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
plugging in the 4 ordered pairs gives this system of equations.
(4,-4,0) $\displaystyle E_1 \rightarrow 4A-4B+D=0$
(5,-6,1) $\displaystyle E_2 \rightarrow 5A-6B+C+D=0$
(6,-8,2) $\displaystyle E_3 \rightarrow 6A-8B +2C+D=0$
(7,-10,3) $\displaystyle E_4 \rightarrow 7A-10B+3C+D=0$
Solving this system of equations will gives you the values of A,B,C,D
I hope this clears it up.
Good luck
B
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.
When I solved the system I got the solution
$\displaystyle A=2t-s$
$\displaystyle B=t$
$\displaystyle C=s$
$\displaystyle D=0$
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 $\displaystyle x+y+z=0 \iff 4+k -4-2k+k=0$ for all k.