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Find the equation of the plane which is ...

Find the equation of the plane which is at a distance of `5sqrt(14)` from the origin and whose normal has d.r.s `(2, -1, 3)`.

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Knowledge Check

  • The equation of the plane which is at a distance of 5 unit from the origin and whose normal has the d.r.'s (6, 12, - 4) is

    A
    3x + 6y – 2z = 5
    B
    6x +12y – 2z = 5
    C
    3x + 6y – 2z = 7
    D
    3x + 6y – 2z = 35

  • The equation of the plane which is at a distance of 2sqrt(3) units from the origin and whose normal has the d.c's ((1)/(sqrt(3)),(-1)/(sqrt(3)),(1)/(sqrt(3))) is

    A
    `x-y+z=6`
    B
    `x-2y+z=13`
    C
    `3x-12y+4z=26`
    D
    `3x-12y+4z+26=0`

  • The equation of the plane which is at a distance of 5 unit from the origin and whose normal has the dc's (6/7, - 2/7, - 3/7) is

    A
    6x – 2y – 3z = 35
    B
    6x + 2y – 3z = 50
    C
    3x – 12y + 4z = 26
    D
    `2x + y +z = 3sqrt11 `

    • Similar Questions

    Explore conceptually related problems

    Find the coordinates of a point on y axis which are at a distance of 5sqrt(2) from the point P(3,-2,5)

    Find the equation of the line whose perpendicular distance from the origin is 4 units and the angle which the normal makes with positive direction of x - axis is 15^(@) .

    The equation of the plane in cartesian form, which is at a distance of (6)/(sqrt(29)) from the origin and its normal vector drawn from the origin being 2 hat(i)-3hat(j)+4hat(k) , is

    A : The vector equation of the plane which is at a distance of 5 unit from origin and perpendicular to 2i - j + 2k is r. (2i - j + 2k) = 15 R : The vector equation of the plane which is at distance of p from origin and perpendicular to the unit vector n is r.n =p

    The equation of the lines passing through the point (1,0) and at distance sqrt3//2 from the origin is

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