The given line is parameterized by
x(t) = 1 + 7t
y(t) = 2 + 2t
z(t) = 3 + 4t
and points in the same direction as the vector
d/dt (x(t), y(t), z(t)) = (7, 2, 4)
So, the line we want has parameteric equations
x(t) = -1 + 7t
y(t) = -3 + 2t
z(t) = 5 + 4t
Solve for t when one of x, y, or z is equal to 0 - this will tell you for which value of t the line cross a given plane. Then determine the other coordinates of these intersections.
• x = 0, which corresponds to the y-z plane:
0 = -1 + 7t ⇒ 7t = 1 ⇒ t = 1/7
y(1/7) = -3 + 2/7 = -19/7
z(1/7) = 5 + 4/7 = 39/7
⇒ intersection = (0, -19/7, 39/7)
• y = 0 (x-z plane):
0 = -3 + 2t ⇒ 2t = 3 ⇒ t = 3/2
x(3/2) = -1 + 21/2 = 19/2
z(3/2) = 5 + 12/2 = 11
⇒ intersection = (19/2, 0, 11)
• z = 0 (x-y plane):
0 = 5 + 4t ⇒ 4t = -5 ⇒ t = -5/4
x(-5/4) = -1 - 35/4 = -39/4
y(-5/4) = -3 - 10/4 = -11/2
⇒ intersection = (-39/4, -11/2, 0)
