Find parametric equations for the line. (enter your answers as a comma-separated list of equations. let x, y, and z be functions of t.) the line in the direction of the vector 5 i + 5 j − 6k and through the point (−4, 4, −2).

The parametric equations are:

x  =  -4  +  5t

y  =  4  +  5t

z  =  -2  - 6t

The given direction vector is:

The direction vector can also be written as:

The point X₀ = (x₀, y₀, z₀) =  (-4, 4, -2)

The parametric equation is of the form:

This is:

The parametric equations are therefore:

x  =  -4  +  5t

y  =  4  +  5t

z  =  -2  - 6t

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x=5t-4 , y=5t+4 , z=-6t-2

Step-by-step explanation:

So we are going to use (-4,4,-2) as an initial point, p.  

The direction vector is v=5i+5j-6k or <5,5,-6>.

The vector equation is r=vt+p.

That means we have r=<5,5,-6>t       +      <-4,4,-2>.

So the parametric equations are

x=5t-4

y=5t+4

z=-6t-2

the degrees would remain the same ( the number to the left of the decimal point.

the multiply the decimal by 60 to get the minutes and use the numbers to the left of the decimal point.

then multiply the decimal portion of that by 60 for the seconds and round off as needed.

39 .956652

the 39 remains the degrees.

the minutes = 0.956652 x 60 = 57.39912

the minutes = 57

the seconds = 0.39912 x 60 = 23.95 = 24 ( rounded to nearest tenth)

so 39.956652 = 39 degrees, 57 minutes and 24 seconds

-75.159809

the -75 remains the degrees

the minutes = 0.159809 x 60 = 9.58854

the minutes = 9

the seconds = 0.58854 x 60 = 35.31, rounded to nearest tenth = 35.3 seconds

so -75.159809 = -75 degrees, 9 minutes and 35.3 seconds.

the first coordinate is north or south, since it it positive it is north.

the second number is east or west, because it is negative it is west.

39 degrees, 57 minutes and 24 seconds n

-75 degrees, 9 minutes and 35.3 seconds w