Question

If l sin theta +m cos theta+n= 0 and l'sin theta+m'cos theta +n'=0. Show that (mn'-m'n)^2 +(nl'-n'l)^2=(lm'-l'm)^2

Answer 1

l sin∅ + m cos∅ + n = 0
l' sin∅ + m' cos∅ + n' = 0

Let x = sin∅ and y = cos∅
then, equation will be
lx + my + n = 0
l'x + m'y + n' = 0
now use Cramer rule to solve it.
x/(mn' - m'n) = -y/(n'l - nl') = 1/(lm' - l'm)
then, x = (mn' - m'n)/(lm' - l'm)
y = -(n'l - nl')/(lm' - l'm) = (nl' - n'l)/(lm' - l'm)

because x = sin∅ and y = cos∅
and we know, sin²∅ + cos²∅ = 1
so, x² + y² = 1
=>{(mn' - m'n)/(lm' - l'm)}² + {(nl' - n'l)/(lm' - l'm)}² = 1
=> (mn' - m'n)² + (nl' - n'l)² = (lm' - l'm)²

hence, proved that ,
(mn' - m'n)² + (nl' - n'l)² = (lm' - l'm)²

Answer 2

Answer:

Step-by-step explanation: