Problem 5.6

จาก \(\vec{A}\perp \vec{B}\) จะได้ \(\vec{A}\cdot \vec{B} = 0\)

\(4x - 3y = 0\)

\(y=\frac{4}{3}x\cdots \cdots (1)\)

จาก \(|\vec{A}| = 3\)

\(|\vec{A}|^{2} = 3^{2}\)

\(x^{2} + y^{2} = 9\)

\(x^{2}+(\frac{4}{3}x)^{2}=9; (1)\)

\(x = \pm \frac{9}{5}\)

ที่\(x=\frac{9}{5}\) จะได้\(y=\frac{12}{5}\) และที่\(x=-\frac{9}{5}\) จะได้\(y=-\frac{12}{5}\)

จาก \(\vec{A}\cdot \vec{C} > 0\)

\(-5x + 5y > 0\)

\(y > x\)

∴เลือกกรณี\(x=\frac{9}{5}\) จะได้\(y=\frac{12}{5}\)

จะได้\(x+y=\frac{9}{5}+\frac{12}{5}=\frac{21}{5}\;\blacksquare\)