Vectors help - Slicing Corner problem

[man0005]

Homework Statement

On sheet

Homework Equations

The Attempt at a Solution

I have no idea...i know you guys arent allowed to help if i don't show some working out, but i seriously have no clue how to even start this question :|
please help me

 

[tiny-tim]

hi man0005!

tell us what formulas you know for the area of a triangle (with or without cross product )

 

[HallsofIvy]

Since there is a reference to using a cross product, can we presume that you know that the area of a parallelogram, having $\vec{u}$ and $\vec{v}$ as adjacent sides, has area $|\vec{u}\times\vec{v}|$? And, of course, a triangle is half a parallelogram.

 

[man0005]

okay I've done that
how would i go about solving the area of D now?

 

[tiny-tim]

done what?

show us! ​

 

[man0005]

lol sorry
ive worked out that the areas of the three right angled faces are
(1/2)sqr root(a^2b^2)
(1/2)sqr root(a^2c^2)
(1/2)sqr root(b^2c^2)

now how do i go about finding D
in order to proof the equation?

what I've tried is to find the length of each side of the triangle and then the height
but I keep getting messy results with lots of square roots :/

 

[tiny-tim]

hi man0005!

(have a square-root: √ and try using the X² icon just above the Reply box )

(1/2)sqr root(a^2b^2)
(1/2)sqr root(a^2c^2)
(1/2)sqr root(b^2c^2)

erm …

isn't that just ab/2, ac/2, bc/2 ? ​

ok, now use the cross product method to find area D …

what do you get? ​

 

[man0005]

oh yeah! sorry lol

umm how would i go about doing that? which points do i use? :S

 

[tiny-tim]

uhh?

let's go back to my original question …

tell us what formulas you know for the area of a triangle (with or without cross product )

 

[man0005]

i only know 1/2bh
but using that for this would be too messy yeah?

 

[tiny-tim]

i only know 1/2bh
but using that for this would be too messy yeah?

it won't work at all …

it only works for right-angled triangles

(but i don't understand … where did you get those square-roots from, if that's the only formula you know? )

you need to go back to your book or lecture notes (or the internet), and find a general formula for the area of a triangle

 

[man0005]

Is this right for Area D?

i made the line from 0,b,0 to a,0,0 as AB
and the line from 0,b,0 to 0,0,c as AC

so AB = (-a, b, 0)
AC = (0, b , -c)
then using cross product
= (-bc, -ac, -ab)

so area = 1/2 √ (b²c² + a²c²+ a²b²)

 

[tiny-tim]

Is this right for Area D?

i made the line from 0,b,0 to a,0,0 as AB
and the line from 0,b,0 to 0,0,c as AC

so AB = (-a, b, 0)
AC = (0, b , -c)
then using cross product
= (-bc, -ac, -ab)

so area = 1/2 √ (b²c² + a²c²+ a²b²)

ohh, so you did know the cross product formula?

yes, that's correct … that finishes the question, doesn't it?

 

[man0005]

LOL nah <3 google
YES I GOT IT ty ty ty

now for b) :P
is the answer triangle?

 

[tiny-tim]

now for b) :P
is the answer triangle?

is there a word shortage where you are?

well, i think you have the right answer, but based on just one word, it's a little difficult to tell!

 

[man0005]

hmm what do you mean?
should i expand?

plane counterpart of a slicing corner would be a triangle? or triangular prism?
whats a slicing plane?

 

[tiny-tim]

hmm what do you mean?
should i expand?

for 5 marks :tongue2: ? YES!

anyway, part (b) is asking for an equation, isn't it?

 

[man0005]

whatt how do you know? D:
I thought i just had to complete the sentences...

sigh I am sorry if I am asking too many questions

 

[tiny-tim]

whatt how do you know? D:

because the question starts "What is the 2-dimensional counterpart of this 3-dimensional result?" …

that's what most of the 5 marks will be for, and it's asking for an equation

 

[man0005]

so i'd just list all the components of the triangular prism?

 

[man0005]

okay
so would i say:
cube - square
sliced corner - triangle
sliced plane - line?

then find equation of line?

 

[tiny-tim]

then find equation of line?

you need to find an equation equivalent to the original equation for areas

 

[man0005]

hmmmm any hints? :P

 

[man0005]

hmm does this work?

if i split the 3 dimensional slice
into four 2-dimensional parts
the 3 right angle faces and the other one?
then use Area =1/2bh for each one?

 

[tiny-tim]

okay
so would i say:
cube - square
sliced corner - triangle
sliced plane - line?

yes …

next, the 3D equation was about a sum of areas squared …

so what would be 2D equivalent of that be?

 

[man0005]

A + B + C = D?

can i just state that or do i need to show it as well?

 

[tiny-tim]

A + B + C = D?

can i just state that or do i need to show it as well?

you need to explain it as well …

i have no idea what you mean …

what are A B C and D?​

 

[man0005]

since the 3D equation is A²+B²+C²= D²
then 2d equation is A + B + C = D?
isnt that what you meant?

 

[tiny-tim]

since the 3D equation is A²+B²+C²= D²
then 2d equation is A + B + C = D?
isnt that what you meant?

D'oh!

did you read my question …

what are A B C and D?

 

[man0005]

oh you mean the actual values?
A= ab/2
B= ac/2
C = bc/2

D = 1/2(ab+bc+ac)?

 

[tiny-tim]

oh you mean the actual values?
A= ab/2
B= ac/2
C = bc/2

D = 1/2(ab+bc+ac)?

Now I'm completely lost …

what parts of the square are A B C D a b and c ?

 

[man0005]

what square?
Im talking about the sliced part...
i'm confused now lol..

 

[tiny-tim]

your square …

cube - square
sliced corner - triangle
sliced plane - line?

 

[man0005]

yes …

next, the 3D equation was about a sum of areas squared …

so what would be 2D equivalent of that be?

how do i do this?