What is the distance of ABC to the centroid of DEF from this truncated prism?

Question

A truncated prism ABC.DEF with AD=8 cm, BE=10 cm, and CF= 12 cm.

ABED is formed 45 degree with ABC, while angle DAB is 45.

Anonim · Accepted Answer

For a prism, edges AD,CF,BE are parallel, say in unit direction d. If position vectors of vertices are represented by corresponding small letters then d=a+8d, e=b+10d, f=c+12e.

Centroid of DEF is at ⅓(d+e+f) = ⅓(a+b+c)+10d

So line joining the centroids of ABC & DEF is also in direction d.

Required distance is projection of 10d onto normal to plane ABC.

To calculate this let A,B,C lie in xy plane with A=O, B on +y-axis.

AD makes angle 45 with y-axis so d₂=cos45=1/√2

ABED forms 45 with xy plane so AD is equally inclined to OX & OZ   →   d₁=d₃

Since d₁²+d₂²+d₃²=1 this means that d=( ½, 1/√2, ½ )

Unit normal to ABC is (0,0,1) so projection of 10d onto this = 10d•(0,0,1) = 5

What is the distance of ABC to the centroid of DEF from this truncated prism?

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