Two parallel planes and a traverse plane

The distance between the two parallel planes is given by

$d_1 = \dfrac{| 1 - 7 |}{ \sqrt{4^2 + (-3)^2 + (7)^2}} = \dfrac{6}{\sqrt{74}}$

If $\theta$ is the angle between one of the two parallel planes and the cutting plane, then the distance between the two lines of intersection is,

$d_2 = \dfrac{d_1}{ | \sin \theta | }$

Now, using the inner product of the two normal vectors, we can calculate $\cos \theta$ , as follows

$\cos \theta = \dfrac{ n_1 \cdot n_2 }{| n_1| |n_2|} = \dfrac{ (4)(-2)+(-3)(5)+(7)(1) }{\sqrt{ 4^2 + (-3)^2 + 7^2} \sqrt{ (-2)^2 + 5^2 + 1^2} } = \dfrac{-16}{\sqrt{ 74} \sqrt{ 30 } }$

Hence, $| \sin \theta | = \sqrt{ 1 - \dfrac{16^2}{(74)(30)}} = \dfrac{\sqrt{1964}}{ \sqrt{74} \sqrt{30} }$

Therefore, the required distance is $d_2 = \dfrac{6}{\sqrt{74}} \dfrac{\sqrt{74}\sqrt{30}}{\sqrt{1964}} =\sqrt{\dfrac{1080}{1964}} = \sqrt{\dfrac{270}{491} }$ . And this makes the answer $270 + 491 = \boxed{761}$

Combining $4x - 3y + 7z = 1$ with $-2x + 5y + z = 4$ gives $y = -\frac{9}{7}z + \frac{9}{7}$ . Letting $z = t$ , $y = -\frac{9}{7}t + \frac{9}{7}$ , and substituting these equations back into $-2x + 5y + z = 4$ gives $-2x + 5(-\frac{9}{7}t + \frac{9}{7}) + t = 4$ which solves to $x = -\frac{19}{7}t + \frac{17}{14}$ . Therefore, the line $L_1$ resulting in the intersection of $4x - 3y + 7z = 1$ and $-2x + 5y + z = 4$ is $(x, y, z) = (\frac{17}{14}, \frac{9}{7}, 0) + (-\frac{19}{7}, -\frac{9}{7}, 1)t$ .

Combining $4x - 3y + 7z = 7$ with $-2x + 5y + z = 4$ gives $y = -\frac{9}{7}z + \frac{15}{7}$ . Letting $z = s$ , $y = -\frac{9}{7}s + \frac{15}{7}$ , and substituting these equations back into $-2x + 5y + z = 4$ gives $-2x + 5(-\frac{9}{7}s + \frac{15}{7}) + s = 4$ which solves to $x = -\frac{19}{7}s + \frac{47}{14}$ . Therefore, the line $L_2$ resulting in the intersection of $4x - 3y + 7z = 7$ and $-2x + 5y + z = 4$ is $(x, y, z) = (\frac{47}{14}, \frac{15}{7}, 0) + (-\frac{19}{7}, -\frac{9}{7}, 1)s$ .

Let $P$ be a point on $L_1$ at $t = 0$ . Then $P = (\frac{17}{14}, \frac{9}{7}, 0)$ .

Let $Q$ be a point on $L_2$ such that $\vec{PQ}$ is perpendicular to $L_1$ and $L_2$ . Since $Q$ is on $L_2$ , $Q = (\frac{47}{14} - \frac{19}{7}s, \frac{15}{7} - \frac{9}{7}s, s)$ , and $\vec{PQ} = (\frac{15}{7} - \frac{19}{7}s, \frac{6}{7} - \frac{9}{7}s, s)$ , and since $\vec{PQ}$ is perpendicular to $L_1$ and $L_2$ , $\vec{PQ} \bullet (-\frac{19}{7}, -\frac{9}{7}, 1) = 0$ , or $(\frac{15}{7} - \frac{19}{7}s)(-\frac{19}{7}) + (\frac{6}{7} - \frac{9}{7}s)(-\frac{9}{7}) + s(1) = 0$ which solves to $s = \frac{339}{491}$ .

Substituting $s = \frac{339}{491}$ into $\vec{PQ}$ gives $\vec{PQ} = (\frac{132}{491}, -\frac{15}{491}, \frac{339}{491})$ , and the distance $d$ between the two lines is $d = |\vec{PQ}| = \sqrt{(\frac{132}{491})^2 + (-\frac{15}{491})^2 + (\frac{339}{491})^2} = \sqrt{\frac{270}{491}}$ . Therefore, $a = 270$ , $b = 491$ , and $a + b = \boxed{761}$ .

A direction vector for either of the intersection lines is $\vec{r}=(19,9,-7)$ This can easily be verified, because both the expressions $4x-3y+7z$ and $-2x-5y+z$ remain unchanged when adding $\vec{r}$ to them.

From the system of equations, the x- and y- coordinate can both be isolated as a function of z, by elimination. Support vectors to lines 1 and 2 are found using an arbitrary value for z, say z=0. We find respectively $\vec{p_1} =(\frac{17}{14}, \frac{9}{7}, 0) \text{ and } \vec{p_2} =( \frac{47}{14}, \frac{15}{7}, 0)$ so $\vec{Δp} = \vec{p_2} -\vec{p_1}=(\frac{15}{7}, \frac{6}{7}, 0)$ Here $\vec{Δp}$ is a vector connecting both lines.  The distance vector between the lines then is given by subtracting the projection of $\vec{Δp}$ onto $\vec{r}$ : $\vec{d}=\vec{Δp}-\frac{\vec{Δp} \cdot \vec{r}}{r^2} \vec{r}$

Here $\vec{Δp} \cdot \vec{r} = \frac{1}{7}(15×19+6×9+0)= \frac{339}{7}$ and $r^2=19^2+9^2+7^2=491$

$\vec{d}=\frac{1}{7×491}(924,-105, 2373)$

$d=|\vec{d}|=\frac{\sqrt{6495930}}{7×491} =\frac{\sqrt{132570}}{491} =\sqrt{\frac{270}{491}}$

$a+b=270+491=\boxed{761}$