Let the position vectors of three vertices of a triangle be , and . If the position vectors of the orthocenter and the circumcenter of the triangle are and respectively, then is equal to: [2025]

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Solution

Q.
Let the position vectors of three vertices of a triangle be $4 \vec{p} + \vec{q} - 3 \vec{r}$ , $- 5 \vec{p} + \vec{q} + 2 \vec{r}$ and $2 \vec{p} - \vec{q} + 2 \vec{r}$ . If the position vectors of the orthocenter and the circumcenter of the triangle are $\frac{\vec{p} + \vec{q} + \vec{r}}{4}$ and $α \vec{p} + β \vec{q} + γ \vec{r}$ respectively, then $α + 2 β + 5 γ$ is equal to: [2025]

1 1

2 3

3 6

4 4

Ans.
(2)

Given Orthocenter $= \frac{\vec{p} + \vec{q} + \vec{r}}{4}$

Circumcenter $= α \vec{p} + β \vec{q} + γ \vec{r}$

Centroid $= \frac{(4 \vec{p} + \vec{q} - 3 \vec{r}) + (- 5 \vec{p} + \vec{q} + 2 \vec{r}) + (2 \vec{p} - \vec{q} + 2 \vec{r})}{3}$

$= \frac{\vec{p} + \vec{q} + \vec{r}}{3}$

Since, centroid divides the line joining orthocenter and the circumcenter in the ratio of 2 : 1,

So, $2 \cdot (α \vec{p} + β \vec{q} + γ \vec{r}) + 1 \cdot (\frac{\vec{p} + \vec{q} + \vec{r}}{4}) = 3 \cdot (\frac{\vec{p} + \vec{q} + \vec{r}}{3})$

$\Rightarrow 8 (α \vec{p} + β \vec{q} + γ \vec{r}) = 3 (\vec{p} + \vec{q} + \vec{r})$

$\Rightarrow 8 α \vec{p} + 8 β \vec{q} + 8 γ \vec{r} = 3 \vec{p} + 3 \vec{q} + 3 \vec{r}$
On comparing, we get

$α = \frac{3}{8}, β = \frac{3}{8} and γ = \frac{3}{8}$

$∴ α + 2 β + 5 γ = \frac{3}{8} + 2 \times \frac{3}{8} + 5 \times \frac{3}{8} = \frac{3}{8} + \frac{6}{8} + \frac{15}{8} = \frac{24}{8} = 3$ .

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