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← JEE Advanced 2025 · Paper 1
MathematicsNumerical+4Question 9 of 48

JEE Advanced 2025 · Paper 1, Mathematics question 9

For any two points MM and NN in the XYXY-plane, let MN→\overrightarrow{MN} denote the vector from MM to NN, and 0⃗\vec 0 denote the zero vector. Let PP, QQ and RR be three distinct points in the XYXY-plane. Let SS be a point inside the triangle ΔPQR\Delta PQR such that

SP→+5 SQ→+6 SR→=0⃗.\overrightarrow{SP} + 5\,\overrightarrow{SQ} + 6\,\overrightarrow{SR} = \vec 0.

Let EE and FF be the mid-points of the sides PRPR and QRQR, respectively. Then the value of

length of the line segment EFlength of the line segment ES\frac{\text{length of the line segment } EF}{\text{length of the line segment } ES}

is _____.

Answer and worked solution

Official answer: 1.15

Position of S. With position vectors, P⃗+5Q⃗+6R⃗=12S⃗\vec P+5\vec Q+6\vec R=12\vec S.

EF. E=P+R2E=\frac{P+R}{2} and F=Q+R2F=\frac{Q+R}{2}, so EF=12∣Q⃗−P⃗∣EF=\frac12|\vec Q-\vec P|.

ES.
S⃗−E⃗=P+5Q+6R−6P−6R12=512(Q⃗−P⃗)  ⟹  ES=512∣Q⃗−P⃗∣.\vec S-\vec E=\frac{P+5Q+6R-6P-6R}{12}=\frac{5}{12}(\vec Q-\vec P)\implies ES=\frac5{12}|\vec Q-\vec P|.

EFES=1/25/12=65=1.2.\frac{EF}{ES}=\frac{1/2}{5/12}=\frac65=1.2.