0
$\begingroup$

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.

  • 0
    Let one of the variables be $t$, and then solve for the others in terms of $t$. Then you should be left with the equation of a line in the form $(p_x,p_y,p_z) + t (d_x,d_y,d_z)$.2012-05-30

2 Answers 2

4

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 $

  • 0
    didn't work. I substituted 0 for each of each number and came up with different results for the other two variables each time.2017-09-18
3

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 $A+t\,\overrightarrow{AB}\,,\,t\in\mathbb{R}$