How many points are common to both the planes x - y + z = 1 and y = z - 2??
Take any two points lying on both the planes . Obtain the equation of the line passing through these points. Does the line you obtained lie on both the planes? Justify your answer
Let the two planes be p1: x – y + z = 1 and p2: y = z – 2.
The normal vectors for p1 and p2 are (1, -1, 1) and (0, 1, -1) respectively.
Since these two normals are not parallel, then the two planes must intersect.
Since the planes intersect, then they have a common line of intersection.
Any common line of intersection will have an infinite number of points
Let there be two points P and Q lying on both planes
The points are: P(a, b, c) and Q(h, k, j)
The equation of the line joining P and Q is r = P + t(QP)
where QP = Q – P = (h, k, j) - (a, b, c) = (h – a, b – k, c – j)
r = (a, b, c) + t(h – a, b – k, c – j)
In coordinate form,
r = {a + t(h – a), b + t(k – b), c + t(j – c)}
Since the coordinates of the points P and Q satisfy the equations of the planes p1 and p2, then we can write,
For p1: x – y + z = 1 -> a – b + c = 1 and h – k + j = 1 --------------------------- (1)
For p2: y = z – 2 -> b – c = -2 and k – j = -2 ---------------------------------------- (2)
If the line r lies on both planes, p1 and p2, then it will satisfy the equations of both planes
Using the coordinate form of r,
p1: x – y + z = 1:
x – y + z = a + t(h – a) – b – t(k – b) + c + t(j – c)
x – y + z = (a – b + c) – t(a – b + c) + t(h – k + j)
x – y + z = 1 – t + t using (1)
x – y + z = 1, which is correct.
p2: y = z – 2:
y – z = b + t(k – b) – c – t(j – c)
y – z = b – c + t(k – b – j + c)
y – z = (b – c) + t{(k – j) – (b – c)}
y – z = = -2 + t{-2 – (-2)} using (2)
y – z = -2, which is correct
Since the coordinate values of the line r satisfy the equations of both planes, p1 and p2, then we have shown that the line r lies on both planes.
From earlier, we have also shown that there are an infinite number of points common to both planes.