maths C2 URGENT

[halosh92]

plz can anyone solve this!!!!!!!!!!!!!! my head broke!!!!!!!!!!!


the fixed point A has coordinates (8,-6,5) and the variable point P has coordiantes ( t, t , 2t)
a) show that AP^2= 6t^2 - 24t + 125
b) hence find the value of "t" for which the distance AP is least
c) determine this least distance

[nid404]

Can u tell which paper this is from...cuz something seems wrong...
the middle term in AP² should be 12t and not -24t...
if i knw the paper i can cross check

[halosh92]

Can u tell which paper this is from...cuz something seems wrong...
the middle term in AP² should be 12t and not -24t...
if i knw the paper i can cross check

it is a pastpaper question but they put in our C2 skool textbook the year is not written but i think its pretty old.

[astarmathsandphysics]

Is f the question correct?

[halosh92]

Is f the question correct?

i rili have no idea

[nid404]

ok...do u have ans at the back of your book?

If yes...then i can proceed my way and check if the other ans are right

[astarmathsandphysics]

Should be home in 2 hours.

[Saladin]

ppl, this question is a screwed one.

A point cannot have 3 co-ordinates. hehe

[nid404]

ppl, this question is a screwed one.

A point cannot have 3 co-ordinates. hehe

well spotted 

This is completely nuts

[halosh92]

well spotted 

This is completely nuts

i totally agree ! i cant believe wat kind of people sit and make up such questions
yea i have the answers but there is nothing for the a) part :
b) t=2
c) square root 101 = 10.0

[Phosu]

oh yah i saw that question in the text book...didn't know how to solve so i skipped
but a point can have 3 coordinate, if you're considering the z axis, but then again we haven't been taught into such details
im sure its something simple, that we're just not getting it.

anyways i have one more question
could someone plz solve C2 question 8 Jun05,

[nid404]

assuming they are vectors A & P and using their first eqn...which i also think is wrong
U can solve the quadratic...
x=(-b +/- root of (b²-4ac))/ 2a
u get t as 2

AP²= 6(4)-24(2)+125=101
AP=root of 101= 10..

strange...

[astarmathsandphysics]

the fixed point A has coordinates (8,-6,5) and the variable point P has coordiantes ( t, t , 2t)
a) show that AP^2= 6t^2 - 24t + 125
b) hence find the value of "t" for which the distance AP is least
c) determine this least distance
a)AP=sqrt((t-8)^2+(t--6)^2+(5-2t)^)
=sqrt(t^2-16t+64+t^2+12t+36+25-20t+4t^2)=sqrt(6t^2-24t+125)
b)AP^2=6t^2-24t+125
diffferentiate and put=0 and solve
12t-24=0 so t=2
c)AP^2 =6*2^2-24*2+125=24-48+125=101
least distance=sqrt(101)

[astarmathsandphysics]

Which board is this? Not cie?

[halosh92]

the fixed point A has coordinates (8,-6,5) and the variable point P has coordiantes ( t, t , 2t)
a) show that AP^2= 6t^2 - 24t + 125
b) hence find the value of "t" for which the distance AP is least
c) determine this least distance
a)AP=sqrt((t-8)^2+(t--6)^2+(5-2t)^)
=sqrt(t^2-16t+64+t^2+12t+36+25-20t+4t^2)=sqrt(6t^2-24t+125)
b)AP^2=6t^2-24t+125
diffferentiate and put=0 and solve
12t-24=0 so t=2
c)AP^2 =6*2^2-24*2+125=24-48+125=101
least distance=sqrt(101)

thanks alooot nid and astar 
yea its is cie board!