1. ## Intersecting Planes

I seem to be having a problem here finding the intersection of the 2 planes and I know these planes intersect at a line. Here are the equations of the 2 planes:

$3x + 5y + 6z = 21$

$2x - y + 4z = 14$

----------------------

Now, the answer I get is:

$x = 7 - 2t$

$y = 0$

$z = \frac{7}{4} + 2t$

----------------------

This is apparently incorrect. Could someone please show me how to get the correct answer - which is:

$x = 3 - 2t$

$y = 0$

$z = 2 + t$

====================================

Thank you. All help is appreciated.

2. Are you sure that answer is correct?

The line certainly lies in the second plane but not necessarily the first. Example, set t = 2:

$3(3-2(2)) + 5(0) + (2 + (2)) = 1 \neq 21$

Works for the second plane though: $2(3-2(2)) - (0) + 4(2+(2)) = 14$

3. Originally Posted by o_O
Are you sure that answer is correct?

The line certainly lies in the second plane but not necessarily the first. Example, set t = 2:

$3(3-2(2)) + 5(0) + (2 + (2)) = 1 \neq 21$

Works for the second plane though: $2(3-2(2)) - (0) + 4(2+(2)) = 14$
I thought the given answer was a little strange because it didn't seem correct...

4. In any case, let's try to find the correct answer.

The normals of both planes are: $\vec{n}_{1} = (3,5,1)$ and $\vec{n}_{2} = (2, -1, 4)$.

Now, the cross product of these two normals will give the direction vector of the line. Any vector perpendicular to the normal of the first plane will lie in the first plane and any vector perpendicular to the normal of the second plane will lie in the second plane. Since the cross product of these two normals is perpendicular to both, it will lie in the line of intersection of both planes.

So, $(3, 5, 1) \times (2, -1, 4) = (a,b,c)$ (Leaving it up to you to find a, b, and c). This will be our direction vector.

Now we have to find a point on this line (recalling that a line can be defined by a point and a direction vector; we have the latter). One way to do this is set one of the variables to a specific constant, say y = 0, and solve for the remaining system of equations:

$\begin{array}{ccccc} 3x & + & z & = & 21 \\ 2x & + & 4z& = &14\end{array}$

After solving for the remaining your variables, you will have a point (call it $(x_{0}, y_{0}, z_{0})$).

Now put it together and we have your equation of a line: $\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} x_{0} \\ y_{0} \\ z_{0} \end{pmatrix} + t\begin{pmatrix} a \\ b \\ c\end{pmatrix}$

5. Originally Posted by sqleung
I seem to be having a problem here finding the intersection of the 2 planes and I know these planes intersect at a line. Here are the equations of the 2 planes:

$3x + 5y + z = 21$

$2x - y + 4z = 14$

----------------------

Now, the answer I get is:

$x = 7 - 2t$

$y = 0$

$z = \frac{7}{4} + 2t$

----------------------

This is apparently incorrect. Could someone please show me how to get the correct answer - which is:

$x = 3 - 2t$

$y = 0$

$z = 2 + t$

====================================

Thank you. All help is appreciated.
Ah darn...looks like I made a slight mistake:

It should be $3x + 5y + 6z = 21$, NOT $3x + 5y + z = 21$

So the equations are:

$3x + 5y + 6z = 21$
$2x - y + 4z = 14$

---------------------------

6. Originally Posted by sqleung
Ah darn...looks like I made a slight mistake:

It should be $3x + 5y + 6z = 21$, NOT $3x + 5y + z = 21$

So the equations are:

$3x + 5y + 6z = 21$
$2x - y + 4z = 14$

It doens't change much of o_O's setup.

It changes the outcome for $\bold n_1\times \bold n_2=\left$..

The system of equations would become

$3x+6z=21$
$2x+4z=14$

the values you get here for x and z, and the point y=0 will give you the point on the line $P(x_0,y_0,z_0)$

Thus, the line will have the form $\left+t\left$

Does this make sense?

--Chris

7. The original question was to find the equation of the plane through $A(3,0,2)$ , $B(1,0,3)$ , $C(2,-3,5)$ and from that, I got $3x + 5y + 6z = 21$. I got the direction vector as $[3,5,6]$ which is DEFINITELY correct.

Next I'm supposed to find where that plane meets with another plane given the equation of the other plane is $2x - y + 4z = 14$

And I seem to get the wrong answer...does there seem to be something wrong with the equation I got?

8. I'm guessing you got $(3,5,6)$ from taking the cross product of AB and AC or something similar (this is the normal of the plane btw)? Well unfortunately, early on in your steps:

$\vec{AB} \times \vec{AC} = (-2, 0, 1) \times (-1, -3, 3) = \ldots = (3, {\color{red}-5}, 6)$

9. Originally Posted by o_O
I'm guessing you got $(3,5,6)$ from taking the cross product of AB and AC or something similar (this is the normal of the plane btw)? Well unfortunately, early on in your steps:

$\vec{AB} \times \vec{AC} = (-2, 0, 1) \times (-1, -3, 3) = \ldots = (3, {\color{red}-5}, 6)$
No offence but -5 does not seem correct:

$[3 , {\color{red}1(-1) - (-2)(3)} , 6]$

$[3 , {\color{red}5} , 6]$

I guess I can just reach a conclusion that the answers provided to me on the sheet are wrong...

10. Originally Posted by o_O
$(-2, 0, 1) \times (-1, -3, 3) = \left( \left| \begin{matrix}0 & 1 \\ -3 & 3\end{matrix}\right|, {\color{red}-}\left| \begin{matrix}-2 & 1 \\ -1 & 3\end{matrix}\right|, \left| \begin{matrix}-2 & 0 \\ -1 & -3\end{matrix}\right|\right)$

$= \bigg( (0)(3) - (1)(-3), \: \:{\color{red}-[}(-2)(3) - (-1)(1){\color{red}]}, \: \: (-2)(-3) - (-1)(0)\bigg) \: = \: (3, {\color{red}5}, 6)$
You forgot the negative...

--Chris

11. Ah rookie mistake

12. Originally Posted by sqleung
I seem to be having a problem here finding the intersection of the 2 planes and I know these planes intersect at a line. Here are the equations of the 2 planes:
$3x + 5y + 6z = 21$
$2x - y + 4z = 14$
If this is the correct question, the the line of intersection is:
$\left\{ {\begin{array}{lll} {x = 7 + 2t} \\ {y = 0} \\
{z = - t} \\\end{array}} \right.$