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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.

Respuesta :

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

[tex]QM= -\frac{1}{2} PQ = a-\frac{3}{2}b[/tex]

[tex]OM=OQ+QM = 3b + a-\frac{3}{2}b = a+\frac{3}{2}b[/tex]

[tex]OM+MB=OB=>MB = b-(a+\frac{3}{2}b) = -a-\frac{1}{2}b[/tex]

(i) AB = b-a

(ii) PQ = 3b-2a

(iii) CQ = 3b-c

(iv) QM = [tex]a-\frac{3}{2}b[/tex]

(v) MB =  [tex]-a-\frac{1}{2}b[/tex]

(vi) OM = [tex]a+\frac{3}{2}b[/tex]

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