Recent content by brotherbobby

I am the OP. Please refer to my post #16 above. I will work out the details if you need me to.

The symbols ##\lambda## and ##\mu## are fractions. Thus, what I mean is, ##\vec {AP} = \mu \vec{AE}##, for instance, or that ##\vec {BP} = \lambda\vec{BD}##. I have shown that ##\lambda = 1/3##. You are correct in that I didn't show the same for the rest of the diagonal BD. However, a similar...

Attempt : I attempt the problem making A as the origin, written with (O) in bold above A, in the diagram. Thus, ##\vec e = \dfrac{\vec b+\vec c}{2}\Rightarrow \vec{AP} = \dfrac{\mu}{2}(\vec b+\vec c)##. The vector ##\vec {AB} = \vec b## and the vector ##\vec {BP} = \lambda(\vec d-\vec b)##...

I have already done it that way, integrate with respect to ##x## and faced no problems. (Please see at the beginning of my post #1, where I start attempting a solution to the problem) Question is, what happens when you integrate with respect to ##y## and maintain your direction of motion along...

I agree with you, but there's a basic discrepancy that bothers me here. If you want to find out the arc length ##s## integrating with respect to ##x##, you run ##x## from ##x = 0\rightarrow a##, you are going clockwise. If you continue running clockwise but integrate with respect to ##y##...

Statement of the problem : Calculate the arc length ##s## of the quadrant in the figure shown alongside. Attempt : (I have been able to solve the problem integrating with respect to ##x##. However, when I integrate relative to ##y##, I get a minus ##(-)## sign. Where I am going wrong? I furnish...

Of course not. Indeed I could have chosen a point on the parallelogram like A as you suggested. I would do it presently, but first, there's something important to ask you. Yes, your second option is indeed the one I have in mind, but the first, which was mine, should work too. How do you think...

I choose an arbitrary origin O and draw the relavant vectors ##\vec a, \vec b, \vec c## by hand to the three vertices of the parallelogram A, B and C. Clearly, the vector ##\vec a\nparallel b## and therefore the vector ##\vec c## can be written as a linear combination of them : ##\vec c = m\vec...

Look at the coefficients of the vector ##\vec a## in his two equations. He (@BvU) is right and it was me who was wrong. It gives me hope that my answers are correct making my second post titled "changes" a second mistake on my part. Indeed, the diagonal BD and the lines AE and AF trisect each...

Changes : The three equations involving the coefficients ##\lambda,\mu##above, assuming linear independence of vectors ##\vec a, \vec b, \vec c## (see equation 5 above), are ##\lambda-\mu=1,\quad 2\lambda+\mu/2=1,\quad \mu=2\lambda##. The last two of these reduce to ##\lambda = 1/3, \mu=2/3##...

Statement of the problem : I copy and paste the statement of the problem as it appears in the text to the right. Additionally, I prove a statement related to it. [Davis, V.T. (2026). Vectors in Physics, p. 33. Springer] Attempt : I begin by drawing the image of the problem to the right...

Yes I was aware of this, but didn't succed when I tried to do the same problem your way. Let me show you the little I could do. Problem Statement : Solve for ##\boldsymbol{x}## : ##\quad\boldsymbol{\sqrt{3}\cos x+\sin x = \sqrt 2}##. Given : ##\cos x = \dfrac{e^{ix}+e^{-ix}}{2}## and...

Thank you all, especially at @haruspex and @pasmith. I have solved the problem. It involved really a mistake that I made in my calculations.

Problem statement : I copy and paste the problem as it appeared in the text to the right. Attempt : ##\quad \begin{matrix} \sqrt{3}\cos x +\sin x &=& \sqrt{2}\\ \Rightarrow\dfrac{\sqrt{3}}{2}\cos x + \dfrac{1}{2}\sin x &=& \dfrac{1}{\sqrt{2}}\\...

I have drawn a figure of what I meant to the right. A ball is projected vertically from a point A on the ground with a velocity ##v_0##. It travels to a maximum point B at a height ##h## from the ground and falls back to the ground. Imagine a point C intermediate between A and B at a height of...