Express each vector from the centroid to the vertices in terms of position vectors: GA = a - g, GB = b - g, and GC = c - g. Adding them gives ( a + b + c) - 3 g. Substituting 3 g = a + b + c, the sum simplifies directly to 0.

Express each vector from the centroid to the vertices in terms of position vectors: GA = a - g, GB = b - g, and GC = c - g. Adding them gives ( a + b + c) - 3 g. Substituting 3 g = a + b + c, the sum simplifies directly to 0.

Using position vectors, we write each segment: AP = p - a, BQ = q - b, and CR = r - c. Summing these together groups the terms of Delta PQR and Delta ABC separately.

Regroup the expression as ( p + q + r) - ( a + b + c). Substituting p + q + r = 3 g_1 and a + b + c = 3 g, we get 3 g_1 - 3 g = 3( g_1 - g). Since g_1 - g = GG_1, this establishes AP + BQ + CR = 3 GG_1.

Regroup the expression as ( p + q + r) - ( a + b + c). Substituting p + q + r = 3 g_1 and a + b + c = 3 g, we get 3 g_1 - 3 g = 3( g_1 - g). Since g_1 - g = GG_1, this establishes AP + BQ + CR = 3 GG_1.

3. If $G$ is the centroid of $\Delta ABC$ , prove that $\vec{GA} + \vec{GB} + \vec{GC} = 0$ . Further if $G_1$ be the centroid of another $\Delta PQR$ , show that $\vec{AP} + \vec{BQ} + \vec{CR} = 3\vec{GG_1}$ .

Answer: Hence proved that $\vec{GA} + \vec{GB} + \vec{GC} = \vec{0}$ and $\vec{AP} + \vec{BQ} + \vec{CR} = 3\vec{GG_1}$ .

Step-by-step solution

Step 1: Define position vectors and express the centroid $G$

Let $O$ be the origin of reference. Let $\vec{a}$ , $\vec{b}$ , and $\vec{c}$ be the position vectors of vertices $A$ , $B$ , and $C$ respectively. Since $G$ is the centroid of $\Delta ABC$ , its position vector $\vec{g}$ is the arithmetic mean of the position vectors of its vertices.

Step 2: Prove that $\vec{GA} + \vec{GB} + \vec{GC} = \vec{0}$

Express each vector from the centroid to the vertices in terms of position vectors: $\vec{GA} = \vec{a} - \vec{g}$ , $\vec{GB} = \vec{b} - \vec{g}$ , and $\vec{GC} = \vec{c} - \vec{g}$ . Adding them gives $(\vec{a} + \vec{b} + \vec{c}) - 3\vec{g}$ . Substituting $3\vec{g} = \vec{a} + \vec{b} + \vec{c}$ , the sum simplifies directly to $\vec{0}$ .

Step 3: Express the centroid $G_1$ of $\Delta PQR$

Let $\vec{p}$ , $\vec{q}$ , and $\vec{r}$ be the position vectors of vertices $P$ , $Q$ , and $R$ of $\Delta PQR$ . Since $G_1$ is the centroid of $\Delta PQR$ , its position vector is $\vec{g}_1 = \frac{\vec{p} + \vec{q} + \vec{r}}{3}$ , which implies $\vec{p} + \vec{q} + \vec{r} = 3\vec{g}_1$ .

Step 4: Express and simplify $\vec{AP} + \vec{BQ} + \vec{CR}$

Using position vectors, we write each segment: $\vec{AP} = \vec{p} - \vec{a}$ , $\vec{BQ} = \vec{q} - \vec{b}$ , and $\vec{CR} = \vec{r} - \vec{c}$ . Summing these together groups the terms of $\Delta PQR$ and $\Delta ABC$ separately.

Step 5: Substitute the centroid relations to complete the proof

Regroup the expression as $(\vec{p} + \vec{q} + \vec{r}) - (\vec{a} + \vec{b} + \vec{c})$ . Substituting $\vec{p} + \vec{q} + \vec{r} = 3\vec{g}_1$ and $\vec{a} + \vec{b} + \vec{c} = 3\vec{g}$ , we get $3\vec{g}_1 - 3\vec{g} = 3(\vec{g}_1 - \vec{g})$ . Since $\vec{g}_1 - \vec{g} = \vec{GG_1}$ , this establishes $\vec{AP} + \vec{BQ} + \vec{CR} = 3\vec{GG_1}$ .

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Step-by-step solution

Step 1: Define position vectors and express the centroid GGG

Step 2: Prove that GA⃗+GB⃗+GC⃗=0⃗\vec{GA} + \vec{GB} + \vec{GC} = \vec{0}GA+GB+GC=0

Step 3: Express the centroid G1G_1G1​ of ΔPQR\Delta PQRΔPQR

Step 4: Express and simplify AP⃗+BQ⃗+CR⃗\vec{AP} + \vec{BQ} + \vec{CR}AP+BQ​+CR

Step 5: Substitute the centroid relations to complete the proof

Step 1: Define position vectors and express the centroid $G$

Step 2: Prove that $\vec{GA} + \vec{GB} + \vec{GC} = \vec{0}$

Step 3: Express the centroid $G_1$ of $\Delta PQR$

Step 4: Express and simplify $\vec{AP} + \vec{BQ} + \vec{CR}$