Question

If a^2 + b^2 + c^2 = 1, then which of the following cannot be a value of (ab + bc + ca)?
A. 0
B. 1/2
C. -1/4
D. -1​

Answer 1

Answer

(a+b+c)² = a²+b²+c²+2(ab +bc+ca)

(a+b+c) >= 0

for any real value of a , b and c

a²+b²+c²+2(ab+bc+CA) >= 0

1 + 2(ab + bc + ca ) >=0

ab + BC + ca >= -1/2.

or

(a-b)² +(b-c)² + (c-a)² >=0

2(a²+b²+c²-ab-bc-ac) >=0

2(1-ab+bc+ca) >=0

ab+bc+ca<=1

so answer Will be (-1/2 or 1 )

Answer 2

a² + b² + c² = 1

___________ [ GIVEN ]

• We have to find the value of (ab + bc + ca)

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We have a² + b² + c² = 1.

ab + bc + ca = ?

We know that..

→ (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Take 2 common

→ a² + b² + c² + 2(ab + bc + ca)

Now put value of a² + b² + c² above

→ (1) + 2(ab + bc + ca)

→ 1 + 2(ab + bc + ca)

→ 2(ab + bc + ca) = - 1

→ ab + bc + ca = -1/2

But the given options are

A. 0

B. 1/2

C. -1/4

D. -1

It means that none of the option is correct.

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ab + bc + ca = -1/2

_______ [ ANSWER ]

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