Question

Let P and Q be the points in R^3 (Real Numbers on a 3-dimensional plane) with position vectors:
p = (3, -1, 1)
q = (7, 1, 3)
respectively.
Let the origin be (0, 0 ,0)

a) Find the vector form of the equation of the line l through P and Q.

b) Determine the position vector, a, of the point A on l such that OA is perpendicular to PQ.

Here is my answer for a:

(3 + 4k, -1 + 2k, 1 + 2k).

I'm struggling with b. I'm aware of the conditions that need to be met for 2 vectors to be perpendicular, but the language of the question is tripping me up.

Answer 1

a) If t is a real variable, then the line through the endpoints of p and q is

(p - q) t + p

That is, p - q is the vector pointing from the tip of q to the tip of p (recall the "triangle law" for vector addition). Then (p - q) t is a line through the origin parallel to p - q. Adding p to this line translates it so that it will pass through p.

So one formulation of the line's equation is

r(t) = ((3, -1, 1) - (7, 1, 3)) t + (3, -1, 1)

… = (-4, -2, -2) t + (3, -1, 1)

… = (-4t + 3, -2t - 1, -2t + 1)

which is identical to the equation you found, with t = -k.

b) OA is the line segment connecting the origin O and the point A on the line we found in part (a).

a is then the vector pointing from the origin to A. If OA is perpendicular to the line, then the dot product of a with the direction vector p - q must be zero.

a • (p - q) = 0

(a₁, a₂, a₃) • (-4, -2, -2) = 0

-4a₁ - 2a₂ - 2a₃  =0

2a₁ + a₂ + a₃ = 0

Let s(t) denote the line containing a. Then

s(t) = (a₁, a₂, a₃) t

r(t) and s(t) intersect exactly once for some value of t such that

(a₁, a₂, a₃) t = (-4, -2, -2) t + (3, -1, 1)

(a₁ + 4, a₂ + 2, a₃ + 2) t = (3, -1, 1)

Now we solve for the components of a such that the conditions we found are all met:

$\begin{cases}2a_1 + a_2 + a_3 = 0 \\ (a_1 + 4)t = 3 \\ (a_2 + 2)t = -1 \\ (a_3 + 2)t = 1\end{cases}$

and you would find the two lines intersect when t = 1/2, and so

a = (2, -4, 0)