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

Given

.OABC is a tetrahedron

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

point P and Q are such that OA =AP and 2OB = BQ

point M is a midpoint of PQ.

Find AB ,PQ, CQ ,QM ,MB ,OM in terms of a, b, and c.

To proof

we proof this by using the Triangle law of vector addition

when two vectors are represented by two sides of a triangle in magnitude and direction taken in same order then third side of the third side of that triangle represents in magnitude and direction the resultant of the vectors.

by using the diagram as shown below

 $\vec{OA}=\vec{a}, \vec{OB}=\vec{b}\\\vec{OP}=2\vec{a},\vec{OQ}=3\vec{b}$

$\vec{OA}+\vec{AB}=\vec{OB}\\\vec{AB}=\vec{b}-\vec{a}$

$\vec{OP}+\vec{PQ}=\vec{OQ}\\\vec{PQ}=3\vec{b}-2\vec{a}$

$\vec{OC}+\vec{CQ}=\vec{OQ}\\\vec{CQ}=3\vec{b}-2\vec{c}\\\vec{QM}=-\frac{1}{2}\vec{PQ}$

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

$\vec{OM}=\vec{a}+\frac{3}{2}\vec{b}\\\vec{MB}=-\vec{a}-\frac{1}{2}\vec{b}$

hence proved