Equation of plane passing through 2 points and perpendicular to a plane?

Question

Hey, I'm really stuck on this question.

I need to find the equation of a plane passing through points (1,1,1) and (2,0,3) and perpendicular to the plane x+2y-3z=0.

It's had my brain twisted all day.

Thanks.

Answer 1

The given plane is:

x + 2y - 3z = 0

Its normal vector n1 is:

n1 = <1, 2, -3>

The desired plane contains the points points P(1,1,1)
and Q(2,0,3). Hence it contains the line thru P and Q also. The line's directional vector u is:

u = PQ = = <2-1, 0-1, 3-1> = <1, -1, 2>

The vector u lies in the desired plane. The normal vector n2, of the desired plane is normal to both u and n1. Therefore take the cross product of the vectors u and n1.

n2 = u X n1 = <1, -1, 2> X <1, 2, -3> = <-1, 5, 3>

With the normal vector n1 and a point in the plane--let's choose the point P(1,1,1), we can write the equation of the desired plane.

-1(x - 1) + 5(y - 1) + 3(z - 1) = 0
-x + 1 + 5y - 5 + 3z - 3 = 0
-x + 5y + 3z - 7 = 0

Answer 2

Observe that if the plane has points (1, 1, 1) and (2, 0, 3), then it contains the line that passes through (1, 1, 1) and (2, 0, 3). This line has direction vector

(2,0,3) - (1,1,1) = (1, -1, 2).

The plane x + 2y - 3z = 0 has a normal vector (1, 2, -3) (the coefficients of x, y and z in the equation; you should've been taught this at some point), and this must be parallel to the plane we're seeking (since it's perpendicular to the given plane). Hence, since (1, 2, -3) is not parallel to (1, -1, 2), we can write the equation for the plane in vector form:

r(s, t) = (1, 1, 1) + s (1, -1, 2) + t (1, 2, -3).

Then, if (x, y, z) is a point on this plane, we have

x = 1 + s + t, (i)
y = 1 - s + 2t, (ii)
z = 1 + 2s - 3t. (iii)

(i) + (ii) and 2(i) - (ii) yield

x + y = 2 + 3t,
2x - y = 1 + 3s,

and so, substituting for s and t into (iii), we see that

3z = 3 + 6s - 9t = 3 + 2(2x - y - 1) - 3(x + y - 2) = 7 + x - 5y,

ie. x - 5y - 3z = -7.

Checking this, we see that it contains both (1, 1, 1) and (2, 0, 3), and is perpendicular to x + 2y - 3z = 0 (since their normals, (1, -5, -3) and (1, 2, -3) have zero scalar product).

Answer 3

Equation From 2 Points

Answer 4

You re full of ******* ****

Answer 5

you rock!

Answer 6

@hemant.... thanks a lot for explaining.... really so beautifully explained... :D