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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.

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    $\begingroup$ It is usually recommended that you give your thoughts on the matter, so that your fellow users might be directed properly when answering your inquiry. It is also considered ungentle to use the imperative. $\endgroup$ – Pedro Tamaroff May 29 '12 at 21:59
  • $\begingroup$ Alright so I tried it. I found the cross product of the 2 planes to find a vector, which is (6,-6,18). Im kind of stuck though... $\endgroup$ – Nick May 30 '12 at 0:06
  • $\begingroup$ Shouldn't you get solution as in $ax+by=c$? $\endgroup$ – Pedro Tamaroff May 30 '12 at 0:08
  • $\begingroup$ How do I find a point on the plane? If i set x = 0, and isolate for y or z in 1 eq and plug it into the other eqn, does it work? $\endgroup$ – Nick May 30 '12 at 0:09
  • $\begingroup$ Not really. Note that the two equations you have impart a condition on what $x$ and $y$ should be. Multiply the first one through $3$ and subtract the other equation multiplied by $2$. The $z$ will cancel out. $\endgroup$ – Pedro Tamaroff May 30 '12 at 0:14
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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:

$$x = -20 - 6t \\y = 31 - 6t \\z = 84 + 18t $$

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  • $\begingroup$ didn't work. I substituted 0 for each of each number and came up with different results for the other two variables each time. $\endgroup$ – Goodwin Lu Sep 18 '17 at 23:51
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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 $$A+t\,\overrightarrow{AB}\,,\,t\in\mathbb{R}$$

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