Question

If l ( my + nz – lx ) = m ( nz + lx – my ) = n ( lx + my – nz ) , prove that $\dfrac{y+z-x}{l}=$ $\dfrac{z+x-y}{l}=$$\dfrac{x+y-z}{l}$.

Answer 1

Hint: In question, we have to prove that $\dfrac{y+z-x}{l}=$ $\dfrac{z+x-y}{l}=$$\dfrac{x+y-z}{l}$ and only know information in question is that l ( my + nz – lx ) = m ( nz + lx – my ) = n ( lx + my – nz ) . we will form given equation into required equation by substitution of other fraction and by dividing whole equation by lmn.

Complete step-by-step solution:
In vector, l, m, and n represent direction cosines of a vector along the x-axis, y-axis, and z-axis respectively, and the relation between direction cosines l, m, and n are represented as ${{l}^{2}}+{{m}^{2}}+{{n}^{2}}=1$.
Now, in question it is given that l , m and n are direction cosines of a vector along x – axis , y – axis and z – axis respectively and let x , and z be any point on vector plane , then there is relation l ( my + nz – lx ) = m ( nz + lx – my ) = n ( lx + my – nz ), then we have to prove that $\dfrac{y+z-x}{l}=$ $\dfrac{z+x-y}{l}=$$\dfrac{x+y-z}{l}$.
Now, we have
l ( my + nz – lx ) = m ( nz + lx – my ) = n ( lx + my – nz ) …………...…..( i )
Now, dividing equation ………..…...( i ) by lmn , we get
\[\dfrac{l\text{ }\left( \text{ }my\text{ }+\text{ }nz\text{ } -\text{ }lx\text{ } \right)\text{ }}{lmn}=\text{ }\dfrac{m\text{ }\left( \text{ }nz\text{ }+\text{ }lx\text{ } -\text{ }my\text{ } \right)\text{ }}{lmn}=\text{ }\dfrac{n\text{ }\left( \text{ }lx\text{ }+\text{ }my -\text{ }\text{ }nz\text{ } \right)}{lmn}\]
On solving, we get
\[\dfrac{\text{ }\left( \text{ }my\text{ }+\text{ }nz\text{ } - \text{ }lx\text{ } \right)\text{ }}{mn}=\text{ }\dfrac{\text{ }\left( \text{ }nz\text{ }+\text{ }lx\text{ } - \text{ }my\text{ }\right)\text{ }}{ln}=\text{ }\dfrac{\text{ }\left( \text{ }lx\text{ }+\text{ }my\text{ }- \text{ }nz\text{ } \right)}{lm}\]
As, all three fractions are same, which means all three fraction are of same ratio on simplification that is if we have $\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{e}{f}$ then $\dfrac{a+c}{b+d}=\dfrac{c+e}{d+f}=\dfrac{e+a}{f+b}$, so we can write equation ( ii ) as ,
\[\dfrac{\text{ }\left( \text{ }my\text{ }+\text{ }nz\text{ }-\text{ }lx\text{ } \right)\text{+}\left( \text{ }nz\text{ }+\text{ }lx\text{ }-\text{ }my\text{ } \right)\text{ }}{mn+nl}=\text{ }\dfrac{\text{ }\left( \text{ }nz\text{ }+\text{ }lx\text{ }- \text{ }my\text{ } \right)\text{+}\left( \text{ }lx\text{ }+\text{ }my\text{ }- \text{ }nz\text{ } \right)\text{ }}{ln+lm}=\text{ }\dfrac{\text{ }\left( \text{ }lx\text{ }+\text{ }my\text{ } -\text{ }nz\text{ } \right)+\left( \text{ }my\text{ }+\text{ }nz\text{ } -\text{ }lx\text{ } \right)}{lm+mn}\]
Taking common factors out, we get
\[\dfrac{\text{ }\left( \text{ }my\text{ }+\text{ }nz\text{ } -\text{ }lx\text{ } \right)\text{+}\left( \text{ }nz\text{ }+\text{ }lx\text{ } - \text{ }my\text{ } \right)\text{ }}{n(m+l)}=\text{ }\dfrac{\text{ }\left( \text{ }nz\text{ }+\text{ }lx\text{ } -\text{ }my\text{ } \right)\text{+}\left( \text{ }lx\text{ }+\text{ }my\text{ } \text{ }nz\text{ } \right)\text{ }}{l(n+m)}=\text{ }\dfrac{\text{ }\left( \text{ }lx\text{ }+\text{ }my\text{ } -\text{ }nz\text{ } \right)+\left( \text{ }my\text{ }+\text{ }nz\text{ } -\text{ }lx\text{ } \right)}{m(l+n)}\]
Solving brackets in numerator, we get
\[\dfrac{\text{ }\left( \text{ 2}nz \right)\text{ }}{n(m+l)}=\text{ }\dfrac{\text{ }\left( \text{ 2}lx \right)\text{ }}{l(n+m)}=\text{ }\dfrac{\text{ }\left( \text{2}my \right)}{m(l+n)}\]
On simplifying fractions, we get
\[\dfrac{\text{ }\left( \text{ 2}z \right)\text{ }}{(m+l)}=\text{ }\dfrac{\text{ }\left( \text{ 2}x \right)\text{ }}{(n+m)}=\text{ }\dfrac{\text{ }\left( \text{2}y \right)}{(l+n)}\]
Adding and subtracting the fractions we get,
\[\dfrac{\text{ }\left( \text{ 2}z \right)\text{+}\left( \text{2}y \right)-\left( \text{ 2}x \right)\text{ }}{(m+l)+(l+n)-\left( n+m \right)}=\text{ }\dfrac{\text{ }\left( 2z \right)+\left( \text{ 2}x \right)-\left( 2y \right)\text{ }}{(m+l)+(n+m)-(l+n)}=\text{ }\dfrac{\left( 2x \right)+\left( \text{2}y \right)-\left( 2z \right)}{(l+n)+(n+m)-(l+m)}\]
On solving numerator and denominator of all three fractions, we get
\[\dfrac{2(z+y-x)}{2l}=\text{ }\dfrac{\text{ 2}\left( z+x-y \right)\text{ }}{2m}=\text{ }\dfrac{2\left( x+y-z \right)}{2n}\]
On simplifying fractions, we get
\[\dfrac{(z+y-x)}{l}=\text{ }\dfrac{\text{ }\left( z+x-y \right)\text{ }}{m}=\text{ }\dfrac{\left( x+y-z \right)}{n}\]
Hence, if l ( my + nz – lx ) = m ( nz + lx – my ) = n ( lx + my – nz ) , prove that $\dfrac{y+z-x}{l}=$ $\dfrac{z+x-y}{l}=$$\dfrac{x+y-z}{l}$.

Note: In these types of questions we need to modify fractions in such a way that the resulting fraction in the simplest form and required form. Sometimes, the relation between l, m, and n that is ${{l}^{2}}+{{m}^{2}}+{{n}^{2}}=1$is required so all concept must be remembered.