JEE Advanced · Mathematics · 30. Vector Algebra
For any two points \(M\) and \(N\) in the \(X Y\)-plane, let \(\overrightarrow{M N}\) denote the vector from \(M\) to \(N\), and \(\overrightarrow{0}\) denote the zero vector. Let \(P, Q\) and \(R\) be three distinct points in the \(X Y\)-plane. Let \(S\) be a point inside the triangle \(\triangle P Q R\) such that \(\overrightarrow{S P}+5 \overrightarrow{S Q}+6 \overrightarrow{S R}=\overrightarrow{0}\). Let \(E\) and \(F\) be the mid-points of the sides \(P R\) and \(Q R\), respectively. Then the value of \(\frac{\text { length of the line segment } E F}{\text { length of the line segment } E S}\) is ________
- A 1.5
- B 1.2
- C 1
- D 2
Answer & Solution
Correct Answer
(B) 1.2
Step-by-step Solution
Detailed explanation

\(\begin{aligned}
&\because \overrightarrow{\mathrm{SP}}+5 \overrightarrow{\mathrm{SQ}}+6 \overrightarrow{\mathrm{SP}}=\overrightarrow{0}\\
&\text { (Let P.V. of } P \text { be } \vec{p}, Q \text { be } \vec{q} \text { and } R \text { be } \vec{r} \text { ) }\\
&\begin{aligned}
& \Rightarrow(\vec{p}-\vec{s})+5(\vec{q}-\vec{s})+6(\vec{r}-\vec{s})=\overrightarrow{0} \\
& \Rightarrow \vec{S}=\frac{\vec{p}+5 \vec{q}+6 \vec{r}}{12} \\
& \Rightarrow \overrightarrow{E F}=\left(\frac{\vec{q}-\vec{p}}{2}\right) \\
& \Rightarrow \overrightarrow{E S}=\left(\frac{5 \vec{q}-5 \vec{p}}{12}\right)=\frac{5}{12}(\vec{q}-\vec{p}) \\
& \therefore \frac{|\overrightarrow{E F}|}{|\overrightarrow{E S}|}=\frac{6}{5}=1.2
\end{aligned}
\end{aligned}\)
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