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.
Best Answer
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 $7x + 3y + 5z = 10$.