Parametric equations of a perpendicular line through a point

Find parametric equations for the line passing through the point (0, 2, 3) that is perpendicular to the line:

x = 1 + t, y = 2 − t, z = 4t and intersects this line.

Solution. This problem is in three dimensions, so it’s kind of hard to draw a picture. For the sake of having a visual reference, though, I am going to draw the picture as if it were in two dimensions. So here’s the set up: we have a point P = (0, 2, 3) and a line L given by the parametric equations above:

L P

What we’re looking for is the equation of the dashed line, which passes through P and intersects L :

L P

To find the equation of a line, we need a point and a direction vector. We already have the point (P ), so all we need now is a vector in the direction of the dashed line. The first step is to pick any two points on the line L, which we will call R and S. We can do this by plugging in two different t values into the parametric equations for L . Let’s use t = 0 and t = 1:

t = 0 ⇒ x = 1 + 0, y = 2 − 0, z = 2(0) gives us the point R = (1, 2, 0), and

t = 1 ⇒ x = 1 + 1, y = 2 − 1, z = 2(1)

1 give us the point S = (2, 1, 2). Then, we can construct the vectors −→ a = RP = P − R = (0, 2, 3) − (1, 2, 0) = h−1, 0, 3i and −→ b = RS = S − R = (2, 1, 2) − (1, 2, 0) = h1, −1, 2i

Here’s the picture so far:

L L P P S b a

R

The next step is to project the a vector onto the b vector:

L P

b a

projba

Calculating this gives us the following: a · b h−1, 0, 3i · h1, −1, 2i proj a = b = h1, −1, 2i b b · b h1, −1, 2i · h1, −1, 2i −1 + 0 + 6 = h1, −1, 2i 1 + 1 + 4 5 5 5 = , − , 6 6 3

Next, we can calculate the vector a − projba. This connects the endpoints of the two vectors, and is exactly the direction vector we want!

a b proj L P − a b a

projba

2 Calculating this vector gives:

5 5 5 a − proj a = h−1, 0, 3i − , − , b 6 6 3  11 5 4 = − , , 6 6 3

This is our direction vector! Let’s call it v, so we don’t have to keep writing a − projba. Now, remember that the vector equation for a line looks like:

r(t) = p + tv where p is a point on the line and v is a direction vector. We have a point, namely, p = h0, 2, 3i, and now 11 5 4 we have a direction vector, v = − 6 , 6 , 3 . Therefore, the vector equation of the line we want is:  11 5 4 r(t) = h0, 2, 3i + t − , , 6 6 3

To get a set of parametric equations, we can distribute the t and add the vectors together:

 11 5 4  r(t) = h0, 2, 3i + − t, t, t 6 6 3  11 5 4  = − t, 2 + t, 3 + t 6 6 3

So that: 11 5 4 x = − t, y = 2 + t, z = 3 + t 6 6 3

3