Question

OABC is a tetrahedron and OA=a, OB=b, and OC=c. The point P and Q are such that OA =AP and 2OB = BQ. The point M is a midpoint of PQ. Find (i)AB (ii)PQ (iii)CQ (iv)QM (v)MB (vi)OM in terms of a, b, and c.

Answer 1

Solution:

Let OA=a, OB=b, and OC=c

Then OP=2a and OQ=3b

OA+AB=OB ⇒ AB=b-a

OP+PQ=OQ ⇒ PQ = 3b-2a

OC+CQ = OQ ⇒ CQ = 3b-c

$QM= -\frac{1}{2} PQ = a-\frac{3}{2}b$

$OM=OQ+QM = 3b + a-\frac{3}{2}b = a+\frac{3}{2}b$

$OM+MB=OB=>MB = b-(a+\frac{3}{2}b) = -a-\frac{1}{2}b$

(i) AB = b-a

(ii) PQ = 3b-2a

(iii) CQ = 3b-c

(iv) QM = $a-\frac{3}{2}b$

(v) MB =  $-a-\frac{1}{2}b$

(vi) OM = $a+\frac{3}{2}b$