Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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) $

But my provided answer is:

$ \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?

share|improve this question
    
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
1  
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

2 Answers 2

up vote 1 down vote accepted

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

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

instead of

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

Your solution will work fine then.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.