plane of intersection

[mathstresser]

Where does the line through (1,0,1) and (4,-2,2) intersect the plane x+y+z=6?
I got the equation of the line between the two points to be
x=1+3t
y=-2t
z=1+t

So, I try to change the form of the equation it's in, I'm just not sure exactly how to do that....

Is it

(x-1)/3=(y+2)/1=(z-1)/1

But, that still doesn't help me much...

So, is this the same thing (because I changed the form again)?

3(x-1)+1(y+2)+1(z-1)=0
3x-3+y+2+z-1=0
=3x+y+z=2



Now what do I do? I have 3 variables with only 2 equations...

32

 

[soroban]

Hello, mathstresser!

You started off great . . .

Where does the line through P(1,0,1) and Q(4,-2,2) intersect the plane x + y + z = 6 ?

I got the equation of the line between the two points to be: \(\displaystyle \,\begin{array}{ccc}x\:=\:1\,+\,3t \\ y\:=\:-2t \\ z\:=\:1\,+\,t\end{array}\)

So, I try to change the form of the equation it's in. . . . but why?

The intersection of a line and a plane is the simplest of all the problems.

Substitute the parametric equations of the line into the equation of the plane.

\(\displaystyle \;\;(1\,+\,3t) \,+\,(-2t)\,+\,(1\,+\,t)\:=\:6\;\;\Rightarrow\;\;t\,=\,2\)

Hence: \(\displaystyle \,(x,y,z)\:=\7,-4,3)\)