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

2 Answers
Jul 27, 2016

See the Proof given in Explanation Section.

Explanation:

Let vecA=(l,1,0). vecB=(0,m,1) and vecC=(1,0,n)A=(l,1,0).B=(0,m,1)andC=(1,0,n)

We are given that vecAxxvecB, and, vecBxxvecCA×B,and,B×C are parallel.

We know, from Vector Geometry, that

vecxx || vecy iff (vecx)xx(vecy)=vec0y(x)×(y)=0

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!

Jul 28, 2016

L M N + 1 = 0

Explanation:

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.