Question

a+b+c=3 a^2+b^2+c^2=6 1/a+1/b+1/c=1 find values of a , b and c Pls no spam here..

Answer 1

Answer:

The scalar triple product of three vectors a, b, and c is (a×b)⋅c. It is a scalar product because, just like the dot product, it evaluates to a single number. (In this way, it is unlike the cross product, which is a vector.) The scalar triple product is important because its absolute value |(a×b)⋅c| is the volume of the parallelepiped spanned by a, b, and c (i.e., the parallelepiped whose adjacent sides are the vectors a, b, and c).

Answer 2

Answer:

a ≈ 2.6474

b ≈ 0.17629 - 0.73179 i

c ≈ 0.17629 + 0.73179 i

Step-by-step explanation:

a + b + c = 3

a² + b² + c² = 6

1/a + 1/b + 1/c = 1

First and foremost :

1/a +1/b + 1/c = 1

⇒ ab + bc + ca = abc ...........( 1 )

Now :

( a + b + c )² = 3²

⇒ a² + b² + c² + 2 ( ab + bc + ca ) = 9

⇒ 6 + 2 ( ab + bc + ca ) = 9

⇒ ab + bc + ca = 3/2 = 1.5

ab + bc + ca = abc = 1.5

Now :

( 1/a + 1/b + 1/c )² = 1

⇒ 1/a² + 1/b² + 1/c² + 2 ( 1/ab + 1/bc + 1/ac ) = 1

⇒ ( a²b² + b²c² + a²c² )/a²b²c² + 2 ( a + b + c )/abc = 1

⇒ (  a²b² + b²c² + a²c² )/( 1.5 )² + 2 × 3/1.5 = 1

⇒ (  a²b² + b²c² + a²c² ) / 2.25 = 1 - 4 = - 3

⇒  a²b² + b²c² + a²c² = - 3 × 2.25

We cannot come to any conclusion if we go on like this .

So let us assume that :

x³ - ( a + b + c ) x² + ( ab + bc + ca ) x - ( abc ) = 0

Inserting all values we get :

⇒ x³ - 3 x² + 3/2 x - 3/2 = 0

⇒ 2 x³ - 6 x² + 3 x - 3 = 0

By calculating the roots in polynomial calculators we get :  

x ≈ 2.6474

x ≈ 0.17629 - 0.73179 i

x ≈ 0.17629 + 0.73179 i

Replace x with a , b and c

a ≈ 2.6474

b ≈ 0.17629 - 0.73179 i

c ≈ 0.17629 + 0.73179 i