Determine the point of intersection of the plane $[x,y,z]=[6,-2,-3]+s[1,3,0]+t[2,2,-1]$ and the $z$-axis.

Question

Determine the point of intersection of the plane $[x,y,z]=[6,-2,-3]+s[1,3,0]+t[2,2,-1]$ and the $z$-axis.

I'm kind of lost with this question; do I start off solving for $s$ by elimination first?

Answer 1

The $z$-axis is where $x=y=0$. So set $6+s+2t=-2+3s+2t=0$ and solve for $s$ and $t$.

Then plug in those values for $s$ and $t$ to find the $z$-coordinate. (I got $2$).

The $x$- and $y$-coordinates of the intersection point should be $0$ (again, since it's on the $z$-axis).

Answer 2

Hint: You need $6+s+2t=0$ and $-2+3s+2t=0$. That's a system of two equations in two unknowns.

Then $z=-3-t$.

Answer 3

One method that I encourage you to employ when it comes to solving for the intersection of a line and a plane in $\Bbb R^3$ is to take the parametric equations of the line and substitute them into the scalar equation of the plane.

In this problem, the line has the parametric equations $$l:\begin{cases}x=0\\y=0\\z=t\end{cases}$$

and the plane has the normal equation $$3x-y+4z-8=0.$$

Substituting, we get $$4t-8=0\implies t=2.$$

To find the point of intersection we can now substitute $t=2$ into $l$ to get the coordinates $(0,0,2)$.