Question

Find an equation of the plane. The plane that passes through the line of intersection of the planes x − z = 3 and y + 2z = 3 and is perpendicular to the plane x + y − 4z = 6

Answer 1

Answer:

x-y-3=0

Step-by-step explanation:

Choose two common points on the line of intersection of the planes x − z = 3 and y + 2z = 3:

• 1st point: if x=0, then z=-3 and y=3-2z=3+6=9, so the 1st point is (0,-3,9);
• 2nd point: if x=3, then z=0 and y=3-2z=3-0=3, so the 2nd point is (3,0,3).

The perpendicular plane to the plane x+y-4z=6 is parallel to the vector with coordinates (1,1,-4) (normal vector of the given plane).

Hence, the equation of needed plane is

$\left\|\begin{array}{ccc}x-0&y-(-3)&z-9\\3-0&0-(-3)&3-9\\1&1&-4\end{array}\right\|=0\Rightarrow \left\|\begin{array}{ccc}x&y+3&z-9\\3&3&-6\\1&1&-4\end{array}\right\|=0$

Thus,

$\left\|\begin{array}{ccc}x&y+3&z-9\\3&3&-6\\1&1&-4\end{array}\right\|=0\Rightarrow \\\\-12x-6(y+3)+3(z-9)-3(z-9)+6x+12(y+3)=0\\ \\-6x+6(y+3)=0\\ \\-x+y+3=0\\ \\x-y-3=0$