Question 62
If $$a^2 + b^2 + c^2 =21$$, and $$a + b + c = 7$$, then $$(ab + bc + ca)$$ is equal to:
Solution
As per the question, it is given that
$$a^2 + b^2 + c^2 =21$$
$$a + b + c = 7$$
We know that $$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$$
Now, substituting the values,
$$\Rightarrow 7^2=21+2(ab+bc+ca)$$
$$\Rightarrow 2(ab+bc+ca)=49-21$$
$$\Rightarrow (ab+bc+ca)=\dfrac{49-21}{2}=\dfrac{28}{2}=14$$
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