If $$a^2 + b^2 + c^2 = 14$$ and a + b + c = 6, then the value of (ab + bc + ca) is,
Given : $$a^2 + b^2 + c^2 = 14$$ and $$a+b+c=6$$ ----------(i)
Using, $$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$$
Substituting values from equation (i),
=> $$(6)^2=14+2(ab+bc+ca)$$
=> $$2(ab+bc+ca)=36-14=22$$
=> $$ab+bc+ca=\frac{22}{2}=11$$
=> Ans - (A)
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