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The equation of the plane passing throug...

The equation of the plane passing through the point 1,1,1) and perpendicular to the planes `2x+y-2z=5a n d3x-6y-2z=7,` is `14 x+2y+15 z=3` `14 x+2y-15 z=1` `14 x+2y+15 z=31` `14 x-2y+15 z=27`

A

`14x+2y+15x=31`

B

`14x+2y-15z=1`

C

`14x+2y+15x=3`

D

`14x-2y+15z=27`

Text Solution

Verified by Experts

The correct Answer is:
A
(a) given that required plane is perpendicular to the given two planes ,
therefore , the ormal vector of required plane I sperpendicular to the normal to the given planes .
therefore , the normal vector of required plane is parallel to the vector :
`|{:(hati,hatj,hatk),(2,1,-2),(3,-6,-2):}|=-14hati-2hatj-15hatk`
thus , the equation of required of required plane passing through (1,1,1) will be :
`-14(x-1)-2(y-1)-15(z-1)=0`
`implies 14x+2y+15z=31`
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Knowledge Check

  • The equation of the plane passing through the point (1,1,1) and perpendicular to the planes 2x+y-2z=5 and 3x-6y-2z=7 is

    A
    `14x+2y-15z=1`
    B
    `14x-2y+15z=27`
    C
    `14x+2y+15z=31`
    D
    `-14x+2y+15z=3`

  • Equation of the plane passing through the point (1,1,1) and perpendicular to each of the planes x+2y+3z=7 and 2x-3y+4z=0 l is

    A
    `17x-2y+7z=12`
    B
    `17x+2y-7z=12`
    C
    `17x+2y+7z=12`
    D
    `17x-2y-7z=12`

  • The equation of the plane passing through the point (1, 2, –3) and perpendicular to the planes 3x + y – 2z = 5 and 2x – 5y – z = 7, is

    A
    3x – 10y – 2z + 11 = 0
    B
    6x – 5y – 2z – 2 = 0
    C
    11x + y + 17z + 38 = 0
    D
    6x – 5y + 2z + 10 = 0

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