Question

Find a vector equation and parametric equations for the line. (Use the parameter t.)

The line through the point (4, -9, 6) and parallel to the vector <1,5,-2/3>
r(t) =
(x(t), y(t), z(t)) = ( )

Answer 1

To find the vector and parametric equations for the line, we use the point (4, -9, 6) and the direction vector <1, 5, -2/3>. The vector equation is ρ(t) = <4, -9, 6> + t<1, 5, -2/3>. The parametric equations are x(t) = 4 + t, y(t) = -9 + 5t, and z(t) = 6 - (2/3)t.

Step-by-step explanation:

To find a vector equation and parametric equations for a line, we need a point on the line and a direction vector. The line in question passes through the point (4, -9, 6) and is parallel to the vector <1, 5, -2/3>. The vector equation of a line can be written as:

ρ(t) = r₀ + tv

Where ρ(t) is the position vector at time t, r₀ is a position vector to a known point on the line, and v is the direction vector. In this case:

r₀ = <4, -9, 6>

v = <1, 5, -2/3>

Thus, the vector equation of the line is:

ρ(t) = <4, -9, 6> + t<1, 5, -2/3>

The parametric equations for the line can be extracted from the vector equation as follows:

x(t) = 4 + t(1) = 4 + t

y(t) = -9 + t(5) = -9 + 5t

z(t) = 6 + t(-2/3) = 6 - (2/3)t

These equations fully describe the motion along the line with the parameter t.