# Math Help - Intersection question

1. ## Intersection question

Two planes, x + 2y + 3z + 4 = 0 and Ax + By + Cz + D = 0 intersect in the line:

X = 4 + k
Y = -4 - 2k
Z = k

Find the values of A, B, C and D

i don't know where to start

2. Originally Posted by Hasan1
Two planes, x + 2y + 3z + 4 = 0 and Ax + By + Cz + D = 0 intersect in the line:

X = 4 + k
Y = -4 - 2k
Z = k

Find the values of A, B, C and D

i don't know where to start
if you plug in 4 different values for k you will get 4 point on the plane Ax+By+Cz+d=0.

k=0 (4,-4,0) k=1 (5,-6,1) k=2 (6,-8,2) k=3 (7,-10,3)

if you plug these points into the equation for the plane you will get a system of 4 equations with 4 unknowns.

Good luck.

B

3. Im not sure I understand your method for answering this question, also, wouldn't three points be enough just to find the scalar equation of the plane anyways? Why would I have to solve a system of equations?

4. plugging in the 4 ordered pairs gives this system of equations.

(4,-4,0) $E_1 \rightarrow 4A-4B+D=0$

(5,-6,1) $E_2 \rightarrow 5A-6B+C+D=0$

(6,-8,2) $E_3 \rightarrow 6A-8B +2C+D=0$

(7,-10,3) $E_4 \rightarrow 7A-10B+3C+D=0$

Solving this system of equations will gives you the values of A,B,C,D

I hope this clears it up.

Good luck

B

5. When I try to solve this system, the first thing I did was look at the first equation. I isolated D, D = -4A + 4B, and then substituted that back into all of the other equations. When I did this, all of my new equations were multiples of each other, which means that they were all coincident, so there would be infinite solutions.

6. ## System solved

When I solved the system I got the solution

$A=2t-s$
$B=t$
$C=s$
$D=0$

these are true for all values of s and t.

This would generate an entire family of planes that would intersect in the line given. Think about why there are an infinite number of planes.

If you want you can check a few. example let s=1 and t=1

we get A=1, B=1 C=1, D=0 so $x+y+z=0 \iff 4+k -4-2k+k=0$ for all k.

7. oh! I see it now, thanks!