Answer:
The x-coordinate of the intersection point between the line and the plane is -3
Step-by-step explanation:
As clarification of the question we have "line with vector equation [tex]r(t) = <2, 4, 1> + t<2, 3,-2 >[/tex]". Thus to find the equation of a line parallel to the given line equation we can use the directional vector of the line and just use the given point.
Line equation representation.
The vector line equation through the point (1,0,2) that is parallel to the given line, can be found using the directional vector <2, 3, -2>, so we get
[tex]\vec L(t)=<1,0,2> +t<2,3,-2>[/tex]
We can expand the vector equation a bit
[tex]\vec L(t)=<1+2t,3t,2-2t>[/tex]
Finding the intersection point of the line with the plane.
We can use the vector representation of the line
[tex]x(t)=1+2t\\y(t)=3t\\z(t)=2-2t[/tex]
And replace it on the plane equation [tex]x-3y-z=9[/tex] so we get
[tex]1+2t-3(3t)-(2-2t)=9[/tex]
Simplifying and solving for t.
[tex]1+2t-9t-2+2t=9[/tex]
[tex]-5t-1=9 \\-5t=10[/tex]
Dividing both sides by -5 give us
[tex]t=-2[/tex]
Thus the intersection point is:
[tex]x=1+2(-2)=-3\\y=3(-2)=-6\\z=2-2(-2)=6[/tex]
Thus the x-coordinate of the intersection point is -3
