Vectors and the Geometry of Space

Question

Give a vector parametric equation for the line through the point (-1,5,-1) that is parallel to the line ⟨-1-t, 2t-1 ,-4-3t⟩:

L(t) =

Answer 1

x=x₀ + at , y= y_{0 +} bt , z = z₀ + ct

by using these expression we can find out the parametric equation for the points (-1 ,5 ,-1)

we have paramtric points (-1-t, 2t-1 ,-4-3t⟩

hence x = -1-t , y =2t-1 and z = -4-3t

x_{0 =} -1, y₀= 5, z₀= -1

a= -1, b= -1, c= -4

by putting these value in above expressions we get

x = -1-1t , y= 5-1t , z= -1-4t