Question

"" Find a vector equation and parametric equations for the line. (Use the parameter t.) The line through the point (0, 11, −9) and parallel to the line x = −1 + 3t, y = 6 − 3t, z = 3 + 7t

Answer 1

Find the tangent vector T to the given line (t) = 〈 x(t), y(t), z(t) 〉, with parametric equations,

x(t) = -1 + 3t

y(t) = 6 - 3t

z(t) = 3 + 7t

by differentiating (t) component-wise:

T = '(t) = 〈 dx(t)/dt, dy(t)/dt, dz(t)/dt 〉

T = 〈3, -3, 7〉

Scale T by any real number t to get the line through the origin (when t = 0) and the point (3, -3, 7) (when t = 1), which of course runs parallel to '(t). This gives the vector tT. Translate this by any vector v = 〈a, b, c〉 to force this line to pass through the point (a, b, c) instead of the origin, while still remaining parallel. You want the new line to pass through (0, 11, -9), so take v = 〈0, 11, -9〉.

Then the vector equation of the new line is

tT + v = t 〈3, -3, 7〉 + 〈0, 11, -9〉 = 〈3t, 11 - 3t, -9 + 7t〉

with parametric equations

x(t) = 3t

y(t) = 11 - 3t

z(t) = -9 + 7t