Find the distance between two planes: C1:x+y+2z=4 and C2:3x+3y+6z=18 And find the other plane C3≠C1...
Find the distance between two planes:
And find the other plane that has the distance d po the plain
Answer & Explanation
Well, we can rewrite
Two plain have normal
Chose from respectively.
Thus, the distance is
For a plane defined by the normal is said to be (a,b,c). This is a direction, so we can normalise it , which means these two planes are parallel and we can write the normal as .
Now we should find two points on the planes.
Let and , and find the corresponding values. For and for . So we know contains the point and contains the point .
The distance between these points is and the direction is . We now that this is not the shortest distance between these two points as so the direction is not perpendicular to these planes. But this is ok as we can use the dot product between and to work out the proportion of the distance that is perpendicular to the planes.
So the distance between the two planes is .
The last part is to find the plane which is the same distance away from as but in the opposite direction. We know the normal must be the same, . Using this we can write and determine . When we moved from , so if we move the same distance again we go and is on .
Therefore, . So final equation of the plane is: