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The distasnce of the plane 2x-3y+6z+14=0...

The distasnce of the plane `2x-3y+6z+14=0` from the origin is (A) 2 (B) 4 (C) 7 (D) 11

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Find the distance of the plane 2x-3y+4z-6=0 from the origin.

Find the distance of the plane 2x - 3y + 4z = 6 from the origin.

Knowledge Check

  • The distance of the plane 6x-3y+2z-14=0 from the origin is

    A
    2
    B
    1
    C
    14
    D
    8

  • If the perpendicular distance of the plane 2x+3y-z= k from the origin is sqrt(14) units then k=……..

    A
    14
    B
    196
    C
    ` 2 sqrt (14)`
    D
    ` (sqrt(14))/(2)`

  • If d_(1),d_(2),d_(3) denote the distances of the plane 2x-3y+4z=0 from the planes 2x-3y+4z+6=0 4x-6y+8z+3=0 and 2x-3y+4z-6=0 respectively, then

    A
    `d_(1)+8d_(2)+d_(3)=0`
    B
    `d_(1)+16d_(2)=0`
    C
    `8d_(2)=d_(1)`
    D
    `d_(1)-2d_(2)+3d_(3)=sqrt(29)`

    • Similar Questions

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    If theta is the angel between the planes 2x-y+z-1=0 and x-2y+z+2=0 then costheta= (A) 2/3 (B) 3/4 (C) 4/5 (D) 5/6

    Distance of the plane 4x+5y+3z = 6 from the point (-2,-1,1) is (3^a+b)/(25sqrt(2)) , then (b-a) is.

    The angle between the planes 2x-y+z=6 and x+y+2z=7 is (A) pi/4 (B) pi/6 (C) pi/3 (D) pi/2

    The direction cosines of a normal to the plane 2x-3y-6z+14=0 are (A) (2/7,(-3)/7,(-6)/7) (B) ((-2)/7,3/7,6/7) (C) ((-2)/7,(-3)/3,(-6)/7) (D) none of these

    What is the distance of the point (2,3,4) from the plane 3x-6y + 2z + 11 = 0

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