Question

find the point(s) of intersection (if any) of the plane and the line. 2x + 3y=-5, (x - 1)/4=y/2=(z - 3)/6. determine whether the line lies in the plane. the line select lie in the plane. (x,y,z)=( )

Explanation:

Step1: Write the parametric equations of the line

The symmetric - form of the line is given by $\frac{x - 1}{4}=\frac{y}{2}=\frac{z - 3}{6}=t$. Then $x=4t + 1$, $y = 2t$, $z=6t+3$.

Step2: Substitute the parametric equations into the plane equation

The plane equation is $2x + 3y=-5$. Substitute $x = 4t+1$ and $y = 2t$ into it:
\[

$$\begin{align*} 2(4t + 1)+3(2t)&=-5\\ 8t+2 + 6t&=-5\\ 14t&=-7\\ t&=-\frac{1}{2} \end{align*}$$

\]

Step3: Find the coordinates of the intersection point

Substitute $t =-\frac{1}{2}$ into the parametric equations of the line.
$x=4\times(-\frac{1}{2})+1=-2 + 1=-1$.
$y=2\times(-\frac{1}{2})=-1$.
$z=6\times(-\frac{1}{2})+3=-3 + 3=0$.

To check if the line lies in the plane, we need to verify if all points on the line satisfy the plane equation. Since we have found a single - point of intersection, the line does not lie in the plane.

Answer:

$(x,y,z)=(-1,-1,0)$; The line does not lie in the plane.