An equation of the plane that passes through the line of intersection of $x − z = 1$ and $y + 4z = 1$, and is perpendicular to $x + y − 2z = 2$

Question

Find an equation of the plane. The plane that passes through the line of intersection of the planes $x − z = 1$ and $y + 4z = 1$ and is perpendicular to the plane $x + y − 2z = 2$.

I keep getting the answer of $7x-y+5z=6$ and I am told that it is wrong. I do not understand what I am doing wrong. I have the 1st normal vector to be $\langle 1,0,-1\rangle$ and the second normal vector to be $\langle0,1,4\rangle$ and when I did a cross product on them, I got $\langle1,-4,1\rangle$ to be the direction of the line. I then got a 2nd vector parallel to the desired plane as $\langle1,1,-2\rangle$ since its perpendicular to $x+y-2z=2$ I got a normal plane by the cross product of the 2 normal vectors and the result was $\langle 7,-1,5\rangle$ Then I plugged $\langle7,-1,5\rangle$ into the scalar equation of the plane for $\langle a,b,c\rangle$ and used the point $(1,1,0) $ and for my final answer I got $7x-y+5z=6$ I need help figuring out where I went wrong.

score 2 · Answer 1 · answered Sep 16 '13 at 04:14

2

$$\langle 1,-4,1\rangle\times\langle 1,1,-2\rangle=\langle 7,3,5\rangle$$

answered Sep 16 '13 at 04:14

Cameron Buie

102,994

Thanks for helping me figure out where I made my mistake. – Diana Sep 16 '13 at 06:39

score 2 · Accepted Answer · answered Sep 16 '13 at 04:15

2

You made a small error in your last cross product. The result is $(7, 3, 5)$. So the equation is of the form $7x + 3y + 5z = c$. Using the point $(1, 1, 0)$, we get $c = 10$. So the equation of the plane is $7x + 3y + 5z = 10$.

answered Sep 16 '13 at 04:15

Pratyush Sarkar

3,734

Thank you this was very helpful. I can now see where I made my mistake. I had left off a negative sign when I did the cross product. – Diana Sep 16 '13 at 06:34

score 0 · Answer 3 · answered Sep 16 '13 at 05:05

This is actually a 2 line problem which can be solved as follows: Equation of the plane passing through the planes x−z-1=0 and y+4z-1=0 can be written as x−z-1 + k(y+4z-1) = 0 for some constant k (which is required to be found out). i.e., x + ky + (4k-1)z - k - 1 = 0 .... (1) Now since this plane is perpendicular to the plane x+y−2z=2 so, 1.1 + k.1 + (4k-1).(-2) = 0 which gives k = 3/7. Putting in (1) we get the required equation of the plane as 7x+3y+5z=10.

An equation of the plane that passes through the line of intersection of $x − z = 1$ and $y + 4z = 1$, and is perpendicular to $x + y − 2z = 2$

3 Answers3