Let the position vectors of points 'A' and 'B' be i + j + k and 2i + j + 3k, respectively. A point '

Q: Wait, why do we use the section formula instead of just finding the coordinates of P directly?

That is a fantastic question! While you *could* find the coordinates $(x, y, z)$ of $P$ and plug them into the equation, it is a trap. It leads to messy arithmetic and a high chance of a calculation error. By keeping the problem in vector form, we leverage the algebraic properties of dot and cross products (like $\vec{a} \times \vec{a} = 0$), which makes the solution elegant and much faster. In JEE Advanced, the method is often more important than the calculation!

Q: Why does the cross product term simplify so much? I thought it would be a nightmare.

It looks intimidating, doesn't it? But remember: $\vec{a} \times (\lambda \vec{b} + \vec{a}) = \lambda(\vec{a} \times \vec{b}) + (\vec{a} \times \vec{a})$. Since the cross product of any vector with itself is the zero vector, the second term vanishes entirely! This is a classic JEE 'hidden simplification.' Always look for these symmetries before you start grinding through the algebra.

Q: Is there a geometric interpretation of the term $|\vec{OA} \times \vec{OP}|^2$?

Absolutely! The magnitude of the cross product $|\vec{OA} \times \vec{OP}|$ is equal to twice the area of the triangle formed by the origin $O$, point $A$, and point $P$. So, you are essentially working with the square of the area of that triangle. Recognizing this can sometimes give you a geometric shortcut, though in this specific problem, the algebraic approach is the most direct path to the answer.

Question

Let the position vectors of points 'A' and 'B' be $\hat{i} + \hat{j} + \hat{k}$ and $2\hat{i} + \hat{j} + 3\hat{k}$, respectively. A point 'P' divides the line segment $AB$ internally in the ratio $\lambda : 1 (\lambda > 0)$. If $O$ is the origin and $\vec{OB} \cdot \vec{OP} - 3|\vec{OA} 	imes \vec{OP}|^2 = 6$, then $\lambda$ is equal to

Sigma Percentile · Accepted Answer

### The Vector Landscape: A Journey into Elegance

Welcome, fellow traveler of the JEE path. Today, we are not just solving a problem; we are peeling back the layers of a vector equation to reveal the beautiful simplicity hidden underneath.

When you first look at an equation like $\vec{OB} \cdot \vec{OP} - 3|\vec{OA} 	imes \vec{OP}|^2 = 6$, it is natural to feel a spike of adrenaline—or perhaps a bit of dread. But I want you to take a deep breath. In the world of vectors, complexity is often just a mask for symmetry.

### Phase 1: The Bridge of the Section Formula

Imagine you are standing at the origin $O$. You have two points, $A$ and $B$, defined by their position vectors $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} + 3\hat{k}$.

A point $P$ lies on the line segment $AB$, dividing it in the ratio $\lambda : 1$. This is the classic section formula scenario. We define the position vector of $P$ as:

$$\vec{OP} = \frac{\lambda \vec{b} + \vec{a}}{\lambda + 1}$$

This vector $\vec{OP}$ is our key. It is the bridge between the geometry of the line segment and the algebra of our equation. Do not think of this as just a fraction; think of it as a weighted average. As $\lambda$ changes, $P$ slides along the segment $AB$.

### Phase 2: The Dot Product and the Art of Simplification

Now, let us tackle the first term: $\vec{OB} \cdot \vec{OP}$. We substitute our expression for $\vec{OP}$ into this dot product:

$$\vec{b} \cdot \left( \frac{\lambda \vec{b} + \vec{a}}{\lambda + 1} ight) = \frac{\lambda |\vec{b}|^2 + \vec{a} \cdot \vec{b}}{\lambda + 1}$$

Look at how clean that is! We have successfully isolated the constants. We know $\vec{a} \cdot \vec{b} = (1)(2) + (1)(1) + (1)(3) = 6$, and $|\vec{b}|^2 = 2^2 + 1^2 + 3^2 = 14$.

Substituting these, we get $\frac{14\lambda + 6}{\lambda + 1}$. This is the first pillar of our solution.

### Phase 3: The Cross Product 'Zero' Trick

This is where the magic happens. We have the term $3|\vec{OA} 	imes \vec{OP}|^2$. Let us substitute $\vec{OP}$ again:

$$3 \left| \vec{a} 	imes \left( \frac{\lambda \vec{b} + \vec{a}}{\lambda + 1} ight) ight|^2$$

When we distribute the cross product, we encounter $\vec{a} 	imes (\lambda \vec{b})$ and $\vec{a} 	imes \vec{a}$. As we discussed, $\vec{a} 	imes \vec{a} = 0$. This is the 'JEE Secret'—a property that turns a potentially terrifying expression into a simple one.

We are left with:

$$3 \left| \frac{\lambda (\vec{a} 	imes \vec{b})}{\lambda + 1} ight|^2 = \frac{3\lambda^2}{(\lambda + 1)^2} |\vec{a} 	imes \vec{b}|^2$$

Calculating $|\vec{a} 	imes \vec{b}|^2$ is straightforward. The cross product $\vec{a} 	imes \vec{b}$ results in $2\hat{i} - \hat{j} - \hat{k}$, and its squared magnitude is $2^2 + (-1)^2 + (-1)^2 = 6$. Thus, our second term becomes $\frac{18\lambda^2}{(\lambda + 1)^2}$.

### Phase 4: The Algebraic Finale

We have arrived at the final showdown. Combining our terms, the equation becomes:

$$\frac{14\lambda + 6}{\lambda + 1} - \frac{18\lambda^2}{(\lambda + 1)^2} = 6$$

Instead of cross-multiplying and creating a massive cubic equation, let us be clever. We can rewrite $\frac{14\lambda + 6}{\lambda + 1}$ as $\frac{6(\lambda + 1) + 8\lambda}{\lambda + 1} = 6 + \frac{8\lambda}{\lambda + 1}$.

Substituting this back, the $6$ on both sides cancels out beautifully:

$$\frac{8\lambda}{\lambda + 1} - \frac{18\lambda^2}{(\lambda + 1)^2} = 0$$

By setting $t = \frac{\lambda}{\lambda + 1}$, we get $8t - 18t^2 = 0$. Since $\lambda > 0$, $t$ cannot be zero. Solving $8 - 18t = 0$ gives $t = \frac{4}{9}$.

Finally, solving $\frac{\lambda}{\lambda + 1} = \frac{4}{9}$ yields **$\lambda = 0.8$**.

Enter Numerical Value:

Visualized Solution

$P os i t i o n V ec t or s an d S ec t i o n F or m u l a$

$A na l y z in g O B \cdot O P$

$A na l y z in g ∣ O A \times O P ∣^{2}$

$C a l c u l a t in g V ec t or C o n s t an t s$

$S u b s t i t u t in g in t o t h e E q u a t i o n$

$A l g e b r ai c M ani p u l a t i o n$

$S o l v in g f or λ$

$F ina l C o n c l u s i o n$

The Sigma Insight: Vector Product

The Vector Landscape

Phase 1

Phase 2

Phase 3

Phase 4

Enter Numerical Value:

Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing OB⋅OP

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing ∣OA×OP∣2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Vector Constants

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting into the Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Algebraic Manipulation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for λ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Conclusion