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 $\displaystyle ax+by+cz+d$ passes through $\displaystyle (-3,4,2)$ so
$\displaystyle -3a+4b+3c\ =\ d\quad\ldots\quad\fbox1$
And every point on the line $\displaystyle (x,y,z)=(3+t,-2-2t,-1+2t)$ lies on the plane. Take any two points on the line, say $\displaystyle (2,0,-3)$ and $\displaystyle (0,4,-7),$ and plug the values into the plane equation:
$\displaystyle 2a-3c\ =\ d\quad\ldots\quad\fbox2$
$\displaystyle 4b-7c\ =\ d\quad\ldots\quad\fbox3$
Solve the simultaneous equations $\displaystyle \fbox1-\fbox3$ for $\displaystyle a,\,b,\,c$ in terms of $\displaystyle d.$
You should get $\displaystyle a=d,\ b=\frac56d,\ c=\frac13d;$ hence the equation of the plane is $\displaystyle x+\frac56y+\frac13z=1.$