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?

Jul 27, 2016

See the Proof given in Explanation Section.

Explanation:

Let $\vec{A} = \left(l , 1 , 0\right) . \vec{B} = \left(0 , m , 1\right) \mathmr{and} \vec{C} = \left(1 , 0 , n\right)$

We are given that $\vec{A} \times \vec{B} , \mathmr{and} , \vec{B} \times \vec{C}$ are parallel.

We know, from Vector Geometry, that

$\vec{x}$ $| |$ $\vec{y} \iff \left(\vec{x}\right) \times \left(\vec{y}\right) = \vec{0}$

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

$\left(\vec{A} \times \vec{B}\right) \times \left(\vec{B} \times \vec{C}\right) = \vec{0.} \ldots \ldots \ldots \ldots \ldots . . \left(1\right)$

Here, we need the following Vector Identity :

$\vec{u} \times \left(\vec{v} \times \vec{w}\right) = \left(\vec{u} \cdot \vec{w}\right) \vec{v} - \left(\vec{u} \cdot \vec{v}\right) \vec{w}$

Applying this in $\left(1\right)$, we find,

$\left\{\left(\vec{A} \times \vec{B}\right) \cdot \vec{C}\right\} \vec{B} - \left\{\left(\vec{A} \times \vec{B}\right) \cdot \vec{B}\right\} \vec{C} = \vec{0.} . . \left(2\right)$

Using $\left[\ldots , \ldots , \ldots\right]$ Box Notation for writing the Scalar Triple Product appearing as the first term in $\left(2\right)$ above, and, noticing that the second term in $\left(2\right)$ vanishes because of $\vec{A} \times \vec{B} \bot \vec{B}$, we have,

$\left[\vec{A} , \vec{B} , \vec{C}\right] \vec{B} = \vec{0}$

$\Rightarrow \left[\vec{A} , \vec{B} , \vec{C}\right] = 0 , \mathmr{and} , \vec{B} = \vec{0}$

But, $\vec{B} \ne \vec{0}$, (even if m=0), so, we must have,

$\left[\vec{A} , \vec{B} , \vec{C}\right] = 0$

$\Rightarrow$ $| \left(l , 1 , 0\right) , \left(0 , m , 1\right) , \left(1 , 0 , n\right) | = 0$

$\Rightarrow l \left(m n - 0\right) - 1 \left(0 - 1\right) + 0 = 0$

$\Rightarrow l m n + 1 = 0$

Q.E.D.

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

Jul 28, 2016

L M N + 1 = 0

Explanation:

$A X B = \left(L , 1 , 0\right) X \left(0 , M , 1\right) = \left(1 , - L , L M\right)$

$B X C = \left(0 , M , 1\right) X \left(1 , 0 , N\right) = \left(M N , 1 , - M\right)$

These are parallel, and so, $A X B = k \left(B X C\right)$, for any constant k.

Thus, $\left(1 , - L , L M\right) = k \left(M N , 1 , - M\right)$

$k = \frac{1}{M N} = - L$. So,

L M N + 1 = 0.