Give a vector parametric equation for the line through the point (−2,−4,0) that is parallel to the line ⟨−2−3t,1−4t,−5t⟩:

The tangent vector for the given line is

T(t) = d/dt ⟨-2 - 3t, 1 - 4t, -5t⟩ = ⟨-3, -4, -5⟩

On its own, this vector points to a single point in space, (-3, -4, -5).

Multiply this vector by some scalar t to get a whole set of vectors, essentially stretching or contracting the vector ⟨-3, -4, -5⟩. This set is a line through the origin.

Now translate this set of vectors by adding to it the vector ⟨-2, -4, 0⟩, which correspond to the given point.

Then the equation for this new line is simply

L(t) = ⟨-3, -4, -5⟩t + ⟨-2, -4, 0⟩ = ⟨-2 - 3t, -4 - 4t, -5t⟩

step-by-step explanation:

given that ab is divided into equal parts as

ac, cd, de, ef, fg, gh, hi, ib.

from this we find that f is the midpoint of ab.

so using mid point formula we find

f = (3,7)

similarly considering fb, we get

h = mid point of fb = ((6,6) and g, mid point of fh is

g = (9/2, 13/2)

i is mid point of hb

so i =(15/2,11/2)