The curve r(t) is defined as:
r(t) = (3t+1)i + (t³+ 2t +2)j + (t² + 2t -1)k
The tangent vector of the curve at point (1, 2, -1) is obtained by taking the derivative of r(t) with respect to t and evaluating it at t=1:
r’(t) = 3i + 5j + 4tk
r’(1) = 3i + 5j + 4k
The line tangent to the curve at point (1, 2, -1) is parallel to the tangent vector r’(1). Therefore, the symmetric equation of the line can be written as:
x-1/3 = y-2/5 = z+1/4 = t
where x, y, and z are the coordinates of any point on the line and t is a parameter that varies over the real numbers.
I hope this helps!