NIMCET 2014 — Mathematics PYQ

Concept: Equation of a Line

The vector equation for a line passing through the points with position vectors $\vec{a}$ and $\vec{b}$ is given by:

Calculation

Let $A$ be the origin $(0)$. Let the position vectors of $B$ and $C$ be $\vec{b}$ and $\vec{c}$ respectively.

1. Since $D$ is the midpoint of $BC$, its position vector is:

   $$\vec{D}=\frac{\vec{b}+\vec{c}}{2}$$

2. Since $E$ is the midpoint of $AD$, its position vector is:

   $$\vec{E}=\frac{0+\frac{\vec{b}+\vec{c}}{2}}{2}=\frac{\vec{b}+\vec{c}}{4}$$

3. Equation of line BF:

   $$\vec{r}=\vec{b}+\lambda \left(\frac{\vec{b}+\vec{c}}{4}-\vec{b}\right)$$

4. Equation of line AC:

   $$\vec{r}=0+\mu \vec{c}=\mu \vec{c}$$

5. For the point of intersection F, we equate the two equations:

   $$\vec{b}+\lambda \left(\frac{\vec{b}+\vec{c}}{4}-\vec{b}\right)=\mu \vec{c}$$
   $$\left(1-\frac{3\lambda }{4}\right)\vec{b}+\frac{\lambda }{4}\vec{c}=\mu \vec{c}$$

6. Equating the coefficients of $\vec{b}$ and $\vec{c}$:

   • For $\vec{b}$: $1-\frac{3\lambda }{4}=0⟹\lambda =\frac{4}{3}$

   • For $\vec{c}$: $\frac{\lambda }{4}=\mu ⟹\mu =\frac{4/3}{4}=\frac{1}{3}$

7. Therefore, the position vector of $F$ is:

   $$\vec{F}=\mu \vec{c}=\frac{\vec{c}}{3}$$

Since $\vec{F}=\frac{1}{3}\overrightarrow{AC}$, we have:

This implies $AF:AC=1:3$.

To find $AF:FC$, if $AF$ is $1$ unit and $AC$ is $3$ units, then $FC=3-1=2$ units.

Hence, $AF:FC=1:2$.

Correct Option: 2