Sliding Planes

At any given $t$ the intersection is a straight line along $n_1 \times n_2$ , and given by:

$p = s (n_1 \times n_2) + x_1 n_1 + y_1 n_2$

where $s$ is arbitrary, and $x_1$ and $y_1$ are a linear function of $t$ . To see this substitute the above into the two equations

$n_1^T (p - p_1(t) ) = 0$

$n_2^T (p - p_2(t) ) = 0$

where $p_1(t) = r_1 + w_1 t$ and $p_2(t) = r_2 + w_2 t$ .

We can assume WLOG that $n_1$ and $n_2$ have been normalized. It follows that

$x_1 + y_1 (n_1 \cdot n_2) = n_1 \cdot p_1(t)$

$x_1(n_1 \cdot n_2) + y_1 = n_2 \cdot p_2(t)$

(note that $| n_1.n_2 | \lt 1$ ) This system can be solved for $x_1$ and $y_1$ for any given $t$ . Taking the derivative with respect to $t$ , we obtain

$\dfrac{dx_1}{dt} + \dfrac{dy_1}{dt} (n_1 \cdot n_2) = n_1 \cdot w_1$

$(n_1.n_2) \dfrac{dx_1}{dt} + d\frac{dy_1}{dt} = n_2 \cdot w_2$

where $w_1$ and $w_2$ are defined above.

Solving for the two derivatives we get,

$\dfrac{dx_1}{dt} = ( n_1 \cdot w_1 - (n_1 \cdot n_2) (n_2 \cdot w_2)/ ( 1 - (n_1 \cdot n_2)^2 )$

$\dfrac{dy_1}{dt} = (-(n_1 \cdot n_2)(n_1 \cdot w_1) + n_2 \cdot w_2 )/( 1 - (n_1 \cdot n_2)^2 )$

Thus $(x_1, y_1)$ traces a line in the $(n_1, n_2)$ plane. The direction vector of this line is

$d = ( \dfrac{dx_1}{dt}) n_1 + (\dfrac{dy_1}{dt}) n_2$

Therefore, the normal to the traced plane is obtained by the cross product of $( n_1 \times n_2 )$ and the vector $d$ :

$(n_1 \times n_2) \times d = (n_1 \times n_2) \times ( \dfrac{dx_1}{dt} n_1 + \dfrac{dy_1}{dt} n_2 )$

$= -(\dfrac{dx_1}{dt}) \left[ (n_1 \cdot n_2) n_1 - n_2 \right] - (\dfrac{dy_1}{dt}) \left[ n_1 - (n_1 \cdot n_2) n_2 \right]$

$= - n_1 \left[ (n_1 \cdot n_2) \dfrac{dx_1}{dt} + \dfrac{dy_1}{dt} \right] + n_2 \left[ \dfrac{dx_1}{dt} + (n_1 \cdot n_2) \dfrac{dy_1}{dt} \right]$

Substituting the given values , we obtain a normal vector along $[8, -3, 5]^T$ , making the answer $8 + 3 + 5 = \boxed{16}$