Question

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).

Answer 1

Answer:

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

Answer 2

The parametric equations are:

x  =  -4  +  5t

y  =  4  +  5t

z  =  -2  - 6t

The given direction vector is:

$\bar{V} = 5i + 5j - 6k$

The direction vector can also be written as:

$\bar{V} = <a, b, c> = <5, 5, -6>$

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

The parametric equation is of the form:

$X = X_{0} + \bar{V}t$

This is:

$\left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}x_0\\y_0\\z_0\end{array}\right] + \left[\begin{array}{ccc}a\\b\\c\end{array}\right]t$

$\left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}-4\\4\\-2\end{array}\right] + \left[\begin{array}{ccc}5\\5\\-6\end{array}\right]t$

The parametric equations are therefore:

x  =  -4  +  5t

y  =  4  +  5t

z  =  -2  - 6t

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