Question

Someone pls answer these questions as quick as possible.

WARNING ⚠️ = No Silly Answers. Don’t answer if you’re not sure too.
THANKS

Answer 1

Let's do the unit cube and worry about the side length later.

Let's assign F(0,0,0), H(1,1,0), A(0,1,1), C(1,0,1)

Let's call the beetle positions P and Q, functions of time t.

P = F + 2t(H-F) = tH = 2t(1,1,0) = (2t,2t,0)

Q = A + t(C-A) = (0,1,1)+t(1,-1,0)=(t, 1-t, 1)

The squared distance is

PQ^2 = (2t - t)^2 + (2t -(1-t))^2 + 1 = t^2 + (3t-1)^2 + 1 = 10t^2-6t+2

\dfrac{d\ PQ^2}{dt} = 20t - 6 = 0

t = 6/20 = 3/10

PQ^2 = 10(3/10)^2 - 6(3/10) + 2 = 11/10

\sqrt{11/10} \times 40 \sqrt{110} = 440

Answer: 440

Answer 2

Let's do the unit cube and worry about the side length later.

Let's assign F(0,0,0), H(1,1,0), A(0,1,1), C(1,0,1)

Let's call the beetle positions P and Q, functions of time t.

P = F + 2t(H-F) = tH = 2t(1,1,0) = (2t,2t,0)

Q = A + t(C-A) = (0,1,1)+t(1,-1,0)=(t, 1-t, 1)

The squared distance is

`PQ^2 = (2t - t)^2 + (2t -(1-t))^2 + 1 = t^2 + (3t-1)^2 + 1 = 10t^2-6t+2`

`(d\ PQ^2)/(dt) = 20t - 6 = 0`

`t = 6/20 = 3/10`

`PQ^2 = 10(3/10)^2 - 6(3/10) + 2 = 11/10`

`√(11/10) * 40 √(110) = 440`

Answer: 440