Intersection between two planes and a line?

Question

What is the coordinates of the point where the planes: $3x-2y+z-6=0$ and $x+y-2z-8=0$ and the line: $(x, y, z) = (1, 1, -1) + t(5, 1, -1)$ intersects with eachother?

I've tried letting the line where the two planes intersect eachother be equal to the given line, this results in no solutions.

I have tried inserting the lines x, y and z values into the planes equations, this too, results in no solutions.

According to the answer sheet the correct solution is: $\frac{1}{2}(7,3,-3)$

Answer 1

Isolate a variable in the planar equations and set the resulting expressions equal to each other (because they intersect):

$z= 6+2y-3x$, and $z=\frac{x+y-8}{2}$

so $6+2y-3x=\frac{x+y-8}{2}$.

Solving for x yields:

$\frac{20+3y}{7}=x$

and setting $y$ as the parameter ($y=s$), and substituting back into the original equation we have the equation of the line of intersection:

$x = \frac{20+3s}{7}$

$y=s$

$z=6+2s-\frac{60+9s}{7}$

Now, if the lines intersect the formulas for x and y must be equal so:

$s=1+t$

$\frac{20+3s}{7}=1+5t$

Solving yields $s=1.5$, which when plugged into our formulas for x,y, and z in terms of s yields $(3.5,1.5,-1.5)$, as desired.

Answer 2

Method #$1$:

$$\left(\begin{array}{ccc|c} 3 & -2 & 1 & 6 \\ 1 & 1 & -2 & 8\end{array}\right) \sim \left(\begin{array}{ccc|c} 1 & 0 & -3/5 & 22/5 \\ 0 & 1 & -7/5 & 18/5\end{array}\right)$$

Therefore the line of intersection of the two planes is $(x,y,z) = (22/5,18/5,0) + s(3,7,5)$.

Now equate the two lines: $(22/5,18/5,0) + s(3,7,5) = (1, 1, -1) + t(5, 1, -1)$ $$\implies \begin{cases} 22+15s=5+25t \\ 18 + 35s=5+5t \\ 5s=-1-t\end{cases} \implies \begin{cases} s=-3/10 \\ t=1/2 \end{cases}$$

So the point of intersection is $(1,1,-1) + \dfrac 12 (5,1,-1) = (7/2,3/2,-3/2)$.

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Method #$2$:

$u=(3,-2,1)$ is a vector normal the surface of the first plane. $v=(1,1,-2)$ is a vector normal to the surface of the second plane. Thus $u\times v = (3,7,5)$ is in the direction of the line of intersection. Then we just need one point on that line:

We're given $\begin{cases} 3x-2y+z-6=0 \\ x+y-2z-8=0\end{cases}$. Here we've only got $2$ equations, but we have $3$ unknowns, so let's just choose one arbitrary values for $x$ and see if we can then solve for $y$ and $z$. $x=0$ seems like a pretty good value. Then $\begin{cases} -2y+z-6=0 \\ y-2z-8=0\end{cases}$, which implies $\begin{cases}y = -\frac {20}3\\ z=-\frac {22}3\end{cases}$. So one point on the line of intersection of the two planes is $(0,-\frac {20}3,-\frac {22}3)$.

Thus another (equally valid) equation for the line of intersection is $(x,y,z) = (0,-\frac{20}3,-\frac {22}3) + s(3,7,5)$.

Then you equate the two equations of lines as in method #$1$.

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Method #$3$:

What we want to do here is find a vector equation for the two planes. To do that, we'll need to find two vectors parallel to each of our planes and a single point on each of them.

So let's look at the first plane. A normal to that plane is clearly $(3,-2,1)$. So if we found two vectors orthogonal to that vector which are not collinear, then we'd have our two vectors. To find the first, let's just take the dot product $(3,-2,1)\cdot (a,b,c) = 3a-2b+c=0$. Anything which solves this will do, so let's choose $(a,b,c) = (0,1,2)$ (I just chose two easy values and solved for the third -- when you do this, make sure you don't choose both $0$, or else you'll just get the $0$ vector, which won't work for us).

Now that we have $1$ vector orthogonal to our normal, to find another we can just do the cross product: $(3,-2,1)\times (0,1,2) = (-5,-6,3)$.

To find a single point on the plane, let's just set $x=y=0$ and solve $3x-2y+z-6=0$ for $z$. We get $z=6$. So with these three vectors we can construct a vector equation of our plane: $$\vec r_1 = (0,0,6) + s_1(0,1,2) + s_2(-5,-6,3)$$

Then we do exactly the same method to get the equation of the second plane which is: $$\vec r_2 = (0,0,-4) + u_1(1,-1,0) +u_2(-2,-2,-2)$$

Now just equate all three of our vector equations and solve : $$(1, 1, -1) + t(5, 1, -1) = (0,0,6) + s_1(0,1,2) + s_2(-5,-6,3) = (0,0,-4) + u_1(1,-1,0) +u_2(-2,-2,-2)$$

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Method #$4$:

Just plug in the variables $(x,y,z) = (1+5t,1+t,-1-t)$ from your linear equation into $3x-2y+z-6=0$ and $x+y-2z-8=0$. If, not only do you find a solution, but that solution is the same $t$ then your line does cross the two planes at their line of intersection. And the point of intersection is $(x,y,z) = (1+5t,1+t,-1-t)$ with the $t$ you solved for.

Answer 3

From the line equation you know that $x$ (as well as $y$) is a function of $z$: when $z = -1-t$, then $x = 1+5t$ and hence $x = -4-5z$. This gives you the third equation you need: \begin{align} 3x-2y+\phantom{1}z-6&=0\\ \phantom{1}x-\phantom{1}y-2z-8&=0\\ \phantom{1}x+0y+5z+4&=0. \end{align}

Answer 4

I get $(3.5,1.5,-1.5)^T$ as well.

The yellow line is the intersection of the two planes, the big red dot the intersection with the given line (black).

$$ E_1: 3x - 2y + z = 6 \wedge E_2: x + y - 2z = 8 $$ Solving for $z$ gives $$ z = 6 - 3x + 2y = -4 + x/2 + y/2 \\ 10 = 7/2 x - 3/2 y \iff y = 7/3 x - 20/3 $$ E.g.choosing $x=3$ and $x=4$ then $a = (3, 1/3, 6 - 9 + 2/3)^T = (3, 1/3, -7/3)^T$ and $b = (4, 8/3, 6-12+16/3)^T = (4, 8/3, -2/3)^T$ are part of the intersection $E_1 \cap E_2$.

From this we generate the line \begin{align} g &: (3, 1/3,-7/3)^T + ((4, 8/3,-2/3)^T - (3, 1/3,-7/3)^T) s \\ &= (3, 1/3,-7/3)^T + (1, 7/3, 5/3)^T s \end{align} and intersect it with the given line $$ f: (1,1,-1)^T + (5,1,-1)^T t $$ $g(s) = f(t)$, which gives the system: $$ \left( \begin{matrix} 1 & -5 \\ 7/3 & -1 \\ 5/3 & 1 \end{matrix} \right) \left( \begin{matrix} s \\ t \end{matrix} \right) = \left( \begin{matrix} -2 \\ 2/3 \\ 4/3 \end{matrix} \right) $$ This has the solution $(s, t) = (1/2, 1/2)$.

Using for example $t=1/2$ with $f$, we get $$ P = (3.5, 1.5, -1.5)^T $$