**Type I** 1. The position vector of four points A,B,C,DA, B, C, D are a⃗,b⃗,2a⃗+3b⃗\vec{a}, \vec{b}, 2\vec{a} + 3\vec{b} and a⃗−2b⃗\vec{a} - 2\vec{b} respectively. Express the vectors AC⃗,DB⃗,BC⃗\vec{AC}, \vec{DB}, \vec{BC} and CA⃗\vec{CA} in terms of a⃗\vec{a} and b⃗\vec{b}. 2. If AD,BEAD, BE and CFCF be the medians of a ΔABC\Delta ABC, prove that AD⃗+BE⃗+CF⃗=0\vec{AD} + \vec{BE} + \vec{CF} = 0 3. If GG is the centroid of ΔABC\Delta ABC, prove that GA⃗+GB⃗+GC⃗=0\vec{GA} + \vec{GB} + \vec{GC} = 0. Further if G1G_1 be the centroid of another ΔPQR\Delta PQR, show that AP⃗+BQ⃗+CR⃗=3GG1⃗\vec{AP} + \vec{BQ} + \vec{CR} = 3\vec{GG_1}. 4. Five forces AB⃗,AC⃗,AD⃗,AE⃗,AF⃗\vec{AB}, \vec{AC}, \vec{AD}, \vec{AE}, \vec{AF} act at the vertices of a regular hexagon ABCDEFABCDEF. Prove that their resultant is 6AO⃗6\vec{AO}, where OO is the centroid of the regular hexagon. 5. If ABCDEFABCDEF is a regular hexagon, prove that AC⃗+AD⃗+EA⃗+FA⃗=3AB⃗\vec{AC} + \vec{AD} + \vec{EA} + \vec{FA} = 3\vec{AB}

Answer: AC⃗=a⃗+3b⃗,DB⃗=−a⃗+3b⃗,BC⃗=2a⃗+2b⃗,CA⃗=−a⃗−3b⃗\vec{AC} = \vec{a} + 3\vec{b}, \quad \vec{DB} = -\vec{a} + 3\vec{b}, \quad \vec{BC} = 2\vec{a} + 2\vec{b}, \quad \vec{CA} = -\vec{a} - 3\vec{b}

Step-by-step solution

Step 1: State the Given Position Vectors

Let OO be the origin of reference. The position vectors of points A, B, C, and DD with respect to the origin OO are given directly by the problem statement.

Step 2: Find the Vector AC⃗\vec{AC}

The vector AC⃗\vec{AC} is obtained by subtracting the position vector of AA from the position vector of CC. Combining like terms gives a⃗+3b⃗\vec{a} + 3\vec{b}.

Step 3: Find the Vector DB⃗\vec{DB}

The vector DB⃗\vec{DB} is given by OB⃗−OD⃗\vec{OB} - \vec{OD}. Distribute the negative sign over (a⃗−2b⃗)(\vec{a} - 2\vec{b}) to get b⃗−a⃗+2b⃗=−a⃗+3b⃗\vec{b} - \vec{a} + 2\vec{b} = -\vec{a} + 3\vec{b}.

Step 4: Find the Vector BC⃗\vec{BC}

Subtract the position vector of BB from that of CC. Subtracting b⃗\vec{b} from 2a⃗+3b⃗2\vec{a} + 3\vec{b} yields 2a⃗+2b⃗2\vec{a} + 2\vec{b}.

Step 5: Find the Vector CA⃗\vec{CA}

The vector CA⃗\vec{CA} is the negative of AC⃗\vec{AC}, which can also be calculated directly as OA⃗−OC⃗=a⃗−(2a⃗+3b⃗)=−a⃗−3b⃗\vec{OA} - \vec{OC} = \vec{a} - (2\vec{a} + 3\vec{b}) = -\vec{a} - 3\vec{b}.

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