Entered Answer Preview Result 5 2t 5-2 t correct 2 3t 2-3 t Correct 1 2t 1-2 t c
ID: 2880622 • Letter: E
Question
Entered Answer Preview Result 5 2t 5-2 t correct 2 3t 2-3 t Correct 1 2t 1-2 t correct correct incorrect incorrect At least one of the answers above is NOT correct. (1 point) (A) Find the parametric equations for the line through the point P (5, 2, 1) that is perpendicular to the plane -2a 3y 22 1. Use "t" as your variable, t E 0 should correspond to P, and the velocity vector of the line should be the same as the standard normal vector of the plane. x 5-2t y 2-3t 1-2t (B) At what point Q does this line intersect the yZ-plane? Q E 0Explanation / Answer
A) we have given point P(5,2,1) and plane -2x-3y-2z=1.
a normal vector n to the plane is
n=<-2,-3,-2>
the vector equation of the line is L = < 5,2,1 > + t <-2,-3,-2>
r(t)=<5-2t,2-2t,1-2t>
the parametric equations are
x=5-2t
y=2-3t
z=1-2t
B) the yz-plane intersect the line x=0
5-2t=0
2t=5 implies t=5/2
r(t)=<5-2t,2-2t,1-2t>
r(5/2)=<5-2*(5/2),2-2*(5/2),1-2*(5/2)> =<0,-3,-4>
this line intersect the yz-plane at Q=<0,-3,-4>
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