BFMracing
General Category => General Board => Homework Haven => Topic started by: Miser on September 18, 2009, 12:00:16 PM
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Well, my first quiz in this subject is Monday and I'm having trouble with two types of problems. I'm hoping that some of the mathematicians here could help me a bit :)
1) Find the point in which the line with parametric equations X=2 - t, Y=1 + 3t, Z= 4t intersects the plane 2X - Y + Z = 2.
I've checked my book and I can only find how to find the point where two lines intersect. If anyone has any suggestions, I'd like to hear them. Thanks all :)
2) Find the distance from the origin to the line X = 1+t, Y = 2-t, Z = -1+2t
Not really even sure how to approach this one. Thought about trying to find the reciprocal of the slope then basing that on the point (0,0,0) but I have no idea if that's even close.
Thanks everyone, I appreciate your help.
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For the first problem, you have a line with a parametric equation, which essentially states that for any value of t, x=2-t, y=1+3t, and z=4t. So with this, you have values for x, y, and z all as a function of one variable, which you can use in the equation for the plane to find the value of t at which both the equations for the plane and the line are satisfied.
For the second way, there are different routes you could take. Being more of a Calculus person, I'd personally plug in the values for x, y, and z (the equations that is) into the equation for calculating distance (root of the sum of the squares), then find the point at which the distance function is minimized using derivatives.
Sorry if I gave too much away on how to solve them; it's hard not to do so when answering in this medium.
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Kool, i think i get it... lol
I'll have to play around with those methods but I follow what your saying :)
Thanks for the help :)
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If you solve the equation of the plane for x, y, and z in terms of t, that will get you the equation of the normal line to the plane. That may help with vizualizing these problems.
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Miser, if you want you can post your answers for the two problems, and I can work them out and tell you whether they are correct.
Have you learned about the cross product yet?
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Ok, well I tried both problems and I was able to solve them. I have an answer sheet so I'm fairly certain that I got both right :)
1. Plugged in the line equation into the plane equation, solved for T and put that value back into the line function to find the point.
2. I found an equation for the distance from the line to the origin and using my new spiffy Ti-89, I found the derivative, found the zero and plugged that in to the original equation and finally solving for the distance from that point to the origin.
Thanks for the help everyone. :)
Miser, if you want you can post your answers for the two problems, and I can work them out and tell you whether they are correct.
Have you learned about the cross product yet?
Is there a way to do one of these with the cross product?
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Is there a way to do one of these with the cross product?
Yes, you can take the magnitude of the cross product of the direction vector for the line and the vector from any point on the line to the origin, then divide that by the magnitude of the direction vector to get your final answer. To solve your problem using this method:
Given:
x = 1 + t
y = 2 - t
z = -1 + 2t
The direction vector is <1, -1, 2>. (The numbers are the coefficient of the t's from x,y, and z.)
Now let P be any point on the line and Q(0,0,0) be the origin. Since P can be any point on the line, we should make it easy on ourselves and let t=0, and we get the point P(1,2,-1). It follows that the vector PQ is <-1, -2, 1>. (Just subtract P from Q)
Now, you take the cross product of vector PQ and the direction vector. You should get <-3, 3, 3>.
To get your answer, you take the magnitude of <-3,3,3> and divide it by the magnitude for the direction vector <1,-1,2>. The answer should sqrt(9/2) or 2.12132.
Why does this method work?
Geometrically, the magnitude of the cross product of two vectors will give you the area of the parallelogram. The shortest distance from any point to a line is basically the height of the parallelogram, which is what you are trying to find. The area of a parallelogram is A=bh. The length of the base is just the magnitude of the direction vector. That's why you can get the height (the shortest distance) by dividing the area of the parallelogram by the its base. Don't mean to make it sound very easy, but my university hired me as a Peer Tutor and Mentor, so I have a habit of making things as easy as I can. :P Now make up a few random numbers and practice so you can ace the exam :)
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Is there a way to do one of these with the cross product?
Yes, you can take the magnitude of the cross product of the direction vector for the line and the vector from any point on the line to the origin, then divide that by the magnitude of the direction vector to get your final answer. To solve your problem using this method:
Given:
x = 1 + t
y = 2 - t
z = -1 + 2t
The direction vector is <1, -1, 2>. (The numbers are the coefficient of the t's from x,y, and z.)
Now let P be any point on the line and Q(0,0,0) be the origin. Since P can be any point on the line, we should make it easy on ourselves and let t=0, and we get the point P(1,2,-1). It follows that the vector PQ is <-1, -2, 1>. (Just subtract P from Q)
Now, you take the cross product of vector PQ and the direction vector. You should get <-3, 3, 3>.
To get your answer, you take the magnitude of <-3,3,3> and divide it by the magnitude for the direction vector <1,-1,2>. The answer should sqrt(9/2) or 2.12132.
Why does this method work?
Geometrically, the magnitude of the cross product of two vectors will give you the area of the parallelogram. The shortest distance from any point to a line is basically the height of the parallelogram, which is what you are trying to find. The area of a parallelogram is A=bh. The length of the base is just the magnitude of the direction vector. That's why you can get the height (the shortest distance) by dividing the area of the parallelogram by the its base. Don't mean to make it sound very easy, but my university hired me as a Peer Tutor and Mentor, so I have a habit of making things as easy as I can. :P Now make up a few random numbers and practice so you can ace the exam :)
Wow thanks »X«, thats the exact answer i got when using edisons method. We just learned about the cross product so I think my professor would rather me use your method.
Thanks for the help guys, the way you explained it is so much easier to understand than my book or professor did.
:neckbeard: