Find the parametric equations for the line that contains the point (3,1,-4) and is parallel to the line
r(t)=(1+2t)i+(-3-t)j+(5+5t)k
To find the parametric equations for the line that contains the given point (3, 1, -4) and is parallel to the line r(t) = (1 + 2t)i + (-3 - t)j + (5 + 5t)k, we can follow these steps:
Step 1: Find the parallel vector.
The given line r(t) can be rewritten as a vector equation: r(t) = (1)i + (-3)j + (5)k + (2t)i + (-t)j + (5t)k
By observing the coefficients of t in each component, we can extract the parallel vector: (2)i + (-1)j + (5)k.
Step 2: Write the parametric equations.
Let's assume the parametric equations for the line we are looking for are:
x = 3 + at
y = 1 + bt
z = -4 + ct
where a, b, and c are constants to be determined.
Since the line we want to find is parallel to the given line, the coefficients of t in the parallel vector should be equal to the corresponding coefficients in the parametric equations. Therefore, we have:
a = 2
b = -1
c = 5
So, the parametric equations for the line that contains the point (3, 1, -4) and is parallel to the line r(t) = (1 + 2t)i + (-3 - t)j + (5 + 5t)k are:
x = 3 + 2t
y = 1 - t
z = -4 + 5t.