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 ____.

JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Vector Algebra

For any two points

M

and

N

in the

X Y

-plane, let

M N

denote the vector from

M

N

, and

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

△ P QR

such that

S P + 5 S Q + 6 S R = 0

Let

E

and

F

be the mid-points of the sides

P R

and

QR

, respectively. Then the value of

\frac{length of the line segment E F}{length of the line segment E S}

is ____.

Answer: 1.2

From

S P + 5 S Q + 6 S R = 0

, we get:

s = \frac{p + 5 q + 6 r}{12}

E

is the midpoint of

P R

e = \frac{p + r}{2}

F

is the midpoint of

QR

f = \frac{q + r}{2}

E F = f - e = \frac{q - p}{2}

E S = s - e = \frac{p + 5 q + 6 r}{12} - \frac{p + r}{2} = \frac{5 ( q - p )}{12}

Therefore:

\frac{∣ E F ∣}{∣ E S ∣} = \frac{∣ q - p ∣/2}{5∣ q - p ∣/12} = \frac{12}{10} = 1.2

JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Numerical answer