MedVision ad

math 1002: linear algebra (1 Viewer)

Wyvern

Member
Joined
Feb 3, 2004
Messages
232
Gender
Male
HSC
2002
Originally posted by JUB JUB
use ur answers from 2ii...think...units squared and triangles....hmmmm
im not even sure if my 2ii is correct
 

JUB JUB

Member
Joined
Apr 20, 2003
Messages
154
Gender
Female
HSC
2012
well ur answers should be opposite ...that is if mine's correct....argh just wait till u get to question 2!
 

Saul

*ss
Joined
Jul 25, 2003
Messages
340
Gender
Undisclosed
HSC
2003
ahem

AP^2 = (a^2 + b^2 - 2 a.b)/4

CP^2 = (a^2 + b^2 +2 a.b)/4

if its a right triangle, a.b goes to zero, and AP^2= CP^2 = BP^2 (BP cause AP = BP) therefore all the vertices are equidistant from the midpoint of the hypotenuse.
 

Wyvern

Member
Joined
Feb 3, 2004
Messages
232
Gender
Male
HSC
2002
does anyone still need help with ii) c)? I think i got the right answer
 

JUB JUB

Member
Joined
Apr 20, 2003
Messages
154
Gender
Female
HSC
2012
nooooooooooo 2iv....come on ppl

omg i have finally reached 101 posts.....how embarassing
 
Last edited:

Saul

*ss
Joined
Jul 25, 2003
Messages
340
Gender
Undisclosed
HSC
2003
for 2) iv) just find the plane that goes through C perpendicular to 'l' and equate it with the plane p, and you'll get a line of solutions
 

JUB JUB

Member
Joined
Apr 20, 2003
Messages
154
Gender
Female
HSC
2012
Originally posted by Saul
ahem

AP^2 = (a^2 + b^2 - 2 a.b)/4

CP^2 = (a^2 + b^2 +2 a.b)/4

if its a right triangle, a.b goes to zero, and AP^2= CP^2 = BP^2 (BP cause AP = BP) therefore all the vertices are equidistant from the midpoint of the hypotenuse.
can u do 2 iv? or tell me how
 

JUB JUB

Member
Joined
Apr 20, 2003
Messages
154
Gender
Female
HSC
2012
Originally posted by Saul
for 2) iv) just find the plane that goes through C perpendicular to 'l' and equate it with the plane p, and you'll get a line of solutions
yeh how do i do that? gimme something to start with
=)
 

Saul

*ss
Joined
Jul 25, 2003
Messages
340
Gender
Undisclosed
HSC
2003
to find the plane, all you need is a point on the plane and a vector perpendicular
any line on l is perpendicular, so take AB or AC or some such thing

edit: use C as the point on the plane
 

Carlito

Member
Joined
Feb 4, 2004
Messages
630
Location
Sydney
Gender
Male
HSC
2003
is 3x +3y +2z = -3 the plane?

im thinking its totally wrong
 
Last edited:

Carlito

Member
Joined
Feb 4, 2004
Messages
630
Location
Sydney
Gender
Male
HSC
2003
Originally posted by Carlito
pls pls pls i need help

Can anyone help me with parts (b) and (c) of this question:



and parts (iii) and (iv) of this question:
 

JUB JUB

Member
Joined
Apr 20, 2003
Messages
154
Gender
Female
HSC
2012
Originally posted by Wyvern
particularly 2 iii) and 2 iv...........i mean WTF
i concur

why is there silence....noooooo
 
Last edited:

JUB JUB

Member
Joined
Apr 20, 2003
Messages
154
Gender
Female
HSC
2012
i tried to find a plane with c but it looks so wrong....what eqaution do i use with it exactly? i'm tired
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top