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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