Question

(A) Find the parametric equations for the line through the point P = (-2, -1, 0) that is perpendicular to the plane −5x − 5y + 5z = 1.
Use "t" as your variable, t = 0 should correspond to P, and the velocity vector of the line should be the same as the normal vector to the plane found directly from its equation.

x = ???
y = ???
z = ???

(B) At what point Q does this line intersect the yz-plane?

Q = ( ? , ? , ? )

Answer 1

either this one is very simple like this
(-2,-1,0)+t(-5,-5,5)

BECAUSE N=(-5,-5,5) thats perpendicular to the plane

Answer 2

Find the intersection of the line Q with the coordinate planes.

For the y-z plane, we simply set x = 0 = -2 - 5tyz to find the value of the parameter t for the y-z plane:

tyz = -2/5.

Now we use this value of t in the equations for y and z:

Yyz = -1 - 5tyz = -1 - 5(-2/5) = 1

Zyz = 5*tyz = 5(-2/5) = -2

In other words, Q intersects the y - z coordinate plane at (0,1,-2)

Answer 3

the second part is right but how about A ?! i mean what is the X, Y and Z ?!

Answer 4

X=-2-5t
Y=-1-5t
Z=5t
or (x,y,z)=(-1,-1,0)+t(-5,-5,5)

Answer 5

thanks !