# Thread: find the equation of the plane.

1. ## find the equation of the plane.

Find the equation, in the form of ax + by + cz = d, of the plane going through a point
P = (-3, 4, 2) to a line L: (x, y, z) = (3, -2, -1) + (1, -2, 2)t.

Can you solve this with brief explanation ?

2. Originally Posted by eunse16
Find the equation, in the form of ax + by + cz = d, of the plane going through a point
P = (-3, 4, 2) to a line L: (x, y, z) = (3, -2, -1) + (1, -2, 2)t.

Can you solve this with brief explanation ?
Hi eunse16.

The plane $ax+by+cz+d$ passes through $(-3,4,2)$ so

$-3a+4b+3c\ =\ d\quad\ldots\quad\fbox1$

And every point on the line $(x,y,z)=(3+t,-2-2t,-1+2t)$ lies on the plane. Take any two points on the line, say $(2,0,-3)$ and $(0,4,-7),$ and plug the values into the plane equation:

$2a-3c\ =\ d\quad\ldots\quad\fbox2$

$4b-7c\ =\ d\quad\ldots\quad\fbox3$

Solve the simultaneous equations $\fbox1-\fbox3$ for $a,\,b,\,c$ in terms of $d.$

You should get $a=d,\ b=\frac56d,\ c=\frac13d;$ hence the equation of the plane is $x+\frac56y+\frac13z=1.$

3. Another way to do this: Choose two points on the line. For example, if t= 0 then (x, y, z) = (3, -2, -1) + (1, -2, 2)0= (3, -2, -1) and if t= 1, (x, y, z) = (3, -2, -1) + (1, -2, 2)1= (4, -4, 1). That gives (-3, 4, 2), (3, -2, -1), and (4, -4, 1) as three points in the plane and we can use the standard method of finding the equation of the plane that includes three given points.

Now, the vector from (-3, 4, 2) to (3, -2, -1) is 6i- 6j- 3k and the vector from (-3, 4, 2) to (4, -4, 1) is 7i- 8j- k. The cross product of those two vectors is a normal vector to the plane and its equation can be written A(x-x0)+ B(y-y0)+ C(z-z0)= 0 where Ai+ Bj+ Ck is a normal vector and (x0, y0, z0) is a point in the plane.

4. Here is a general way.
If $\ell :Q + tD\;\& \,P \notin \ell$ then the plane containing $\ell~\&~P$ is:
$\left( {\overrightarrow {QP} \times D} \right) \cdot \left( {\left\langle {x,y,z} \right\rangle - Q} \right) = 0$