The length of the perpendicular from the origin to the plane 3x + 2y &...
Given equation of plane is : 3x + 2y - 6z - 21= 0
the length of perpendicular from a given point
(x' , y', z') on a plane ax + by + cz + d = 0 is given as :-
d = modulus of [{ax' + by' + cz' + d}/{√(a² + b² + c)²}]
so, d = modulus of [{(3*0) + (2*0) + (-6*0) + (-21)}/{√(3² + 2² + (-6)²)}]
d= modulus of (-21/√49) = (-21/7) = 3 units
hence option A is correct....
The length of the perpendicular from the origin to the plane 3x + 2y &...
Solution:
Concept: If a line is perpendicular to a plane, then the direction of the line is parallel to the normal vector of the plane.
Step 1: Finding the normal vector of the plane
The given plane 3x + 2y + 6z = 21 can be written in the form of ax + by + cz = d, where a, b, and c are the components of the normal vector to the plane.
So, the normal vector to the plane is (3, 2, 6).
Step 2: Finding the distance between the origin and the plane
The distance between the origin and the plane is the length of the perpendicular from the origin to the plane.
Let P be a point on the plane such that the perpendicular from the origin meets the plane at P. Let OP be x.
Then, the vector OP is parallel to the normal vector to the plane.
So, the dot product of OP and the normal vector to the plane is zero.
OP . (3, 2, 6) = 0
x(3, 2, 6) . (3, 2, 6) = 0
9x + 4x + 36x = 0
49x = 0
x = 0
So, the distance between the origin and the plane is x = 0, which is equal to 3.
Hence, the length of the perpendicular from the origin to the plane 3x + 2y + 6z = 21 is 3.
Therefore, option (A) is correct.