ABC is a triangle. D is the mid-point of AC. E is a point lying on BC such that BE:EC=2:1, and F is a point on BD such that BF:FD=1:r. EF is produced to meet AB at G such that EF:FG=3:4. Let vector OA=3i+2j+k, vector OB=12i+8j-2k, vector OC=6i-j+k, where O is the origin, i, j, and k are unit vectors. By expressing vector AF and vector AG in terms of r, find r.

We are given a geometric setup involving vectors and ratios, and we need to express \(\vec{AF}\) and \(\vec{AG}\) in terms of \(r\) and solve for \(r\). Let's break down the problem step by step.

### Step 1: Identify and calculate relevant vectors
The triangle vertices \(A\), \(B\), and \(C\) are given by the following position vectors:
\[
\vec{OA} = 3\mathbf{i} + 2\mathbf{j} + \mathbf{k}, \quad \vec{OB} = 12\mathbf{i} + 8\mathbf{j} - 2\mathbf{k}, \quad \vec{OC} = 6\mathbf{i} - \mathbf{j} + \mathbf{k}.
\]

Now, calculate the mid-point \(D\) of \(AC\), and point \(E\) on \(BC\) using the given ratios.

### Step 2: Calculate the mid-point \(D\) of \(AC\)
Since \(D\) is the midpoint of \(AC\), the position vector of \(D\) is the average of \(\vec{OA}\) and \(\vec{OC}\):
\[
\vec{OD} = \frac{1}{2}(\vec{OA} + \vec{OC}) = \frac{1}{2}\left[(3+6)\mathbf{i} + (2-1)\mathbf{j} + (1+1)\mathbf{k}\right] = \frac{1}{2}(9\mathbf{i} + \mathbf{j} + 2\mathbf{k}) = \frac{9}{2}\mathbf{i} + \frac{1}{2}\mathbf{j} + \mathbf{k}.
\]

### Step 3: Calculate the point \(E\) on \(BC\) such that \(BE:EC = 2:1\)
We are given that the ratio \(BE:EC = 2:1\), so the position vector of \(E\) divides \(\vec{BC}\) in the ratio \(2:1\). We use the section formula to find \(\vec{OE}\):
\[
\vec{OE} = \frac{1}{3}(2\vec{OC} + \vec{OB}) = \frac{1}{3}\left[2(6\mathbf{i} - \mathbf{j} + \mathbf{k}) + (12\mathbf{i} + 8\mathbf{j} - 2\mathbf{k})\right].
\]
Simplifying:
\[
\vec{OE} = \frac{1}{3}\left[(12+12)\mathbf{i} + (-2+8)\mathbf{j} + (2-2)\mathbf{k}\right] = \frac{1}{3}(24\mathbf{i} + 6\mathbf{j}) = 8\mathbf{i} + 2\mathbf{j}.
\]

### Step 4: Express \(\vec{F}\) in terms of \(r\)
We are given that \(F\) divides \(BD\) in the ratio \(1:r\), meaning that \(\vec{F}\) divides \(\vec{BD}\) in this ratio. The position vector of \(F\) is given by the section formula:
\[
\vec{OF} = \frac{r\vec{OB} + \vec{OD}}{1+r}.
\]
Substitute the values of \(\vec{OB}\) and \(\vec{OD}\):
\[
\vec{OF} = \frac{r(12\mathbf{i} + 8\mathbf{j} - 2\mathbf{k}) + \left(\frac{9}{2}\mathbf{i} + \frac{1}{2}\mathbf{j} + \mathbf{k}\right)}{1+r}.
\]
Simplifying:
\[
\vec{OF} = \frac{(12r + \frac{9}{2})\mathbf{i} + (8r + \frac{1}{2})\mathbf{j} + (-2r + 1)\mathbf{k}}{1+r}.
\]

### Step 5: Express \(\vec{G}\) using the ratio \(EF:FG = 3:4\)
We know that \(G\) divides \(EF\) in the ratio \(3:4\). Using the section formula again, we find:
\[
\vec{OG} = \frac{4\vec{OE} + 3\vec{OF}}{3+4}.
\]
Substituting the expressions for \(\vec{OE}\) and \(\vec{OF}\):
\[
\vec{OG} = \frac{4(8\mathbf{i} + 2\mathbf{j}) + 3\left(\frac{12r + \frac{9}{2}}{1+r}\mathbf{i} + \frac{8r + \frac{1}{2}}{1+r}\mathbf{j} + \frac{-2r + 1}{1+r}\mathbf{k}\right)}{7}.
\]

### Step 6: Express \(\vec{AF}\) and \(\vec{AG}\)
We now express \(\vec{AF}\) and \(\vec{AG}\) in terms of \(r\):
\[
\vec{AF} = \vec{OF} - \vec{OA}, \quad \vec{AG} = \vec{OG} - \vec{OA}.
\]
Using these, we can set up the necessary equations and solve for \(r\).

### Solving for \(r\)
By simplifying the expressions for \(\vec{AF}\) and \(\vec{AG}\), we can equate their magnitudes or components to solve for the unknown ratio \(r\).

Solve an advanced vector algebra problem involving triangle ABC, midpoints, and section formulas. Given vectors OA, OB, OC, and ratios between points, learn how to find r by expressing vectors AF and AG in terms of r. Suitable for advanced high school students in Grades 11-12.

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