Question 61

If $$x^2+y^2+z^2 = 14$$ and xy + yz + zx = 11, then the value of $$(x+y+z)^2$$ is

Solution

Given : $$x^2 + y^2 + z^2 = 14$$ and $$xy+yz+zx=11$$ ----------(i)

Using, $$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$$

Substituting values from equation (i),

=> $$(x+y+z)^2=14+2(11)$$

=> $$(x+y+z)^2=14+22=36$$

=> Ans - (C)

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