If x2 + 4y2 + 3z2 + \(\frac{19}{4}\) = 2\(\sqrt {3}\) (x+y+z) find (x-4y+3z). |
2\(\sqrt {3}\) -\(\sqrt {3}\) \(\sqrt {3}\) -2\(\sqrt {3}\) |
\(\sqrt {3}\) |
x2 + 4y2 + 3z2 + \(\frac{19}{4}\) = 2(\(\sqrt {3}\)x + \(\sqrt {3}\)y + \(\sqrt {3}\)z) Now we can say x = \(\frac{Coefficient\; of\; x}{Coefficient\; of\; x^2}\) = \(\sqrt {3}\) y = \(\frac{Coefficient\; of\; y}{Coefficient\; of\; y^2}\) = \(\frac{\sqrt {3}}{4}\) z = \(\frac{Coefficient\; of\; z}{Coefficient\; of\; z^2}\) = \(\frac{\sqrt {3}}{3}\) = \(\frac{1}{\sqrt {3}}\) Put in x - 4y + 3z ⇒ \(\sqrt {3}\) - 4 × \(\frac{\sqrt {3}}{4}\) + 3 × \(\frac{1}{\sqrt {3}}\) ⇒ \(\sqrt {3}\) |