Recent content by wonder1

First distance vector (117,90,15) - (20,22,20) = (97,68,-5) Equation AB(t) = (20+97t, 22+68t, 20-5t) X=20+97t Y=22+68t Z=20-5t

Hi all, for the below questions i used the 3D distance formula $$\sqrt{((117t-20)^2+((117−20)^2+(90−22)^2+(15−20))}$$ $$d=(97)^2+(68)^2+(−5)^2$$ $$AB=118.566437$$ to give me an answer of 118.566437 i don't understand what it means by finding formulas for x(t), y(t), and z(t) is it...

;) ;) That's fantastic Thank you Mark. Yes, I can do to the other poles.

I am sorry mark but i am just stuck. I have found the length between the the SW ground corner to at the Pole top of the NE to be 228.03 thinking i could work out a perpendicular length but i don't think its right and i tried using dot-point product but its not making sense.

Hi Mark, Thank you for your help. I have found the coordinates of the top of poles, however I am having a hard time understanding what it means with finding the initial position C(t) of the camera at time t.. because when i substituted the values i got 80 and i know its wrong. If i knew the...

Hi all, i am having a problem with question 3, as its not clear if i should use the Z value for the camera as 15 or 25 m or... Could you suggest me. Cheers The goal of this project is to obtain some understanding of the camera’s motion in space. Also, based on the camera’s motion, we will...