Question:
Given, point is (1, 2, 3)
and equation of plane is: x + y + z = 1
The normal vector of the plane is (1, 1, 1)
Now, the general equation of a plane is
n * (x - x0 , y - y0 , z - z0 ) = 0 {n is the normal}
=> (1, 1, 1) * (x - 1 , y - 2 , z - 3) = 0
=> x - 1 + y - 2 + z - 3 = 0
=> x + y + z - 6 = 0
=> x + y + z = 6
This is the required equation of the plane.
Answer:
Given, point is (1, 2, 3)
and equation of plane is: x + y + z = 1
The normal vector of the plane is (1, 1, 1)
Now, the general equation of a plane is
n * (x - x0 , y - y0 , z - z0 ) = 0 {n is the normal}
=> (1, 1, 1) * (x - 1 , y - 2 , z - 3) = 0
=> x - 1 + y - 2 + z - 3 = 0
=> x + y + z - 6 = 0
=> x + y + z = 6
This is the required equation of the plane.
Not what you are looking for? Go ahead and submit the question, we will get back to you.
