Question

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?`

Answer 1

See the Proof given in Explanation Section.

Explanation:

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!

Answer 2

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.