Question

43. Numerele a , b, c ∈ Q sunt direct proporţionale cu 2, 3, 6. Demonstraţi ca :

2a+3b+6c=7 ⇔ $a^{2} +b^2+c^2=1$

Answer 1

a/2 = b/3 = c/6
a² /4 = b²/9 = c²/36 = (a² + b² + c²)/49 = 1/49
⇒ a = 2/7    b = 3/7    c = 6/7
2a + 3b + 6c = 4/7 + 9/7 + 36/7 = 49/7 = 7

Answer 2

a,b,c     d.p cu
2;3;6

a/2=b/3=c/6=k

a=2k
b=3k
c=6k

2a+3b+6c= 2*2k+3*3k+6*6k=7
                     k(4+9+36)=7
                      k*49=7   => k= 7/49= 1/7

a=2 *1/7= 2/7
b=3 * 1/7= 3/7
c=6*1/7=6/7

(2/7)^2+(3/7)^2+(6/7)^2= (4+9+36)/49= 49/49=1 c.c.t.d.