Short Answer TypeFind equation of the plane parallel to x + 3y – 2z + 7 = 0 and passing through the origin.
Long Answer TypeThe equation of plane is
2x – 3y + 4z – 6 = 0 ...(1)
Direction ratios of the normal to the plane (1) are 2, –3, 4.
Dividing each by 
we get the direction cosines of the normal as 
∴ dividing (1) throughout by 
we get, 
This is of the form l x + m y + n z = p, where p is the distance of the plane from the origin.
distance of plane from the origin = 
.
Let (x1, y1, z1) be the coordinates of the foot of perpendicular drawn from origin 0(0, 0, 0) to the plane (1).
∴ direction ratios of OP are .x1– 0, y, –0, –0 i.e. x1, y1, z1
∴ Direction cosines of OP are
Since direction cosines and direction ratios of a line are proportional
Since P(x1, y1, z1) lies on plane (1)
Short Answer TypeIn each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
z = 2
In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
x + y + z = 1
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