Question

Find parametrizations for the line segments joining the points in Exercises 13–20. Draw coordinate axes and sketch each segment, indicating the direction of increasing t for your parametrization. (0, 0, 0), (1, 1, 3/2).

Answer 1

Answer:

$x=1+t, y=1+t, z=3/2 +3/2t$

Where $-1 \leq t \leq 0$

Step-by-step explanation:

We need a set of parametric equation like this:

$x = x_1 +a t$

$y = y_1 + bt$

$z = z_1 +ct$

We have two points given where the line passes through

(0,0,0) and (1,1,3/2)

So we can find the director vector z as the difference between the two points:

$z = <x-x_1 , y-y_1 , z-z_1>$

And replacing we got:

$z=<1-0, 1-0,3/2 -0>=<1,1,3/2>$

So we have the direction coordinates given by:

$<d,e,f>=<1,1,3/2>$

Now we can use the point <1,1,3/2> and the direction coordinates and we have:

$x = 1 + t$

$y = 1+ 1t$

$z= 3/2 + 3/2 t$

and then we have this for the coordinates x,y,z

$(x,y,z) = <1+t, 1+t, 3/2 +3/2t>$

Since the line passes through (0,0,0)

$1+t =0 , t =-1$

$1+t = 0 , t=-1$

$3/2 + 3/2t = 0, t=-1$

So one solution is t=-1

And for the other point (1,1,3/2) we have this:

$1+t = 1, t=0$

$1+t = 1, t=0$

$3/2 + 3/2t = 3/2 , t=0$

So the other solution its t =0

So we can define the general solution as :

$x=1+t, y=1+t, z=3/2 +3/2t$

Where $-1 \leq t \leq 0$

The figure attached shows the vector when t=0 for example.