Give the parametric equations of the line of intersection of the planes $4x + 2y + 2z = -1$ and $3x + 6y + 3z = 7$
Also, give the equation of the plane that passes through the point $(2,-1,4)$ and is perpendicular to the line found above.
Give the parametric equations of the line of intersection of the planes $4x + 2y + 2z = -1$ and $3x + 6y + 3z = 7$
Also, give the equation of the plane that passes through the point $(2,-1,4)$ and is perpendicular to the line found above.
Hope next time you'll be more gentle when asking.
The cross product of the normal vectors (let's call it $v_3$) of these planes is $(-6, -6, 18)$.
The next step is to find a point in the line of intersection. To do that, let any variable be $0$.
Say, $z=0$: $4x + 2y= −1$ $3x+6y=7$
Using these two new equations, find the values of $x$ and $y$. You'll get $x = -10/9$ and $y = 31/18$.
Using the values of x and y, find z from any of the two original equations, $z=14/3$.
Therefore, $v_0 = (-10/9, 31/18, 14/3)$ in simplest form
$v_0 = (-20, 31, 84)$
The equation of the line of intersection is $ f(t) = (-20, 31, 84) + (-6, -6, 18)t$ and therefore, the parametric equations are:
$x = -20 - 6t \\y = 31 - 6t \\z = 84 + 18t $
Can't you solve the easy, though slightly annoying, linear system? $\left(\begin{array}{cccr}4&2&2&-1\\3&6&3&7\end{array}\right)\to\left(\begin{array}{cccr}1&1/2&1/2&-1/4\\0&9/2&3/2&31/4\end{array}\right)\Longrightarrow$$\Longrightarrow L:=\left\{\left(\frac{-10-3z}{9}\,,\,\frac{31-6z}{18}\,,\,z\right)\,\,/\,\,z\in\mathbb{R}\right\}$ is the intersection line. Take now $\,\,A,B\in L\,\,$, evaluate $\,\,\overrightarrow{AB}\,\,$, and the vector parametric form of $L$ is $A+t\,\overrightarrow{AB}\,,\,t\in\mathbb{R}$