As shown in the figure, BD 1⊥ surface ACB 1 and point P move on the side surface BCC 1B 1 and its boundary, so the trajectory of point P is surface ACB 1 and surface bcc1b/kloc-0.

Solution: As shown in the figure, connect AC, AB 1, B 1C, and in the cube ABCD-A1B1C,

There is a BD 1⊥ surface ACB 1, and the point p moves on the edge of BCC 1B 1 and its boundary.

∴ Therefore, the locus of point P is the intersection line segment CB 1 between surface ACB 1 and surface bcc 1.

Let point p( 1, 2,3) be symmetrical about the origin. q？ ,

So (distance formula between two points) | pq | = 2 | po | = 2 √ (1+3 2+2) = 2 √14.