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[His_Dudeness3]

Hi, I've got these 2 questions left on an advanced Mathematics assignment (due Monday morning :( ) that I've been trying to crack but I'm not sure if what I have done is correct. Any help at all is greatly appreciated.

Question:
(1) (a) According to the Flat Mars Society, Mars is also a plane, given by the equation
4x + 3y − z = −3.
Find a parametric equation of the line, L, in which Earth intersects Mars.
(5 marks)
(b) How far away is Canberra, given by the point (−5, 10, 13), from this line?

You are given that the plane describing Earth is given by the equation
x + y + z = 18

My Answer:
(a) I firstly substituted z = 0 into both plane equations as a point that is on the line where both Earth and Mars intersect, where I got two simultaneous equations:
4x + 3y = -3
x + y = 18

I solved for x and y, and I got the point, P < -57, 75, 0 > that lies on the line of intersection between the two planes. Now, I know that the line that intersects both planes must be perpendicular to both the normal vectors the Earth plane and the normal vector of the Mars plane.

Normal Vector(Earth) = < 1, 1, 1 >
Normal Vector(Mars) = < 4, 3, -1 >

I did the cross product, and ended up getting the vector
L = < 4, -5, 1 >

Thus, using the values I got from earlier, I wrote them in the form:

L = (x - 57)/ 4 = (y + 75)/ -5 = 0

(b) Using the line L = < 4, -5, 1 >, I calculated the distance between L and Canberra

D = ( (x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2 )^(1/2)
I used the points from L as x1,y1 and z1 and I used the points from Canberra as x2,y2 and z2. After crunching the numbers, I got
D = 15((2)^0.5) units as the distance between them.

Question 2.
Two perfectly round pieces of rock are hurtling through space. The first, Superman’s
holiday asteroid, has radius 0.3, and is traveling on the line given by the parametric
equation
x(t) = 2 + t, y(t) = −1 − t, z(t) = t.
The second rock, made of kryptonite and set in motion by one of Superman’s enemies,
has radius 0.1 and is traveling on a line given by the parametric equation
x(s) = 3 − s, y(s) = 1, z(s) = 1 + s.
Calculate the distance between the two lines and use this distance to prove
that Superman has nothing to worry about.

I attempted this question but I got some reaaaaally obscure answer ( I think I got the answer for the distance between Superman and the Asteroid as 2.39x10^27 :s). Anyway, I've exhausted all my options, reread my lecture notes and the textbook but I can't seem to do this question.

Again, ANY help is greatly appreciated.

Thanks.
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution

[dynamicsolo]

For question 2, the distance between these two skew lines would be measured along a line mutually perpendicular to both. So we'll have to construct it.

Pick a point on each line and find the vector between them, say,

< x(t)-x(s), y(t)-y(s), z(t)-z(s) > = < -1+t+s, -2-t, t-s-1 > .

We need the values of t and s where this vector is perpendicular to each of the two lines. So take the dot product of this vector with the vector for each line individually (what are those vectors?) and set the two dot products to zero. You now have two linear equations in t and s. With the values of t and s you find, you can get the coordinates on each line of the points where the two lines are closest. Finally, find the distance between those two points; that is the minimal separation between the two lines. (Ol' Supes will be A-OK...)

[dynamicsolo]

Question 1.
...
My Answer:
(a) I firstly substituted z = 0 into both plane equations as a point that is on the line where both Earth and Mars intersect, where I got two simultaneous equations:
4x + 3y = -3
x + y = 18

I solved for x and y, and I got the point, P < -57, 75, 0 > that lies on the line of intersection between the two planes. Now, I know that the line that intersects both planes must be perpendicular to both the normal vectors the Earth plane and the normal vector of the Mars plane.

Normal Vector(Earth) = < 1, 1, 1 >
Normal Vector(Mars) = < 4, 3, -1 >

I did the cross product, and ended up getting the vector
L = < 4, -5, 1 >

I agree with you to here.


Thus, using the values I got from earlier, I wrote them in the form:

L = (x - 57)/ 4 = (y + 75)/ -5 = 0

Almost: the terms in symmetric equations use (x-xo), etc., and the z-coordinate of the point you chose may be zero, but the z-term has to be

L: ( x - (-57) )/ 4 = ( y - 75 )/ -5 = ( z - 0 )/ 1

or ( x + 57 ) / 4 = ( y - 75 )/ -5 = z .

But the question appears to ask for parametric equations for L, so we should set all of these terms equal to t and write

x = 4t - 57 , y = -5t + 75 , z = t .

For part (b), we could now use a method comparable to what I described for Question 2. We want the perpendicular distance from "Canberra" at (-5, 10, 13) to this line, so we want to find that perpendicular vector, which will have components

< (4t-57) - (-5), (-5t+75) - 10, t - 13 >

and, as it is perpendicular to L, the dot product must be

< 4, -5, 1 > · < (4t-57) - (-5), (-5t+75) - 10, t - 13 > = 0 ,

which gives us one linear equation in t. The value of t will get us the coordinates of the closest point on L to "Canberra"; lastly, we can find the distance between those points, which is the perpendicular distance of "Canberra" from L.

[His_Dudeness3]

For question 2, the distance between these two skew lines would be measured along a line mutually perpendicular to both. So we'll have to construct it.

Pick a point on each line and find the vector between them, say,

< x(t)-x(s), y(t)-y(s), z(t)-z(s) > = < -1+t+s, -2-t, t-s-1 > .

We need the values of t and s where this vector is perpendicular to each of the two lines. So take the dot product of this vector with the vector for each line individually (what are those vectors?) and set the two dot products to zero. You now have two linear equations in t and s. With the values of t and s you find, you can get the coordinates on each line of the points where the two lines are closest. Finally, find the distance between those two points; that is the minimal separation between the two lines. (Ol' Supes will be A-OK...)

I got up to here (bold lettering) but I got two equations (with both variables s and t), as
-4+2s+3t-2s^(2)=0 equation (1)
-2+2s-t+3t^(2)=0 equation (2)
Where do I go from here? I tried making t=2s-2+3t^(2) but no matter which way I do it, I still end up with both variables :s.

[dynamicsolo]

The dot products are to be taken between the "linking" vector and each of the direction vectors for the two skew lines. Thus,

< -1+t+s, -2-t, t-s-1 > · < 1, -1, 1 > = 0

and

< -1+t+s, -2-t, t-s-1 > · < -1, 0, 1 > = 0 .

I'm afraid I don't see how you arrived at your equations. Did you use the entire term of each of the components of the line equations? You only want the linear coefficients, which correspond to the "slopes" of the lines in each of the three dimensions.

[His_Dudeness3]

The dot products are to be taken between the "linking" vector and each of the direction vectors for the two skew lines. Thus,

< -1+t+s, -2-t, t-s-1 > · < 1, -1, 1 > = 0

and

< -1+t+s, -2-t, t-s-1 > · < -1, 0, 1 > = 0 .

I'm afraid I don't see how you arrived at your equations. Did you use the entire term of each of the components of the line equations? You only want the linear coefficients, which correspond to the "slopes" of the lines in each of the three dimensions.

lol oops, I used vectors from a previous question(I've been awake for about 23 hours). Thanks for the clarification dynamicsolo

I ended up using those two dot product equations to get the values, t=2/3 and s=0. I subbed those into the original parametric equations to get:
Superman= <8/3, 1/3, 2/3> and Kryptonite= <3, 1, 1>

I calculated the distance between those two points and I got D= 0.8164 (0.8).Is this correct?

[dynamicsolo]

I found a mistake in my own work from yesterday, but what I posted for the equations is correct. I am getting (a bit surprisingly) is that both s = 0 and t = 0. This makes the closest points on the two lines (2, -1, 0) for the t-line and (3, 1, 1) for the s-line. [The linking vector is then <1, 2, 1>, which is indeed orthogonal to the direction vectors of both lines.] The closest approach of the two lines is then sqrt(1^2 + 2^2 + 1^2) = sqrt(6).

[essecks]

sqrt(6) is correct for this, but you have to take into account the radii of the asteroids too (0.3 and 0.1)

so the separation is then ~2.45-0.3-0.1