Vector Equations of Curves (2 of 3: Midpoint of a chord)

[@aashsyed1277]

barely no one teaches vectors.... BUT eddie is to the rescue!

[@Mixingmudit]

I am happy for giving me such lectures Actually i left maths because i thought it was so much burden but still i love math and study it because of you.

[@chandonroy7442]

Sir i am a student of mathematics.I am from Bangladesh.Few days ago i saw your first video.It was a very good explanation of that math.It was very helpful for me.I am a big fan of you.My dream is also to be a Mathematician.

[@llunaecy]

At around 6:30, you shouldn't be able to just write lambda = -b/2a as the next line immediately, since it's not really giving a solution to the previous line's quadratic. Should be fine if you clarify further what you're doing, like saying "taking the average of the roots" and using a different name for the variable or writing it as (lambda_1 + lambda_2)/2 though, right?

[@bookpattarapong4596]

Thank you very much for coming out and explaining my doubts as well.

[@SoloRenegade]

Beginning: what's going on? Middle: oh.. End: OH! That's cool!

[@magician4058]

For anyone wondering why the axis of symmetry was found instead of the roots of lambda, its because the question wants the midpoint of chord, so this way its a bit faster to get to the vector we want. (I think)

[@AriaHarmony]

My first instinct was to think of the circle as: x=1+3cos(theta), y=-2+sin(theta) and the line as: x=2+lambda, y=1+2lambda Then set them equal and rearrange so cos and sin on one side and everything on the other. Then square both sides, and substitute into cos^2(theta)+sin^2(theta)=1 and solve the resulting quadratic. I actually did the whole calculation with the radical and everything.. and got the same answer (0.6, -1.8) in the end, it felt wrong to substitute lambda (-0.07335...) and (-2.7266499...), and then you just skipped all that decimal work using logic. Just goes to show work smarter not harder lol

[@tasbeats2135]

Hey do you do tutoring ???

[@semiraf]

👍

[@gjermundification]

Now put the circle or make it an eliptical for the fun of it, and have a calculator rotate this around itself without making the calculator heat up. Did this for my HP48GX back in the day; '95/'96 or so, and the screen started fading away after about 15 mins.

[@mathswithmohantysir12]

👍

[@mathswithmohantysir12]

❤️