Line 1's equation is given as r(t) = <t, t, 1-t>. Line 2 passes through the origin and is perpendicular to line 1. Find the equation of line 2.
The Attempt at a Solution
If it is a 2-D case then it would be a lot easier, since m1 x m2 = -1. However, this is a 3-D case.
Anyway, here is my attempt.
I know that the dot product of two vector lines equals 0 iff they are perpendicular.
Line 1, r(t) = <t, t, 1-t> = <0, 0, 1> + t <1, 1, -1>
Line 2, s(t) = <0, 0, 0> + u <a, b, c>.
So I need to find the value of a, b and c.
<1, 1, -1> dot <a, b, c> = 0
a + b - c = 0.
Let a = b = 1, so c = 2.
So the equation of line 2 is s(t) = <0, 0, 0> + u <1, 1, 2>.
The answer and method seems plausible to me.
However, I remember seeing somewhere in the book where in order to solve this kind of problem, we need to find the equation of the plane that contains the first line. Then, find the equation of another plane and the intersection of this 2 planes is the equation of the line 2. This is done by using the cross product of 2 normal lines.