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

y = - 6

step-by-step explanation:

given the point (- 2, y) lies on the graph then it is a solution of it's equation.

substitute x = - 2 into the equation for value of y

y = 3(- 2) = 3 × - 2 = - 6

the point on the graph is (- 2, 6)

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step-by-step explanation:

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