# Find the intersection of these two planes.

Find the intersection of $8x + 8y +z = 35$ and

$x = \left(\begin{array}{cc} 6\\ -2\\ 3\\ \end{array}\right) +$ $\lambda_1 \left(\begin{array}{cc} -2\\ 1\\ 3\\ \end{array}\right) +$ $\lambda_2 \left(\begin{array}{cc} 1\\ 1\\ -1\\ \end{array}\right)$

So, I have been trying this two different ways. One is to convert the vector form to Cartesian (the method I have shown below) and the other was to convert the provided Cartesian equation into a vector equation and try to find the equation of the line that way, but I was having some trouble with both methods.

Converting to Cartesian method:

normal = $\left(\begin{array}{cc} -4\\ 1\\ -3\\ \end{array}\right)$

Cartesian of x $=-4x + y -3z = 35$

Solving simultaneously with $8x + 8y + z = 35$, I get the point $(7, 0, -21)$ to be on both planes, i.e., on the line of intersection.

Then taking the cross of both normals, I get a parallel vector for the line of intersection to be $(25, -20, -40)$.

So, I would have the vector equation of the line to be:

$\left(\begin{array}{cc} 7\\ 0\\ -21\\ \end{array}\right) +$ $\lambda \left(\begin{array}{cc} 25\\ -20\\ -40\\ \end{array}\right)$

$\left(\begin{array}{cc} 6\\ -2\\ 3\\ \end{array}\right)+$ $\lambda \left(\begin{array}{cc} -5\\ 4\\ 8\\ \end{array}\right)$

I can see that the directional vector is the same, but why doesn't the provided answer's point satisfy the Cartesian equation I found?

Also, how would I do this if I converted the original Cartesian equation into a vector equation? Would I just equate the two vector equations and solve using an augmented matrix? I tried it a few times but couldn't get a reasonable answer, perhaps I am just making simple errors, or is this not the correct method for vector form?

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What values of $\lambda_1$ and $\lambda_2$ show that $\langle 7,0,-21\rangle$ is in the second plane? By my mental arithmetic $\langle 7,0,7/3\rangle$ is in that plane, not $\langle 7,0,-21\rangle$, with $\lambda_1=1/3$ and $\lambda_2=5/3$. – Brian M. Scott May 21 '12 at 10:59
Your "cartesian" is off by sign. $(6,-2,3)$ is not on $-4x+y-3z=35$, but if you make that $-35$, you're OK. – Gerry Myerson May 21 '12 at 11:01
I guess I am tired. So, just carelessness for the first method. What about the other method? Is the way I described it correct? Equating the two vector equations and solving the augmented matrix? – stariz77 May 21 '12 at 11:05

It's just a simple sign mistake. The equation should be

$$-4x+y-3z=-35$$

$$-4x+y-3z=35.$$
Note: In this case it might be quicker to "plug in" $$8(6-2\lambda_1+\lambda_2)+8(-2+\lambda_1+\lambda_2)+3(3+3\lambda_1-\lambda_2)=0$$ which means $\lambda_1=-41-13\lambda_2$ and then you plug that into the original line equation.