# A line and a plane in 3 dimensions

Line L is defined as point P(3, 2, -2) with a directional vector v <1, -1, 2>.

Plane S: A $(1, 2, 1)$, B $(2, -1, 2)$, C $(0, -2, 1)$

a) When t = 2, how far away is the particle's position located from the plane?

b) If the source of light is shines normal to the plane, how fast the shadow of the particle is moving on the plane?

First of all, I have figured out the equations for the line and plane. The parametric equation for the line L: x(t) = 3 + t, y(t) = 2 - t, z(t) = -2 + 2t.

The equation for a plane is 4x - y - 7z = - 5 by finding the normal vector N = AB x AC = <4, -1, -7) and plugging-in the point A. 4(x - 1) - (y-2) - 7(z - 1) = 0.

When t = 2, x = 5, y = 0, z = 2 so the new position of the particle on the line L is Q (5, 0, 2).

I know how to solve part a) of this question by finding the distance between a point Q and a plane. However, I am not sure about the part b)...What we want is v = distance / t. I know that t = 2, but how can I find the distance from point P to point Q on the plane? I suspect that I need some kind of projection of the vector PQ on the plane, but I am not really confident about that.

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I agree with the results you show. For part (a), you would find that the point (we'll call it $\ E \$ ) in plane $\ S \$ closest to $\ Q \$ lies at $\ ( \frac{26}{6} , \frac{1}{6} ,\frac{19}{6} ) \ ,$ the distance separating points $\ Q \$ and $\ E \$ being $\ \frac{\sqrt{66}}{6}$ .

There are a few ways to go about finding the rate at which the "shadow" of the particle moving along line $\ L \$ traverses plane $\ S \$ . We will start with a "naive" method and see that we can develop a more sophisticated approach from that.

The straightforward way would be to determine the point $\ D \$ on the plane which is closest to external point $\ P \$ . This point can be thought of as the "shadow" of $\ P \$ , since the closest point on a plane to an external point lies on along a normal line to the plane, which here is also the direction of the "illumination". As was done to locate point $\ E \$ , we set up parametric equations for the normal line to the plane which passes through $\ P \$ ,

$$x \ = \ 3 \ + \ 4v \ \ , \ \ y \ = \ 2 \ - \ v \ \ , \ \ z \ = \ -2 \ - \ 7v \ ,$$

and solve $\ 4x - y - 7z + 5 = 0 \$ for the value of $\ v \$ of the point $\ D \$ in the plane. Passing over the algebra of the calculation, we obtain $\ 66v + 29 = 0 \ \Rightarrow \ v = -\frac{29}{66} \ ,$ giving us $\ D \ ( \frac{82}{66} , \frac{161}{66} ,\frac{71}{66} ) \ .$ Applying the "distance formula", the separation in plane $\ S \$ between $\ D \$ and $\ E \$ is

$$\sqrt{\frac{204^2 \ + \ 150^2 \ + \ 138^2}{66^2}} \ = \ \frac{\sqrt{83160}}{66} \ = \ \sqrt{\frac{1260}{66}} \ \approx \ 4.369 \ .$$

Since this is the distance the particle's shadow moves on the plane in 2 time-units, the speed of the shadow is $\ \frac{\sqrt{83160}}{132} \ = \ \sqrt{\frac{315}{66}} \ \approx \ 2.185 \$ .

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We worked matters out that way first for the purpose of checking our other methods. The next thing we can investigate is the motion of the particle in the plane containing $\ \vec{v} \$ and the two normal lines $\ DP \$ and $\ EQ \$ . In a single time-unit, the particle itself travels a distance given by the length of $\ \vec{v} \$ , which is $\ \sqrt{1^2 + (-1)^2 + (-2)^2} \ = \sqrt{6} \$ . The distance between points $\ D \$ and $\ P \$ is

$$\sqrt{\frac{116^2 \ + \ 29^2 \ + \ 203^2}{66^2}} \ = \ \frac{\sqrt{55506}}{66} \ = \ \frac{29}{\sqrt{66}} \ \approx \ 4.369 \ ,$$

while the separation between $\ E \$ and $\ Q \$ was already found to be $\ \frac{\sqrt{66}}{6} \ = \ \frac{11}{\sqrt{66}} \ .$

Hence, the perpendicular distance of the particle from plane $\ S \$ has diminished by $\ \frac{18}{\sqrt{66}} \$ in 2 time-units, or $\ \frac{9}{\sqrt{66}} \$ per time-unit. Now, this change in distance along the normal direction to the plane forms a right triangle with the change in distance parallel to the plane and the vector $\ \vec{v} \ ,$ which serves as the hypotenuse. So the length of the "leg" parallel to the plane gives the velocity of the particle along the plane, and thus the speed of the particle's shadow is found from

$$|| \ \vec{v} \ ||^2 \ - \ (v_{\perp})^2 \ = \ (v_{\parallel})^2$$

$$\Rightarrow \ (v_{\parallel})^2 \ = \ ( \ \sqrt{6} \ )^2 \ - \ ( \ \frac{9}{\sqrt{66}} \ )^2 \ = \ \frac{315}{66} \ \Rightarrow \ v_{\parallel} \ = \ \sqrt{\frac{315}{66}} \ ,$$

as before.

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This suggests a still more direct way to the "parallel velocity". Still working in the plane of vector $\ \vec{v} \$ and the two normal line segments, we can simply consider the "scalar projection" of $\ \vec{v} \$ onto the normal vector to the plane, $\ \vec{n} \$ , given by

$$\frac{\vec{v} \cdot \vec{n}}{|| \ \vec{n} \ ||} \ = \ \frac{\langle 1, -1, 2 \rangle \cdot \langle 4, -1, -7 \rangle}{ \sqrt{4^2+1^2+7^2} } \ = \ \frac{-9}{\sqrt{66}} ,$$

which, as we have seen, represents the speed at which the particle is approaching the plane along the normal direction. We complete the calculation as we did previously, using $\ (v_{\parallel})^2 \ = \ ( \ \sqrt{6} \ )^2 \ - \ ( \ -\frac{9}{\sqrt{66}} \ )^2 \ .$

Note that, in this last method, we have pretty much wrested the problem free of any discussion at all about the points in the plane $\ S \$ .

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Thanks for your help! I have actually solved this problem by myself using the projection of vector on the plane as you have described in your answer. Nevertheless, it was still beneficial for me to see that there are other approaches to this problem. – KurodaTsubasa Jun 18 '13 at 6:04
I thought it would be of interest to show how a direct examination of the points on the plane can be evolved into a form which only involves direction vectors. I'm glad you found this helpful. – RecklessReckoner Jun 18 '13 at 6:08