Parametric equations of a plane are x=1+s+t, y=2-s+t, z=4+2s-3t. What is the rectangular equation of the plane

Question

Can someone verify if my answer is right? my answer is 3x+15y+6z=17. The answer in the book is wrong, so i want to know if my answer is right. If not, then please show how to get the right answer.

Answer 1

Add the first two equations to get
x + y = 3 + 2t
2* second + third gives
2y+z = 8 - t

Double this and then add to the first one we found:
4y + 2z + x + y = 16 - 2t + 3 + 2t
x + 5y + 2z = 19

Check my work, because I haven't!!

PS. I have now. Here's how you could have checked whether your answer was right or not: Substitute the given expressions for x, y, z into your equation:

3(1+s+t) + 15(2-s+t) +6(4+2s-3t)
=57, so you were nearly right, just had the wrong constant. Notice that with 57 instead of 17, there is a common factor 3, and when you divide by it you get the same answer as I did.

Answer 2

Parametric equations of a plane are

x = 1 + s + t
y = 2 - s + t
z = 4 + 2s - 3t

What is the rectangular equation of the plane?

To find one point on the plane set s = t = 0.

x = 1 + s + t = 1
y = 2 - s + t = 2
z = 4 + 2s - 3t = 4

One point on the plane is P(1,2,4).

We have two vectors that lie in the plane. One is based on s and the other on t.

u = <1,-1,2>
v = <1,1,-3>

Take the cross product of u and v to get the normal vector n to the plane.

n = u X v = i + 5j + 2k

Using P and n we can write the equation of the plane as:

1(x -1) + 5(y - 2) + 2(z - 4) = 0

x - 1 + 5y - 10 + 2z - 8 = 0

x + 5y + 2z - 19 = 0

I see the two fellows above me used a method that is quicker, but we all got the same answer.

Answer 3

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

2y = 4 - 2s + 2t
..z = 4 + 2s - 3t
---------------------
2y+z = 8 - t
t = 8 - 2y - z

x + y = 3 + 2(8 - 2y - z)
x + y = 19 - 4y - 2z
x + 5y + 2z = 19