Question

The lines which intersect the skew lines $y = mx, z = c; y = -mx, z = -c$ and the X-axis lie on the surface :

• A. cz = mxy
• B. cy = mxz
• C. xy = cmz
• D. cx = myz

Answer 1

Answer: (B)

cy = mxz

Answer 2

The two given skew lines are $L_1: y = mx, z = c$ and $L_2: y = -mx, z = -c$. The X-axis is given by $y = 0, z = 0$. Let a transversal line intersect $L_1$ at $P_1(t_1, mt_1, c)$, $L_2$ at $P_2(t_2, -mt_2, -c)$, and the X-axis at $P_X(t_X, 0, 0)$. For these three points to be collinear, the vectors $P_2 - P_1$ and $P_X - P_1$ must be parallel. $P_2 - P_1 = (t_2 - t_1, -m(t_1+t_2), -2c)$ $P_X - P_1 = (t_X - t_1, -mt_1, -c)$ The condition for parallelism is: $\frac{t_2 - t_1}{t_X - t_1} = \frac{-m(t_1+t_2)}{-mt_1} = \frac{-2c}{-c}$ From $\frac{-2c}{-c} = 2$ (assuming $c \neq 0$), and $\frac{-m(t_1+t_2)}{-mt_1} = 2$ (assuming $m \neq 0$), we get $\frac{t_1+t_2}{t_1} = 2$, which implies $t_2 = t_1$. Then, $\frac{t_2 - t_1}{t_X - t_1} = 2$ becomes $\frac{0}{t_X - t_1} = 2$, so $t_X - t_1 = 0$, which means $t_X = t_1$. Thus, $t_1 = t_2 = t_X = t$. The points of intersection are $P_1(t, mt, c)$, $P_2(t, -mt, -c)$, and $P_X(t, 0, 0)$. A general point $(x,y,z)$ on the transversal line passing through these points can be parameterized. For instance, the line through $P_X(t,0,0)$ and $P_1(t,mt,c)$ has direction vector $(0, mt, c)$. The parametric equation is $(x,y,z) = (t,0,0) + \lambda(0, mt, c) = (t, \lambda mt, \lambda c)$. This gives $x=t$, $y=\lambda mt$, $z=\lambda c$. Substituting $t=x$ and $\lambda = z/c$ (assuming $c \neq 0$) into $y=\lambda mt$, we get $y = (z/c)mx$, which rearranges to $cy = mxz$. This is the equation of the surface.