①The 2 planes have the same normal vector. Since p1 is perpendicular (90°) to the normal vector and the normal vector is perpendicular (also 90°) to p2, then it follows that the 2 planes are parallel to one another.


② As for the why the origin is between the two planes :

We know that the line r = (0 0 0) + λ(2 1 -2) , λ ∈ R passes through both planes and is perpendicular to them both since this line is parallel to their normal vector.

In fact, this line and the normal vector are coincident (lie on top of each other).

This is also the line you would have used to solve part i).


We find the coordinates of the intersection point between p2 and the origin first.


r = (0 0 0) + λ(2 1 -2) = (2λ λ -2λ)

Sub this point into the equation of p2.

(2λ λ -2λ)•(2 1 -2) = 9

2λ(2) + λ(1) - 2λ(-2) = 9

9λ = 9

λ = 1

So the coordinates of the intersection between the line and p2 are (2, 1, -2)

Now you would have already found in part i) that the coordinates of intersection between the line and p1 are (-⅔, -⅓, ⅔) . The value of λ for that is -⅓.

The values of λ are opposite in sign. This tells us that the position vectors of the two intersection points are going in opposite directions from the origin.

For the point on p2 , it is (2 1 -2). For the point on p1, it is (-⅔ -⅓ ⅔) = -⅔(2 1 -2) . The vectors are negative scalar multiples of each other.

(Also notice that the x,y and z coordinates of one point are opposite in sign to their respective coordinates on the other point)


Since the position vectors are going in opposite directions from the origin, the two points themselves must lie on opposite ends of the line from the origin.

This also means that the two planes intersect the line on opposite ends of it, away from the origin.

So the origin is between the planes.

The last part is easy. It is just the sum of the perpendicular distances from the origin to each plane. And this is actually just the linear distance between the 2 points.