The equation of the plane which contains the lines L1- x-y-z-6-0-x-z-6 and L2- x-2-y-1-z-6-2 A- x-z-6-0B- x-z-6-0C- x-2 z-12-0D- x -2 z -12-0

The correct option is A x z+6=0D.R.'s of the plane d_1(1,1,1) and d_2(1,0, 1) D.R.'s of the L_1 is d_1× d_2 =( 1,2, 1) D.R.'s of the plane containing the L_1 a ...

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Byju's Answer

Standard XII

Mathematics

Equation of Line: Symmetrical Form

The equation ...

Question

The equation of the plane which contains the lines $L_{1} : x + y + z - 6 = 0 = x - z + 6$ and $L_{2} : \frac{x}{2} = \frac{y}{1} = \frac{z - 6}{2}$ is

x - z + 6 = 0

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x + z - 6 = 0

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x - 2 z + 12 = 0

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x + 2 z + 12 = 0

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Solution

The correct option is A $x - z + 6 = 0$
D.R.'s of the plane $d_{1} (1, 1, 1) and d_{2} (1, 0, - 1)$
D.R.'s of the $L_{1}$ is $d_{1} \times d_{2}$
$= (- 1, 2, - 1)$
D.R.'s of the plane containing the $L_{1} and L_{2}$ is
$(- 5, 0, 5)$
Equation of the plane is
$x - z + 6 = 0$

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The equation of the plane which contains the lines L1:x+y+z−6=0=x−z+6 and L2:x2=y1=z−62 is

The correct option is A x−z+6=0D.R.'s of the plane d1(1,1,1) and d2(1,0,−1) D.R.'s of the L1 is d1×d2 =(−1,2,−1) D.R.'s of the plane containing the L1 and L2 is (−5,0,5) Equation of the plane is x−z+6=0

The equation of the plane which contains the lines $L_{1} : x + y + z - 6 = 0 = x - z + 6$ and $L_{2} : \frac{x}{2} = \frac{y}{1} = \frac{z - 6}{2}$ is

The correct option is A $x - z + 6 = 0$
D.R.'s of the plane $d_{1} (1, 1, 1) and d_{2} (1, 0, - 1)$
D.R.'s of the $L_{1}$ is $d_{1} \times d_{2}$
$= (- 1, 2, - 1)$
D.R.'s of the plane containing the $L_{1} and L_{2}$ is
$(- 5, 0, 5)$
Equation of the plane is
$x - z + 6 = 0$