Question

Consider the line which passes through the point P(−1,-3,5), and which is parallel to the line x=1+7t, y=2+2t, z=3+4t Find the point of intersection of this new line with each of the coordinate planes:

Answer 1

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)