Let OACB be a parallelogram with O at the origin and OC a diagonal. Let D be the midpoint of OA. Usi

Q: Wait, why can't we just use similar triangles or geometry to solve this?

You absolutely can! Geometry is a beautiful way to see the result. However, in JEE Advanced, we use vector methods because they provide a systematic, algebraic framework that works even in three dimensions or when the geometry isn't immediately obvious. Mastering this approach is essential for tackling more complex problems where visual intuition might fail.

Q: Why do we use $\lambda : 1$ instead of $\lambda : \mu$ for the ratio?

Great observation! Using $\lambda : 1$ is a standard algebraic trick. It reduces the number of variables from two ($\lambda$ and $\mu$) to one, making the equation much cleaner. It doesn't change the generality of the answer; it just makes the algebra less prone to errors.

Q: How do we know we can just equate the coefficients of $\vec{a}$ and $\vec{b}$?

This is the core of vector algebra! Because $\vec{a}$ and $\vec{b}$ are non-collinear (they form the sides of a parallelogram), they are linearly independent. This means the only way $x\vec{a} + y\vec{b} = z\vec{a} + w\vec{b}$ is if $x=z$ and $y=w$. It is the fundamental property that allows us to solve for the unknowns.

Question

Let $OACB$ be a parallelogram with $O$ at the origin and $OC$ a diagonal. Let $D$ be the midpoint of $OA$. Using vector methods prove that $BD$ and $CO$ intersect in the same ratio. Determine this ratio.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the origin $O$ of a coordinate system, looking out at a parallelogram $OACB$. We define two fundamental vectors, $\vec{OA} = \vec{a}$ and $\vec{OB} = \vec{b}$, which define the entire structure.

By the parallelogram law of vector addition, the diagonal $OC$ is represented by:
$$ \vec{OC} = \vec{a} + \vec{b} $$

Consider the midpoint $D$ of $OA$. Since $D$ is the midpoint, its position vector is:
$$ \vec{OD} = \frac{1}{2}\vec{a} $$

### The Intersection Point

Let $P$ be the intersection of $OC$ and $BD$. We aim to find the ratio in which $P$ divides these segments.

First, let $P$ divide $OC$ in the ratio $\lambda : 1$. Using the section formula, the position vector of $P$ is:
$$ \vec{p} = \frac{1 \cdot \vec{OC} + \lambda \cdot \vec{O}}{\lambda + 1} = \frac{\vec{a} + \vec{b}}{\lambda + 1} $$

Next, let $P$ divide $BD$ in the ratio $\mu : 1$. Using the section formula again:
$$ \vec{p} = \frac{\mu \cdot \vec{OD} + 1 \cdot \vec{OB}}{\mu + 1} $$

Substituting $\vec{OD} = \frac{1}{2}\vec{a}$ into the expression above, we obtain:
$$ \vec{p} = \frac{\mu(\frac{1}{2}\vec{a}) + \vec{b}}{\mu + 1} = \frac{\frac{\mu}{2}\vec{a} + \vec{b}}{\mu + 1} $$

### The Algebraic Dance

Since both expressions represent the same point $P$, we equate them:
$$ \frac{1}{\lambda + 1}\vec{a} + \frac{1}{\lambda + 1}\vec{b} = \frac{\mu}{2(\mu + 1)}\vec{a} + \frac{1}{\mu + 1}\vec{b} $$

Because $\vec{a}$ and $\vec{b}$ are linearly independent, we equate their respective coefficients. Comparing the coefficients of $\vec{b}$ yields:
$$ \frac{1}{\lambda + 1} = \frac{1}{\mu + 1} \Rightarrow \lambda = \mu $$

Now, comparing the coefficients of $\vec{a}$ and substituting $\lambda = \mu$:
$$ \frac{1}{\mu + 1} = \frac{\mu}{2(\mu + 1)} $$

Canceling the common term $(\mu + 1)$, we find:
$$ 1 = \frac{\mu}{2} \Rightarrow \mu = 2 $$

### Final Calculation

We have determined that $\mu = 2$, and since $\lambda = \mu$, it follows that $\lambda = 2$.

This confirms that the intersection point $P$ divides both $OC$ and $BD$ in the ratio **$2:1$**. This result serves as a testament to the power of vector algebra in uncovering geometric truths.

Animated Solution for Mathematics - Vector Algebra: Let $O A C B$ be a parallelogram with $O$ at the origin and $O C$ a diagonal. Let $D$ be the midpoint of $O A$ . Using vector methods prove that $B D$ and $C O$ intersect in the same ratio. Determine this ratio.

Visualized Solution

Visualizing the Parallelogram $O A C B$

Defining Position Vectors $a$ and $b$

Finding the Diagonal Vector $O C$

Locating the Midpoint $D$

Defining the Intersection Point $P$

Applying Section Formula on $C O$

Applying Section Formula on $B D$

Equating the Position Vectors

Comparing Coefficients of $b$

Solving for the Ratio $μ$

Conclusion and Takeaways

The Sigma Insight: Addition of Vectors

Analyzing the Setup

The Intersection Point

The Algebraic Dance

Final Calculation

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Visualized Solution

Visualizing the Parallelogram .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OACB

Defining Position Vectors .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Finding the Diagonal Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OC

Locating the Midpoint .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D

Defining the Intersection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Applying Section Formula on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CO

Applying Section Formula on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BD