Determine vector, parametric, and if possible, symmetric equations of the line through C(2, -2, 1) and parallel to the line with parametric equations x = -1 + 5t, y = 2 – t, z = 3 – 4t.
from the given parametric equation x = -1 + 5t, y = 2 – t, z = 3 – 4t
you know the direction vector is <5,-1,-4>
so simply edit it so it contains the new point C(2, -2, 1)
The rest is very straight forward.
Let me know what you get.
To find the vector equation of a line parallel to the given line, we can take any point on the line and the direction vector of the given line.
Let's first find the direction vector of the given line:
The direction vector of the line is given by the coefficients of t in the parametric equations.
So, the direction vector of the given line is <5, -1, -4>.
Now, to find the vector equation of the line parallel to the given line, we can use the point-direction form of a line, which is:
r = r₀ + tv
where:
r is the position vector of any point on the line,
r₀ is the position vector of a known point on the line, and
v is the direction vector of the line.
Given that C(2, -2, 1) is a point on the line, and the direction vector <5, -1, -4> as v, the vector equation of the line can be written as:
r = <2, -2, 1> + t<5, -1, -4>
Simplifying,
r = <2 + 5t, -2 - t, 1 - 4t>
This is the vector equation of the line through C(2, -2, 1) and parallel to the given line.
To obtain the parametric equations, we can separate the components:
x = 2 + 5t
y = -2 - t
z = 1 - 4t
These are the parametric equations of the line.
As for the symmetric equations, we can isolate t in each of the parametric equations:
x - 2 = 5t, so t = (x - 2) / 5
y + 2 = -t, so t = -(y + 2)
z - 1 = -4t, so t = (z - 1) / -4
Since each of these expressions represents t, we can equate them:
(x - 2) / 5 = -(y + 2) = (z - 1) / -4
This gives us the symmetric equations:
(x - 2) / 5 = -(y + 2) = (z - 1) / -4
Note that the symmetric equations are not unique; there are multiple ways to express the equations by multiplying each of the expressions by a constant.