3D vectors (1 Viewer)

[hanod]

Hi can someone help me solve the attached question

 

[unofficiallyred12]

u.v/|u||v| = v.w/|w||v| = u.w/|u||w| = ½

w = (a, 0, c)

v = (0, b, c)

u = (a, b, 0)
using these and the first line, we get:

b^2/sqrt(a^2+b^2)sqrt(b^2+c^2) = ½

b^2 = ½sqrt(a^2+b^2)sqrt(b^2+c^2)=1/2(sqrt4b^4)

c^2 = 1/2sqrt(c^2+a^2)sqrt(c^2+b^2) = ½(sqrt4c^4)

a^2 = 1/2sqrt(a^2+b^2)sqrt(a^2+c^2)

= ½(sqrt(4a^4))


1.(a^2+b^2)(a^2+c^2) = 4a^4

2.(a^2+c^2)(b^2+c^2) = 4c^4

3.(a^2+b^2)(b^2+c^2) = 4b^4



taking 1-2, we get A^4 +a^2b^2 + c^2b^2 + c^2a^2 – a^2b^2 – c^2b^2- c^2a^2-c^4 = 4(a^4-c^4)



A^4 – c^4 = 4(a^4-c^4)

So a = c

Then it’s easy to prove b =c

 

[unofficiallyred12]

typing this on word is a bit messy so lemme know if i made a mistake!

 

[s97127]

My solution

 

[s97127]

u.v/|u||v| = v.w/|w||v| = u.w/|u||w| = ½

w = (a, 0, c)

v = (0, b, c)

u = (a, b, 0)
using these and the first line, we get:

b^2/sqrt(a^2+b^2)sqrt(b^2+c^2) = ½

b^2 = ½sqrt(a^2+b^2)sqrt(b^2+c^2)=1/2(sqrt4b^4)

c^2 = 1/2sqrt(c^2+a^2)sqrt(c^2+b^2) = ½(sqrt4c^4)

a^2 = 1/2sqrt(a^2+b^2)sqrt(a^2+c^2)

= ½(sqrt(4a^4))


1.(a^2+b^2)(a^2+c^2) = 4a^4

2.(a^2+c^2)(b^2+c^2) = 4c^4

3.(a^2+b^2)(b^2+c^2) = 4b^4



taking 1-2, we get A^4 +a^2b^2 + c^2b^2 + c^2a^2 – a^2b^2 – c^2b^2- c^2a^2-c^4 = 4(a^4-c^4)



A^4 – c^4 = 4(a^4-c^4)

So a = c

Then it’s easy to prove b =c

I got a similar approach for the first part but for the second i use am-gm: a^2 +b^ >= 2ab

 

[s97127]

Does anyone where this problem is from? Is it from a math textbook?

 

[unofficiallyred12]

Does anyone where this problem is from? Is it from a math textbook?

looks like it ye

 

[hanod]

u.v/|u||v| = v.w/|w||v| = u.w/|u||w| = ½

w = (a, 0, c)

v = (0, b, c)

u = (a, b, 0)
using these and the first line, we get:

b^2/sqrt(a^2+b^2)sqrt(b^2+c^2) = ½

b^2 = ½sqrt(a^2+b^2)sqrt(b^2+c^2)=1/2(sqrt4b^4)

c^2 = 1/2sqrt(c^2+a^2)sqrt(c^2+b^2) = ½(sqrt4c^4)

a^2 = 1/2sqrt(a^2+b^2)sqrt(a^2+c^2)

= ½(sqrt(4a^4))


1.(a^2+b^2)(a^2+c^2) = 4a^4

2.(a^2+c^2)(b^2+c^2) = 4c^4

3.(a^2+b^2)(b^2+c^2) = 4b^4



taking 1-2, we get A^4 +a^2b^2 + c^2b^2 + c^2a^2 – a^2b^2 – c^2b^2- c^2a^2-c^4 = 4(a^4-c^4)



A^4 – c^4 = 4(a^4-c^4)

So a = c

Then it’s easy to prove b =c

Thanks heaps

 

[hanod]

My solution

Thanks heaps

 

[hanod]

Thanks heaps

Can I also trouble you with this question

 

[unofficiallyred12]

I assume you'd be fine with part a, and most of part b except perhaps the finding s part. For s, one way to approach it would be to let the vector = [x, y, z]. Then since it's perpendicular to the face PQR, it is perpendicular to both PQ and PR. We find both of these, then dot [x, y, z] with both PQ and PR and using this, we can find both y and z in terms of x. Then we sub y and z interms of x back into [x, y, z] thus getting x multiplied by a vector that should be in terms of p, q, r. We know s is a scalar multiple of this vector in terms of p, q, r and we also know that s has length = to area of triangle pqr. we can find this area using projections or dot product and 1/2absintheta, so we get s = Area x unit vector of the vector we found in terms of p, q, r and then we're done.