FPM(2023) January Paper 01 que 11 and solution

Solution

Given: Triangle $OAB$ with vectors $\vec{OA} = \mathbf{a}$ and $\vec{OB} = \mathbf{b}$. Point $M$ is the midpoint of $OA$, and point $N$ divides $OB$ in the ratio $ON : NB = 3 : 1$. The lines $AN$ and $BM$ intersect at point $X$.

(a) Find expressions for:

(i) $\vec{AN}$:

Point $N$ divides $OB$ in the ratio $ON : NB = 3 : 1$. Therefore, the position vector of point $N$ is:

$$ \vec{ON} = \frac{3}{3+1}\vec{OB} = \frac{3}{4}\mathbf{b}. $$

The vector $\vec{AN}$ is the vector from $A$ to $N$, so:

$$ \vec{AN} = \vec{ON} - \vec{OA} = \frac{3}{4}\mathbf{b} - \mathbf{a}. $$

Thus, the expression for $\vec{AN}$ is:

$$ \boxed{\vec{AN} = \frac{3}{4}\mathbf{b} - \mathbf{a}} $$

(ii) $\vec{BM}$:

Since $M$ is the midpoint of $OA$, the position vector of $M$ is:

$$ \vec{OM} = \frac{1}{2}\vec{OA} = \frac{1}{2}\mathbf{a}. $$

The vector $\vec{BM}$ is the vector from $B$ to $M$, so:

$$ \vec{BM} = \vec{OM} - \vec{OB} = \frac{1}{2}\mathbf{a} - \mathbf{b}. $$

Thus, the expression for $\vec{BM}$ is:

$$ \boxed{\vec{BM} = \frac{1}{2}\mathbf{a} - \mathbf{b}} $$

(b) Find $AX : XN$ using vector methods:

To solve this, we will use the fact that $\vec{AB} = \vec{AX} + \vec{XB}$, where $X$ divides $AB$ in some ratio.

Let point $X$ divide $AN$ in the ratio $\lambda : 1$. The position vector of $X$ along $AN$ is:

$$ \vec{AX} = \lambda \vec{AN} = \lambda\left(\frac{3}{4}\mathbf{b} - \mathbf{a}\right). $$

Let point $X$ divide $BM$ in the ratio $\mu : 1$. The position vector of $X$ along $BM$ is:

$$ \vec{BX} = \mu \vec{BM} = \mu\left(\frac{1}{2}\mathbf{a} - \mathbf{b}\right). $$

Since the point $X$ is the same in both cases, we use the equation $\vec{AB} = \vec{AX} + \vec{XB}$, where $\vec{AB} = \vec{OB} - \vec{OA} = \mathbf{b} - \mathbf{a}$.

Now, we use the relationship:

$$ \vec{AB} = \vec{AX} + \vec{XB}. $$

Substitute for $\vec{AX}$ and $\vec{XB}$:

$$ \mathbf{b} - \mathbf{a} = \lambda\left(\frac{3}{4}\mathbf{b} - \mathbf{a}\right) - \mu\left(\frac{1}{2}\mathbf{a} - \mathbf{b}\right). $$

Equating the coefficients of $\mathbf{a}$ and $\mathbf{b}$, we solve for $\lambda$ and $\mu$:

For $\mathbf{b}$:

$$ 1 = \lambda\left(\frac{3}{4}\right) + \mu. $$

For $\mathbf{a}$:

$$ -1 = -\lambda - \mu\left(\frac{1}{2}\right). $$

Solving this system of equations gives:

$$ \lambda = \frac{4}{5}, \quad \mu = \frac{2}{5}. $$

Thus, the ratio $AX : XN$ is:

$$ \boxed{AX : XN = 4 : 1} $$