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QJEEM25Ma6877: JEE Main 2025 Physics - Class 11 JEEM25Ma6877 Solution: Let the position vectors of three vertices of a triangle be and …

Let the position vectors of three vertices of a triangle be 4p⃗+q⃗−3r⃗,−5p⃗+q⃗+2r⃗4 \vec{p}+\vec{q}-3 \vec{r},-5 \vec{p}+\vec{q}+2 \vec{r} and 2p⃗−q⃗+2r⃗2 \vec{p}-\vec{q}+2 \vec{r} . If the position vectors of the orthocenter and the circumcenter of the triangle are p⃗+q⃗+r⃗4\frac{\vec{p}+\vec{q}+\vec{r}}{4} and αp⃗+βq⃗+γr⃗\alpha \vec{p}+\beta \vec{q}+\gamma \vec{r} respectively, then α+2β+5γ\alpha+2 \beta+5 \gamma is equal to :

Multiple Choice Options

  • A 4
  • B 3
  • C 1
  • D 6

Step-by-Step Verified Solution

We know that,

JEE Main 2025 (Online) 24th January Evening Shift Mathematics - Vector Algebra Question 2 English Explanation

amp;amp;O (orthocentre) p→+q→+r→4amp;C( circum centre) αp→+βq→+γr→amp;C (centroid) =p→+q→+r→3amp; by relation amp;amp;⇒2(αp→+βq→+γr→)+p→+q→+r→4=3(p→+q→+r→3)amp;⇒8(αp→+βq→+γr→)=3(p→+q→+r→)amp;⇒8α=3,8β=3,8γ=3amp;α=38,β=38,γ=38amp;∴α+2β+3γamp;38+68+158=248=3\begin{aligned}&\begin{aligned}& \mathrm{O} \text { (orthocentre) } \frac{\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}}{4} \\& \mathrm{C}(\text { circum centre) } \alpha \overrightarrow{\mathrm{p}}+\beta \overrightarrow{\mathrm{q}}+\gamma \overrightarrow{\mathrm{r}} \\& \mathrm{C} \text { (centroid) }=\frac{\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}}{3}\end{aligned}\\&\text { by relation }\\&\begin{aligned}& \Rightarrow 2(\alpha \overrightarrow{\mathrm{p}}+\beta \overrightarrow{\mathrm{q}}+\gamma \overrightarrow{\mathrm{r}})+\frac{\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}}{4}=3\left(\frac{\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}}{3}\right) \\& \Rightarrow 8(\alpha \overrightarrow{\mathrm{p}}+\beta \overrightarrow{\mathrm{q}}+\gamma \overrightarrow{\mathrm{r}})=3(\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}) \\& \Rightarrow 8 \alpha=3,8 \beta=3,8 \gamma=3 \\& \alpha=\frac{3}{8}, \beta=\frac{3}{8}, \gamma=\frac{3}{8} \\& \therefore \alpha+2 \beta+3 \gamma \\& \frac{3}{8}+\frac{6}{8}+\frac{15}{8}=\frac{24}{8}=3\end{aligned}\end{aligned}

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