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# Distance between planes

2010 IIT JEE Paper 1 Problem 51 Distance Between Planes. Created by Sal Khan.

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• At , Sal claims that given any point in the first plane, the "closest point" on the second plane will be exactly 6 units away. However, wouldn't that mean that the distance between the two planes is 6? And doesn't the original question state that the distance between the two points is equal to the square root of 6?

Unless I'm mistaken, the "d" in the video is merely the right-hand side of the 1st equation for the plane, which is not the same as the distance between the two planes. Am I right or am I missing something??
• I think that you are right and that Sal messed up in that last part. The equation for the red plane is x-2y+z=-6 and the equation for the blue plane is x-2y+z=0. This means that the planes are parallel with the red one is shifted down. If we calculate the distance between the two planes with those equations we get: (1-4+3-(-6))/sqrt(1+4+1) and that is equal to 6/sqrt(6), if you multiply by sqrt(6)/sqrt(6) you get that the distance between the two planes is sqrt(6), which is what was stated originally.
• How does (x-1)/2 = (y-2)/3 = (z-3)/4 represent a line? Is there any tutorial on this? I can't find one.
• Any equation with highest power of 1 is not a line but rather a plane.
For instance ax + by +cz = D is a plane. A line needs to be defined by intersection of two planes or parametrically using vectors
See Parametric representation of a line video
In this question (x-1)/2 = (y-2)/3 = (z-3)/4 = t represents a line because it is actually giving you in effect three parametric equations, where t is any real number
if you want to define a line in R3 through <x1, y1, z1> and < x2, y2, z2> parametrically, then it will have three equations
x = (x1-X2) + x1 t
y = (y1-y2) + y1 t
z = (z1-z2) + z1 t
the above equations can be arranged as
{(x - (x1-x2)}/x1 = { y- (y1-y2)}/ y1 = {z- (z1-z2)}/ z1 = t
where t is user set parameter
• At , since we just found the normal vector of the blue plane to be (-1, 2, -1), can't we just go ahead and say the equation for the blue plane is -x+2y-z=D. (From the Normal vector from plane equation video).

Plug in a point on the plane (1, 2,3) we get -1+4-3=D=0. So the equation for the blue plane is -x+2y-z=0?
• Yes, we can. 😊

The equation -x+2y-z=D applies to all planes parallel to (and including) the blue plane. Simply plugging a point into this formula lets us specify a particular plane.
• whats with all the i, j, and k's?
• i, j and k are the unit vectors. "i" is the unit vector in the x direction, "j" in the y direction and k in the z direction. For example,
let v be a 3D vector.
v=3i+5j+6k
This means the vector has components 3 in the "x" 5 in the "y" and 6 in the "z"
Hope this helps.
• At Sal says that the planes must be parallel since the distance between them is sqrt(6), but couldn't the planes be skew? Skew planes still have a distance from each other, but the distance is different at each point on the planes.
• The reason he says they must be parallel is because it doesn't say the distance between them is sqrt(6) at a specific point, it just says that that's the distance between them period, meaning at every point.
• I was wondering why 2 planes can't intersect at only one point. What if the tip of one of the planes touched the other?
(1 vote)
• theres no tip when you're talking about planes. theyre not actually rectangles, theyre infinitely large. theyre just visualized as rectangles.
• why sal is taking point x and multiply with n whereas points are already available in plane
• Because you are aiming to result an equation , whereas x,y,z contained.
If you use the give two lines, their dot product will all be scalar, useless then.
Hope you see that.
• Why dont we just get the point of intersection of the 2 given lines,
and then use the "point distance to plane" formula to get the distance??