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.