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BITSAT 2024Paper 2
Exam
BITSAT
Year
2024
Paper
Paper 2
Subject
Mathematics
Chapter
Vector Algebra

BITSAT 2024 · Paper 2 · Mathematics · Vector Algebra

Let ABC be a triangle and be a , b , c the position vectors of A,...

Let ABC be a triangle and be a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} the position vectors of A,B,C\mathrm{A}, \mathrm{B}, \mathrm{C} respectively. Let D divide BCB C in the ratio 3:13: 1 internally and EE divide AD in the ratio 4:14: 1 internally. Let BE meet AC in F. IfE divides BF in the ratio 3:23: 2 internally then the position vector of FF is

Options

  1. Aa⃗+b⃗+c⃗3\frac{\vec{a}+\vec{b}+\vec{c}}{3}
  2. Ba⃗−2b⃗+3c⃗2\frac{\vec{a}-2 \vec{b}+3 \vec{c}}{2}
  3. Ca⃗+2b⃗+3c⃗2\frac{\vec{a}+2 \vec{b}+3 \vec{c}}{2}
  4. Da⃗−b⃗+3c⃗3\frac{\vec{a}-\vec{b}+3 \vec{c}}{3}

Correct answer

D. a⃗−b⃗+3c⃗3\frac{\vec{a}-\vec{b}+3 \vec{c}}{3}

Step-by-step solution

Here we are given that OA→=a→,OB→=b→,OC→=c→\overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{OB}}=\overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{OC}}=\overrightarrow{\mathrm{c}} Now P.V of Di.e OD→=1×OB→+3×OC→1+3\overrightarrow{\mathrm{OD}}=\frac{1 \times \overrightarrow{\mathrm{OB}}+3 \times \overrightarrow{\mathrm{OC}}}{1+3}OD→=b→+3c→4OE→=4OD→+OA→4+1=4(b→+3c→)4+a→5⇒OE→=a→+b→+3c→5 Now, OE→=2OB→+3OF→2+3⇒OF→=5OE→−2OB→3⇒OF→=5(a⃗+b⃗+3c→)5−2b→3⇒OF→=a→−b→+3c→3⇒P.V. of F is a→−b→+3c→3\begin{aligned} & \overrightarrow{\mathrm{OD}}=\frac{\overrightarrow{\mathrm{b}}+3 \overrightarrow{\mathrm{c}}}{4} \\ & \overrightarrow{\mathrm{OE}}=\frac{4 \overrightarrow{\mathrm{OD}}+\overrightarrow{\mathrm{OA}}}{4+1}=\frac{\frac{4(\overrightarrow{\mathrm{ b}}+3 \overrightarrow{\mathrm{c}})}{4}+\overrightarrow{\mathrm{a}}}{5} \\ & \Rightarrow \overrightarrow{\mathrm{OE}}=\frac{\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+3 \overrightarrow{\mathrm{c}}}{5} \\ & \text{ Now, } \overrightarrow{\mathrm{OE}}=\frac{2 \overrightarrow{\mathrm{OB}}+3 \overrightarrow{\mathrm{OF}}}{2+3} \\ & \Rightarrow \overrightarrow{\mathrm{OF}}=\frac{5 \overrightarrow{\mathrm{OE}}-2 \overrightarrow{\mathrm{OB}}}{3} \\ & \Rightarrow \overrightarrow{\mathrm{OF}}=\frac{\frac{5(\vec{a}+\vec{b}+3 \overrightarrow{\mathrm{c}})}{5}-2 \overrightarrow{\mathrm{ b}}}{3} \\ & \Rightarrow \overrightarrow{\mathrm{OF}}=\frac{\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}+3 \overrightarrow{\mathrm{c}}}{3} \Rightarrow P. V. \text{ of } \mathrm{F} \text{ is } \frac{\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}+3 \overrightarrow{\mathrm{c}}}{3}\end{aligned}$$
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