a1x+b1y+c1z=0①
a2x+b2y+c2z=0②
①×c2:a1c2x+b1c2y+c1c2z=0③
②×c1:a2c1x+b2c1y+c1c2z=0④
③-④:(a1c2-a2c1)x+(b1c2-b2c1)y=0
x=(-b1c2+b2c1)/(a1c2-a2c1)*y⑤
将⑤代入①:
a1*(-b1c2+b2c1)/(a1c2-a2c1)*y+b1y+c1z=0
整理得:z=(a2b1-a1b2)/(a1c2-a2c1)*y⑥
由⑤和⑥
x:y:z==(b2c1-b1c2)/(a1c2-a2c1)/(a2b1-a1b2)