The equation of the plane passing through the point -1-2-0 and parallel to the lines L 1- x -3- y -1-0- z -2-1 and L 2- x -1-1-2 y -1-2-z -1-1 isA- x-2 y-3 z-3B- x-2 y-3 z-3-0C- x-2 y-3 z-3D- x-2 y-3 z-3

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Byju's Answer

Standard XII

Mathematics

Equation of Line: Symmetrical Form

The equation ...

Question

The equation of the plane passing through the point $(- 1, 2, 0)$ and parallel to the lines $L_{1} : \frac{x}{3} = \frac{y + 1}{0} = \frac{z - 2}{- 1}$ and $L_{2} : \frac{x - 1}{1} = \frac{2 y + 1}{2} = \frac{z + 1}{- 1}$ is

x + 2 y + 3 z = 3

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x - 2 y + 3 z = 3

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x + 2 y + 3 z + 3 = 0

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x + 2 y - 3 z = 3

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Solution

The correct option is A $x + 2 y + 3 z = 3$
Equartion of the plane through $(- 1, 2, 0)$ is
$a (x + 1) + b (y - 2) + c z = 0$
$∵$ plane is parallel to the lines
$∴ 3 a - c = 0, a + b - c = 0 b = 2 a, c = 3 a$
equation of the plane is $x + 2 y + 3 z = 3$

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The equation of the plane passing through the point (−1,2,0) and parallel to the lines L1:x3=y+10=z−2−1 and L2:x−11=2y+12=z+1−1 is

The correct option is A x+2y+3z=3 Equartion of the plane through (−1,2,0) is a(x+1)+b(y−2)+cz=0 ∵ plane is parallel to the lines ∴3a−c=0,a+b−c=0b=2a, c=3a equation of the plane is x+2y+3z=3

The equation of the plane passing through the point $(- 1, 2, 0)$ and parallel to the lines $L_{1} : \frac{x}{3} = \frac{y + 1}{0} = \frac{z - 2}{- 1}$ and $L_{2} : \frac{x - 1}{1} = \frac{2 y + 1}{2} = \frac{z + 1}{- 1}$ is

The correct option is A $x + 2 y + 3 z = 3$
Equartion of the plane through $(- 1, 2, 0)$ is
$a (x + 1) + b (y - 2) + c z = 0$
$∵$ plane is parallel to the lines
$∴ 3 a - c = 0, a + b - c = 0 b = 2 a, c = 3 a$
equation of the plane is $x + 2 y + 3 z = 3$