Parametric Equations,Equations of planes

Question

(i) Find the parametric equation of the line l through the point (2, 2, −2) and
parallel to the line x = 2 − 7t, y = 3 + 4t, z = −t, t ∈ R.

 

(ii)Calculate the equation of the plane which contains the point (1, 1, 1) and which
is perpendicular to the line x = 2t, y = 45 − t, z = 17 + 5t, t ∈ R

Answer 1

the key to answering this one is to rewrite the second equation in parametric form (2,3,0) + (-7,4,-1)t

the line parallel to this through (2,2,-2) will go through this point and have the same orientation as the first line (-7,4,-1).

So, the equation would be L = (2,2,-2) + (-7,4,-1)t

2) the line should be rewritten as L = (0,45,17) + (2,-1,5)t, its parametric form. You need two vectors that are non-colinear that are perpendicular to the vector (2,-1,5). Two non-colinear vectors fill a plane. The easiest way to identify vectors perpendicular to (2,-1,5) is to set vectors that operate on two variables (of the three) only and yield a dot product of 0 when multiplied by (2,-1,5).

for instance (1,2,0) . (2,-1,5) gives 1x2 + 2x-1 + 0x5 = 2 - 2 + 0 = 0

and (0,5,1) . (2,-1,5) gives 0x2 + 5x-1 + 1x-5 = 0 -5 + 5 =0

 

so a plane that is defined by (1,2,0)u + (0,5,1)v will be perpendicular to the line (0,45,17) + (2,-1,5)t, but it does not contain (1,1,1) unless translate the plane by writing R = (1,1,1) + (1,2,0)u + (0,5,1)v.

 

P.S.  I see by your other question that we use two different conventions for writing parametric equations.  So, using your method:

 

1) line l: x = 2 - 7t, y = 2 + 4t, z = -2 - t

2) plane R: x = 1 + u, y = 1 + 2u + 5v, z = 1 + v