Vector Proofs Help Please

Question

Please help!!

I'm currently studying vector proofs and have no idea how I should even go about proving these questions. My teachers give little explanation and I cannot contact them as they are away. I apologise in advance for any incorrect formatting in my equations -- I've never written a post on here before. Any advice would be GREATLY appreciated, but please may I request that explanations are written in such a way that a numpty like me could understand. Nothing overly-complicated please.

QUESTION 1:

Consider a circle with centre O and radius of OA=a. B and C are points on the circle with BC=b. Prove that the angle ABC subtended by a diameter of the circle is a right angle. They provide us with this diagram:

The only proofs I've seen so far that show the angle is 90 degrees initially define OA, OB, and OC as vectors a, b, and c, and find BC and BA in terms of these. They then show that the dot product of BC and BA is zero -- makes sense.

BUT this question has defined the BC as b, which I initially didn't think would be too much of a problem. I found the OB in terms of OC and BC (i.e. OB = -a-b) and found the BA in terms of OA and OB (i.e. BA = -OB + a = (a + b) + a = 2a + b). I then tried to prove that BC dotted with BA = 0. BUT this is where my difficulties appeared. I figured:

b . (2a + b)

= b . (2a) + b . b

= 2(a.b) + |b|^2

but I could not figure out how to continue so as to prove that the dot product is zero. Maybe I've messed up in my method?

---

QUESTION 2: (please note this is entirely unrelated to question 1)

ABCD is a rectangle where point O is an origin not on the same plane as the rectangle. Prove that OA . OC = OB . OD

I haven't got a clue where to start with this one.

Any help given would be MAJORLY appreciated -- but please may I request once more that any help is explained in such a way that a student like me could understand (I'm not as good as you guys with your fancy proofs lol)

Thank you in advance.

Answer 1

1. Hint: $|OB| = |OA|$ since they are radii of the same circle, so $|\vec{a}| = |-\vec{a}-\vec{b}|$. Square this to get $$\vec{a}\cdot\vec{a} = (\vec{a}+\vec{b})\cdot(\vec{a}+\vec{b}).$$ Then expand the RHS and cancel $\vec{a}\cdot\vec{a}$ on both sides to recover the expression that you've already got for $\vec{BA} \cdot \vec{BC}$.
2. Draw a diagram. Let $\vec{AB}=\vec\ell$ and $\vec{AD}=\vec{w}$. Then \begin{cases} \vec{OA}&=\vec{a} \\ \vec{OB}&=\vec{a}+\vec\ell \\ \vec{OC}&=\vec{a}+\vec\ell+\vec{w} \\ \vec{OD}&=\vec{a}+{w}. \end{cases} Since $ABCD$ is a rectangle, $\vec\ell \cdot \vec{w} = 0$. The rest of the proof should be straightforward, and I left this as exercise.