तर

Let the required line have direction ratios a, b, c
Since the line passes through the origin, its cartesian equation are

`x/a = y/b = z/c`                   ...(1)

This line is perpendicular to the line
x – 1 = y – 2 = z – 1 whose direction ratios are 1, 1, 1.
∴ a + b + c = 0                 ...(2)

The lines `(x - x_1)/a_1 = (y - y_1)/b_2 = (z- z_1)/c_1` intersect, if

`|(x_2 - x_1, y_2 - y_1, z_2 - z_1),(a_1, b_1, c_1),(a_2, b_2, c_2)|` = 0

Applying this condition for the lines

`x/a = y/b = z/c and (x- 1)/(2) = (y + 1)/(3) =  (z - 1)/(4)` we get

`|(1 -0, -1 - 0, 1 - 0),(a, b, c),(2, 3, 4)|` = 0

∴ 1(4b – 3c) + 1(4a –2c) + 1(3a – 2b) = 0
∴ 4b – 3c + 4a – 2c + 3a – 2b = 0
∴ 7a + 2b – 5c = 0         ...(3)

From (2) and (3), we get

`a/|(1, 1),(2, -5)| = b/|(1, 1),(-5, 7)| = a/|(1, 1),(7, 2)|`

∴ `a/(-7) = b/(12) = c/(-5)`

∴ the required line has direction ratios –7, 12, –5.

From (1), cartesian equation of required line are

`x/(-7) = y/(12) = z/(-5)`

i.e. `x/(7) = y/(-12) = z/(5)`.

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पाठ 6: Line and Plane - Miscellaneous Exercise 6 A [पृष्ठ २०८]

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बालभारती Mathematics and Statistics 1 (Arts and Science) [English] Standard 12 Maharashtra State Board
पाठ 6 Line and Plane
Miscellaneous Exercise 6 A | Q 20 | पृष्ठ २०८

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