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Problem involving the equation of a plane

  1. Dec 21, 2009 #1
    vectors question: finding the equation of the plane and the distance to it from the origin..

    The vertices of a quadrelateral OAB and C are (0,0,0) (1,4,1) (3,3,0) and (-1,1,4) respectively.

    forces of magnitudes 3,2 and 4 newtons act along OA OB and OC respectively.

    (a) Express each force as a vector and find the resultant force.

    (b) find the equation of the plane containing A,B and C and show the distance from the origin to the plane is [itex]\frac{15}{\sqrt{29}}[/itex]
    -----------------------------------------------------------------------
    My attempt:

    (a)
    find the unit vectors in the direction of each of the forces:
    [itex]F_1 (OA) = (\frac{1}{\sqrt{18}})(1,4,1)[/itex]

    [itex]F_2 (OB) = (\frac{1}{\sqrt{18}})(3,3,0)[/itex]

    [itex]F_3 (OC) = (\frac{1}{\sqrt{18}})(-1,1,4)[/itex]

    multiply the unit vectors by the magnitude of the respective force:

    [itex]F_1 = (\frac{3}{\sqrt{18}}) (1,4,1)[/itex]

    [itex]F_2 = (\frac{2}{\sqrt{18}}) (3,3,0)[/itex]

    [itex]F_3 = (\frac{4}{\sqrt{18}})(-1,1,4)[/itex]

    Add the forces together to find the resultant:

    [itex]F_r = F_1 + F_2 + F_3[/itex]

    [itex](\frac{1}{\sqrt{18}})(i(3+6-4)+j(12+6+4)+k(3+16))[/itex]

    =[itex]\frac{1}{\sqrt{18}} (5i + 22j + 19k) N[/itex]



    (b)

    cross product 2 of the points to get the normal vector:

    [itex](1,4,1)\times(3,3,0) = n[/itex]

    =[itex]\left[\begin{array}{ccc}i&j&k\\1&4&1\\3&3&0\end{array} \right][/itex]

    [itex]= i \left[\begin{array}{ccc}4&1\\3&0\end{array} \right] - j \left[\begin{array}{ccc}1&1\\3&0\end{array} \right] + k \left[\begin{array}{ccc}1&4\\3&3\end{array} \right][/itex]

    [itex]n = -3i + 3j + 9k[/itex]


    equation of a plane:

    [itex](r-a).n[/itex]

    [itex]-3x +3y +9z = (1\times -3) + (4\times 3) + (1\times 9) = 18[/itex]

    i have gone wrong somewhere because this lot should be equal to 15 i think, anyone know where i screwed up?
     
  2. jcsd
  3. Dec 21, 2009 #2
    Here is the problem

    [itex]n = -3i + 3j + 9k[/itex]

    it must be [itex]-9k[/itex] Not [itex]+9k[/itex]
     
  4. Dec 22, 2009 #3
    i see, just an error on my part, even so the equation for the plane should then look like this:

    [itex]-3i+3j-9k=0[/itex]

    this seems to mean that the origin does not lie on the plane of the quadrelateral, so i should get the vectors of points B and C from point A and cross product them instead. that might work. If it does this question is badly worded.
     
  5. Dec 22, 2009 #4
    ok, so here it is again, instead this time i assumed that the origin does not lie on the plane of A,B and C. So i worked out the normal vector to the plane containing the points A,B and C.
    First find the position vector of [itex]\vec{AB}[/itex] and [itex]\vec{BC}[/itex]
    I have called them [itex]\underline{d}[/itex] and [itex]\underline{e}[/itex] respectively

    [itex]\underline{d}=\underline{b}-\underline{a}=2i-j-k[/itex]
    [itex]\underline{e}=\underline{c}-\underline{b}=-4i-2j+4k[/itex]

    next, i crossed these two vectors to get a normal, [itex]\underline{n}[/itex]

    [itex]\underline{n}=\underline{d}\times\underline{e}[/itex]

    [itex]\underline{n}=\left[\begin{array}{ccc}i&j&k\\2&-1&-1\\-4&-2&4\end{array}\right][/itex]

    [itex]\underline{n}=\underline{i}\left[\begin{array}{ccc}-1&-1\\-2&4\end{array}\right] -\underline{j}\left[\begin{array}{ccc}2&-1\\-4&4\end{array}\right] +\underline{k}\left[\begin{array}{ccc}2&-1\\-4&-2\end{array}\right][/itex]

    n = -6i - 4j - 8k

    now plug these numbers into the equation for a plane:

    [itex](\underline{r}-\underline{a}). \underline{n}=0[/itex]

    [itex]\underline{r}.\underline{n}=\underline{a}.\underline{n}[/itex]

    choosing the point A (1,4,1):

    [itex]-6x-4y-8z=-6-16-8=-30[/itex]

    bit of algebra:

    [itex]3x+2y+4z=15[/itex]

    find the distance from O to the plane:

    [itex]d=\underline{a}.\underline{\hat{n}}[/itex]

    [itex]\underline{a}.\underline{n}=15[/itex]

    [itex]\therefore \underline{a}.\underline{\hat{n}} =\frac{15}{\left|\underline{n}\right|}[/itex]

    [itex]d=\frac{15}{\sqrt{29}}[/itex]

    got there in the end, does this look right?
     
  6. Dec 22, 2009 #5
    If you're saying point A lies on the plane, then shouldn't the distance of the plane from the origin simply be the magnitude of the line segment OA be sufficient to find that distance? I'm not entirely sure what you've done to find the distance.
     
  7. Dec 22, 2009 #6
    to be honest i'm just doing what it says in my notes to get the distance, it seems to work but i have no idea what it means. i have this equation written down:

    [itex]d=\underline{a}.\underline{\hat{n}}[/itex]

    where d is the distance
    a is a point that lies on the plane
    [itex]\underline{\hat{n}}[/itex] is the unit vector in the direction of the normal to the plane.

    i can plug the numbers in fine but i have no idea what this means in the real world.
     
  8. Dec 22, 2009 #7
    I'm not sure if I quite understand how taking the point a and using the inner product with the unit vector of the normal vector of your plane will give you the distance from the origin.

    In my head this is what seems to make the most sense:

    You know that the points A, B, and C lie on your plane. (In fact, they'd better be on the plane because those are the points we used to define said plane!)

    So if you want to find the distance from the origin to your plane why not define a line segement running from the origin to one of the points on the plane and simply solve the magnitude of that line segment.

    Magnitude of the line segements: OA, OB, and OC are all going to be the same because they all lie in the same plane, a quick mental calculation shows this, because you're going to end up with magnitudes of sqr(18) for all of the line segements.

    So if I'm not mistake the distance from the origin to your plane is simply,

    |OA| = |OB| = |OC| = sqr(18) = 3sqr(2)

    Edit: Answers aside,

    This is something you might want to reconsider. If you want to excel through the course and not save yourself alot of extra work in the future, its worth the time it takes now to sit down and hammer this stuff into your head until you UNDERSTAND. Mathematics isn't about memorizing relationships and formulas to get the answer, its about understanding them and knowing why they exist in the first place!
     
  9. Dec 22, 2009 #8
    yeah i know what you mean ihave learned this part backwards. I understand other concepts in vectors such as dot and cross product and what they mean in a system of vectors but i just cant get my head around where the distance from the origin comes into it and how you get it from this formula. i know my answer is right because it was a 'show that' style question and i used the method given to me to solve it. I thik i should ask my lecturer to explain it to me again when I get back to uni.
     
  10. Dec 22, 2009 #9
    A problem that my professor pointed out to me when I had taken linear algebra was that often students only consider the algebraic significance of vectors when in reality they posses excellent geometrical significance! If you take the time to understand the geometrical significance of things, it tends to make the algebraic process flow more smoothly. Why? Because you actually know what's going on and you can see it geometrically!

    If someone could explain to me how the formula he stated above generates the distance from the origin to the plane please by all means.

    The unit vector of the normal vector of the plane is going to allow us to preserve the direction of the norm of this plane but how would taking the dot product with a point on the plane (namely, a) give us the distance. To me, that doesn't seem to make much sense.
     
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