brotherbobby
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- Homework Statement
- Use vector methods to show that two lines from one vertex of a parallelogram to the midpoints of the two opposite sides trisect the diagonal. (Additionally, if I may add, show that the diagonal itself trisects both those lines.
- Relevant Equations
- 1. Position vector of a point Z exactly midway between two points X and Y with position vectors ##\vec x## and ##\vec y## is ##\vec z =\dfrac{\vec x+\vec y}{2}##.
2. The vector ##\vec {AB} = \vec b-\vec a##.
3. Triangle law of vectors : Three vectors, arranged head to tail making up a triangle, add up to zero. Thus ##\vec a+\vec b+\vec c=0##.
4. Linear indepencence of ##n## vectors : Vectors ##\vec{a}_1, \vec {a}_2, \ldots, \vec {a}_n## are said to be linearly independent (L.I) if ##\Sigma_{i=1}^{i=n} \lambda_i a_i = 0\Rightarrow \lambda_i's=0\;\forall i## .
[I could do the problem with ease. It is towards the end, when it comes to the linear independence (L.I.) of vectors that helped me solve it in the first place, that I ran into difficulty. Briefly, the position vectors ##\vec a, \vec b, \vec c## corresponding to the points A, B and C, are not LI, and yet, assuming (wrongly) that they are, helps me get the answer! I furnish the details below.]
Let the position vectors of each point in capitals be given by its equivalent in small letters, with respect to an arbitrary origin chosen anywhere say O. Thus, for instance, the position vector of A is ##\vec a##, etc. Hence, using (1) of the relevant equations (R.E.) above, I find that the position vector of E is ##\vec e = \dfrac{\vec b+\vec c}{2}##. Using (2) from the RE's above, the vector ##\vec{\text{AE}} = \vec e -\vec a = \dfrac{\vec b+\vec c}{2} -\vec a = \dfrac{\vec b+\vec c-2\vec a}{2}##. Thus the vector ##\vec{\text{AP}} = \dfrac{\mu}{2}(\vec b+\vec c-2\vec a)\qquad\mathbf{(1)}##
Likewise, the vector ##\vec{\text{BP}}= \lambda(\vec d-\vec b)\qquad\mathbf{(2)}##.
But since ##\vec{AB}+\vec{BP} = \vec{AP}\Rightarrow \vec b - \vec a + \lambda(\vec d-\vec b)=\dfrac{\mu}{2} (\vec b+\vec c-2\vec a)\qquad\mathbf{(3)}##, upon using (1) and (2) above.
The (position) vectors appearing in (3) are not all independent. Since ABCD is a parallelogram, ##\vec{AB} = \vec{DC}\Rightarrow \vec b-\vec a = \vec c-\vec d\Rightarrow \vec d = \vec a-\vec b+\vec c\qquad\mathbf{(4)}##.
Using (4), (3) simplifies to ##\vec b -\vec a+\lambda(a-2\vec b+\vec c)=\dfrac{\mu}{2}(\vec b+\vec c-2\vec a)\Rightarrow (\lambda-1-\mu)\vec a+(1-2\lambda-\mu/2)\vec b+(\lambda -\mu/2)\vec c=0\qquad\mathbf{(5)}##
If we assume the vectors ##\mathbf{\vec a, \vec b, \vec c}## to be L.I., their coefficients should be zero. Thus ##\lambda-\mu = 1\, , 2\lambda+\mu/2 = 1\; \text{and} \lambda = \mu/2\Rightarrow\mu = 2\lambda##. This yields ##\boxed{\lambda=\dfrac{1}{3}\;\text{and}\;\mu = \dfrac{2}{3}}##, which solves the problem.
The issue : The vectors ##\vec a, \vec b, \vec c## are not L.I. Since the vectors ##\vec a,\vec b## are not parallel, (and they cannot be, or we wouldn't have a parallelogram), a third vector ##\vec c## can be expressed as a linear combination of the two earlier vectors, viz. ##\vec c=\lambda a+\mu b##. Here ##\lambda,\mu\ne 0##, which contradicts the case of linear independence I assumed towards the end.
Request : A resolution of the "paradox" above. Far as I suspect, only two of the three vectors ##\vec a, \vec b, \vec c## can be assumed to be L.I. The point C anywhere and hence its position vector ##\vec c## can be different in each case, depending on where point C is. How do you express the vector ##\vec c## as a linear combination of the two other vectors to further reduce equation (5) above?
Many thanks.