3D line in a 3D plane. Find the intersection of the two.

Question

(I'm new to Math.StackExchange, so if you see any errors, please comment below!)

$\mathcal{P}$ is the plane containing the three points $(-3,4,-2)$, $(1,4,0)$, and $(3,2,-1)$. $\ell$ is the line containing the two points $(2, 4, -3)$ and $(-1, -1, -9)$. What is the intersection of plane $\mathcal{P}$ and line $\ell$?

I honestly have no clue how to solve this, or even where to start. Starter hints would be nice.

Much appreciated.

Answer 1

Find an equation of the plane, and one of the line.

• Plane:

$V_1 := (4,0,2)$ and $V_2 := (-2,2,1)$ are two linearly independent vectors in $P$, so $N := V_1 \times V_2 = (-4,-8,8)$ is a normal vector to $P$. Let $n = (1,2,-2)$ (just for simplicity). Then $n$ is also a normal to $P$, and $P$'s equation is given by:

$$n \cdot (x - 1, y-4, z) = 0$$

Therefore:

$$ P: x + 2y -2z = 9$$

• Line:

$U:= (3,5,6)$ is a direction vector of $\ell$, and $\ell$ passes through $A:= (2,4,-3)$. We know:

$$\ell : \begin{cases} x = x_A + x_U t \\ y = y_A + y_U t \\ z = z_A + z_U t \end{cases}$$

Thus:

$$\ell : \begin{cases} x= 3t + 2 \\ y = 5t + 4 \\ z = 6t - 3 \end{cases}$$

Let $M=(a,b,c)$ be the intersection point between $P$ and $\ell$. $M$ satisfies the equations of each: $P$ and $\ell$, so that for some $t$: $a= 3t + 2$, $b= 5t + 4$ and $c = 6t -3$, and $a+2b -2c = 9$. Hence, $3t + 2 + 10t + 8 -12t + 6 = 9$, so $t = -7$.

Therefore, $M = (-19, -31, -45)$.

Answer 2

Hint: I'll give the steps.

• Recall that given three non-collinear points (which is our case) $A,B$ and $C$ in a plane $\pi$, a normal direction to the plane is given by the vector $\overrightarrow{AB} \times \overrightarrow{AC}$. If you write this last vector as $(a,b,c)$, the equation of the plane is of the form $\pi \colon \,ax+by+cz = d$, where $d$ can be found by plugging any of the points $A,B$ or $C$ in the equation (it doesn't matter which one you choose, the result will be the same).

• Recall that given two points $D$ and $E$ in some line $\ell$, we can write $\ell \colon P(t) = D + t(E-D)$, with $t$ ranging over $\Bbb R$.

• Once you have both equations above, you need to find the special $t_0$ which makes the point $P(t_0)$ (which has three components) satisfy the equation $ax+by+cz=d$. This will be a first degree equation in $t_0$, so it will be pretty easy to solve it. The point $P(t_0)$ is the intersection you seek.