Let $a>0$. Take the hyperbolic paraboloid, denoted $S$, defined as the set of $(x,y,z)\in \mathbb{R}^3$ such that $$ \frac{x^2}{a^2} - \frac{y^2}{a^2}=z. $$ We know that an hyperbolic paraboloid is doubly ruled surface. But in this particular case, one can show that for every point $M$ of the hyperbolic paraboloid, the two lines $D_1$ and $D_2$ that are contained in $S$ and such that $M\in D_1$ and $M\in D_2$ are perpendicular.
Take $(x_0,y_0,z_0)\in S$. I found that the line $D_1$ $$ \left\{ \begin{array}{rcl} x/a - y/a - x_0/a+y_0/a &=& 0\\ (x/a + y/a)(x_0/a-y_0/a)-z&=&0 \end{array} \right. $$ is contained in $S$ and $(x_0,y_0,z_0)\in D_1$. I also found that the vector $$ \vec{v}_1 = \begin{pmatrix} 1/a \\ -1/a \\ 0 \end{pmatrix} \wedge \begin{pmatrix} 1/a(x_0/a-y_0/a) \\ 1/a(x_0/a-y_0/a) \\ -1 \end{pmatrix} = \begin{pmatrix} 1/a \\ 1/a \\ 2/a^2(x_0/a-y_0/a) \end{pmatrix} $$ is a direction vector for $D_1$.
Similarly, the line $D_2$ $$ \left\{ \begin{array}{rcl} x/a + y/a - x_0/a-y_0/a &=& 0\\ (x/a - y/a)(x_0/a+y_0/a)-z&=&0 \end{array} \right. $$ is contained in $S$ and $(x_0,y_0,z_0)\in D_2$. I also found that the vector $$ \vec{v}_2 = \begin{pmatrix} 1/a \\ 1/a \\ 0 \end{pmatrix} \wedge \begin{pmatrix} 1/a(x_0/a+y_0/a) \\ -1/a(x_0/a+y_0/a) \\ -1 \end{pmatrix} = \begin{pmatrix} -1/a \\ 1/a \\ -2/a^2(x_0/a+y_0/a) \end{pmatrix} $$ is a direction vector for $D_2$. So I should get that $$ \vec{v_1}\cdot \vec{v_2} =0 $$ but with my calculations, I get that $$ \vec{v_1}\cdot \vec{v_2} = - 4/a^4(x_0^2-y_0^2) = -2/a^2z_0. $$ So obviously, I'm doing something wrong...Could you help me for this problem?