If P(1,-2,4) is reflected in the plane with equation 2x-3y-4z+66=0, determine the coordinates of its image point, P'. (Note that the plane 2x-3y-4z+66=0 is the right bisector of the line joining P(1,-2,4) with its image).
Hi skeske1234
A is the intersection between the line L (that passes through A and P) and the plane so you need to find the eqution of line L.
I'm pretty sure you've learnt about equation of line. Can you fiind it? Then, have you learnt to find the intersection between line and plane ?
so am i supposed to find the intersection between the line and the plane or am i supposed to find the equation of the line.. sorry confused here
also,
i'm actually not sure how to find the equation of the line if I am only given a point
But this is my attempt or guess so far:
l is the eqtn of the line i am looking for right now (right?)
l: (x,y,z)= (a,b,c) + t(2,-3,-4).
Hi skeske1234
You need to find the equation of the line first, then find the intersection between the line and plane.
Yes, you're on the right track. (a,b,c) is the point that lies on the line L. Do you know what the point is ?l is the eqtn of the line i am looking for right now (right?)
l: (x,y,z)= (a,b,c) + t(2,-3,-4).
l1: (x,y,z)=(1,-2,4)+t(2,-3,-4)
l2: (x,y,z)=(a,b,c)+t(2,-3,-4)
so when i made them equal to each other with their parametric equations, i got A(1,-2,4).
This is the same as l1.. this doesn't make sense. What did I do wrong? can you show me how you get the correct answer?
No
There is only one equation of line, which is L : (x,y,z)=(1,-2,4)+t(2,-3,-4). This is the line that passes through A and P and is perpendicular to plane 2x - 3y -4z + 66 = 0.
Now you need to find the point of intersection between the line and plane. Do you get what I mean ?
so I got
x=-3
y=4
z=12
as my POI, this answer doesn't match the back of the book. Here's my work:
x=1+2t
y=-2-3t
z=4-4t
2(1+2t)-3(-2-3t)-4(4-4t)+66=0
t=-2
x=1+2(-2)
y=-2-3(-2)
z=4-4(-2)
(-3,4,12) << does not match the answer they have in the textbook, where did i go wrong here?