Find Parametric Equations for Line Parallel to Plane in Space

[michaelwiggin]

Homework Statement

Find a set of parametric equations for the line passing through the point (1, 0, 1) that is parallel to the plane given by x + y + z = 5, and perpendicular to the line
x = −2 + t, y = 1 + t, z = 2 − t.

Homework Equations

The section deals with distances betweens points, lines, and planes in space, so perhaps:

D (distance between a point and a line in space) = ||PQ x u||/||u||, wherein u is the direction vector for the line, and P is a point on the line (so PQ is a vector between P and the Q not on the line).

D (distance between a plane and a point in space) = |PQ dot n|/||n||, wherein n is a vector normal to the plane, and P is a point on the plane, and Q is the point in space.

The Attempt at a Solution

Well, this problem is number 128 in the textbook section... its the last problem, which in Larson's Calculus (this is the first unit in my MV calc class, btw) usually means that it requires a lot more visualization and logic than the prior ones. So I don't want to look like I didn't try, but all I really have to show for myself is some brainstorming.. I haven't been able to put pen to paper meaningfully. I know that I could use a line parallel to the given line, but I can't figure out how to find that given the perpendicular line provided. If the given plane was perpendicular instead of parallel, it would be easy to find a vector normal to it (which would be parallel to the line I'm trying to find) and work from there. I'm just having a big brain fart, and I don't have time today to grapple with this for an hour like I usually do :\ any tips or suggestions would be very very very appreciated! :) thanks!

-MichaelEdit: P.S., in the equations, |...| is absolute value, and ||...|| is magnitude of the vector.

 

[mighty2000]

Dear Michael,

Thank you for your post. This is an interesting problem that requires a combination of visualization and logic. Let's break it down step by step.

First, we need to find the direction vector for the line parallel to the given plane. Since the plane is defined by the equation x + y + z = 5, we can use the coefficients of x, y, and z as the components of our direction vector. So, the direction vector would be u = (1, 1, 1).

Next, we need to find a point that lies on the line we are trying to find. Since the line is parallel to the given plane, we can choose any point on the plane and use it as a point on our line. Let's choose the point (1, 0, 0) since it satisfies the equation x + y + z = 5.

Now, we can use the point and direction vector to write the parametric equations for the line. The parametric equations for a line are x = x0 + at, y = y0 + bt, and z = z0 + ct, where (x0, y0, z0) is a point on the line and (a, b, c) is the direction vector. Plugging in our point and direction vector, we get the following parametric equations for our line: x = 1 + t, y = bt, z = ct.

The only thing left to determine is the value of b and c. We know that our line needs to be perpendicular to the given line x = −2 + t, y = 1 + t, z = 2 − t. This means that the direction vector of our line needs to be perpendicular to the direction vector of the given line. Using the dot product, we can set up the following equation: (1, b, c) ⋅ (-1, 1, -1) = 0. Solving this equation, we get b = -1 and c = 1.

Therefore, the final parametric equations for the line passing through the point (1, 0, 0) that is parallel to the plane x + y + z = 5 and perpendicular to the line x = −2 + t, y = 1 + t, z = 2 − t are: x = 1 + t, y = -t, z = t.

I