Vectors #A = ( L, 1, 0 ), B = ( 0, M, 1 ) and C = ( 1, 0, N ). A X B and B X C# are parallel. How do you prove that #L M N + 1 = 0?#

Let #vecA=(l,1,0). vecB=(0,m,1) and vecC=(1,0,n)#

We are given that #vecAxxvecB, and, vecBxxvecC# are parallel.

We know, from Vector Geometry, that

#vecx# #||# #vecy iff (vecx)xx(vecy)=vec0#

Utilising this for our #||# vectors, we have,

#(vecAxxvecB) xx (vecBxxvecC)=vec0..................(1)#

Here, we need the following Vector Identity :

#vecu xx (vecv xx vecw)=(vecu*vecw)vecv-(vecu*vecv)vecw#

Applying this in #(1)#, we find,

#{(vecAxxvecB)*vecC}vecB-{(vecAxxvecB)*vecB}vecC=vec0...(2)#

Using #[..., ..., ...]# Box Notation for writing the Scalar Triple Product appearing as the first term in #(2)# above, and, noticing that the second term in #(2)# vanishes because of #vecA xx vecB bot vecB#, we have,

#[vecA, vecB, vecC]vecB=vec0#

#rArr [vecA, vecB, vecC]=0, or, vecB=vec0#

But, #vecB != vec0#, (even if m=0), so, we must have,

#[vecA, vecB, vecC]=0#

#rArr# #|(l,1,0),(0,m,1),(1,0,n)|=0#

#rArr l(mn-0)-1(0-1)+0=0#

#rArr lmn+1=0#

Q.E.D.

I enjoyed proving this. Didn't you?! Enjoy Maths!

#A X B = ( L, 1, 0 ) X ( 0, M, 1) =( 1, -L, L M)#

# B X C = ( 0, M, 1 ) X ( 1, 0, N )=( M N, 1, -M )#

These are parallel, and so, #A X B = k ( B X C)#, for any constant k.

Thus, #(1, -L, LM ) = k (M N, 1, -M)#

#k =1/( M N ) = -L#. So,

L M N + 1 = 0.