Find the intersection of the lines r=-i+s(cosai+sinaj) and r=i+t(-sinai+cosaj), giving your answer in a simplified form. Interpret your answer geometrically.
This problem can be solved by 2 methods.
Method 1:-
r = -i + s(cosai + sinaj)
or, r = (scosa - 1)i + ssinaj------I
r = i + t(-sinai + cosaj)
or, r = (1 - tsina)i + tcosaj-------II
At the pt. of intersection of st. lines I & II,
(scosa - 1)i + ssinaj = (1 - tsina)i + tcosaj
=>scosa - 1 = 1 - tsina-------III
& ssina = tcosa-------IV
IV=> t = stana-------V
III & V=> scosa - 1 = 1 - s(sin a)^2 * seca
=> s[cosa + (sina)^2/cosa] = 2
=> s = 2cosa-------VI
V & VI=>
t = 2cosa*tana
=> t = 2sina--------VII
I & VI=>
r = [2(cosa)^2 -1]i + 2cosa*sinaj
=> r = cos2ai + sin2aj--------VIII
Therefore, the pt. of intersection is (cos2a,sin2a).
Method 2:-
Put r = xi +yj in both the eqs.
Then, equating the coefficients of i & j for both the eqs. & solving them, we get the eqs. of the 2 st. lines as follows:-
y = tana(x+1)-----A
& y = cota(1-x)-----B
Then, their pt. of intersection can be easily found out.