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
[tex]\vec{OA}=\vec{a}, \vec{OB}=\vec{b}\\\vec{OP}=2\vec{a},\vec{OQ}=3\vec{b}[/tex]
[tex]\vec{OA}+\vec{AB}=\vec{OB}\\\vec{AB}=\vec{b}-\vec{a}[/tex]
[tex]\vec{OP}+\vec{PQ}=\vec{OQ}\\\vec{PQ}=3\vec{b}-2\vec{a}[/tex]
[tex]\vec{OC}+\vec{CQ}=\vec{OQ}\\\vec{CQ}=3\vec{b}-2\vec{c}\\\vec{QM}=-\frac{1}{2}\vec{PQ}\\[/tex]
[tex]\vec{QM}=\vec{a}-\frac{3}{2}\vec{b}\\\vec{OM}=\vec{OQ}+\vec{QM}[/tex]
[tex]\vec{OM}=\vec{a}+\frac{3}{2}\vec{b}\\\vec{MB}=-\vec{a}-\frac{1}{2}\vec{b}[/tex]
hence proved
