Are you sure that answer is correct?
The line certainly lies in the second plane but not necessarily the first. Example, set t = 2:
Works for the second plane though:
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:
Now, the answer I get is:
This is apparently incorrect. Could someone please show me how to get the correct answer - which is:
Thank you. All help is appreciated.
In any case, let's try to find the correct answer.
The normals of both planes are: and .
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, (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:
After solving for the remaining your variables, you will have a point (call it ).
Now put it together and we have your equation of a line:
The original question was to find the equation of the plane through , , and from that, I got . I got the direction vector as 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
And I seem to get the wrong answer...does there seem to be something wrong with the equation I got?