Question 29: Know \( {\frac{{bz – cy}}{a} = \frac{{cx – az}}{b} = \frac{{ay – bx}}{c}}\) (with a, b, c# 0).

We have:

\(\begin{array}{*{20}{l}} {\frac{{bz – cy}}{a} = \frac{{cx – az}}{b} = \frac{{ay – bx }}{c}}\\ { = \frac{{abz – acy}}{{a^2}}} = \frac{{bcx – abz}}{{{b^2}}}}\\ { = \frac{{acy – bcx}}{{{c^2}}}}\\ { = \frac{{abz – acy + bcx – abz + acy – bcx}}{{{a^2} + {b^2} + {c^2}}}}\\ { = \frac{0}{{{a^2} + {b^2} + {c^2}}} = 0} \end{ array}\)

Derived from: \( \frac{{bz – cy}}{a} = 0\),

\( \begin{array}{l} bz = cy \Leftrightarrow \frac{y}{b} = \frac{z}{c}(1)\\ \frac{{cx – az}}{b} = 0 \to cx = az \Leftrightarrow \frac{z}{c} = \frac{x}{a}(2) \end{array}\)

From (1) and (2) it follows that: \( \frac{a}{x} = \frac{b}{y} = \frac{c}{z}.\)

The answer to choose is: A