Let the point A divide the line segment joining the points P(-1,-1,2) and Q(5,5,10) internally in th

Q: Wait, why can't we just find the coordinates of A first?

You absolutely could! However, in JEE Advanced, time is your most precious resource. Calculating coordinates involves more arithmetic steps, which increases the probability of a sign error. Using vector algebra allows you to manipulate the entire expression at once, which is not only faster but also more elegant and less prone to calculation traps.

Q: Why does the cross product term $\vec{OP} \times \vec{OA}$ simplify so easily?

It simplifies because $\vec{OA}$ is a linear combination of $\vec{OP}$ and $\vec{OQ}$. When you distribute the cross product, you get terms like $\vec{OP} \times \vec{OP}$ and $\vec{OP} \times \vec{OQ}$. Since any vector crossed with itself is zero, the $\vec{OP} \times \vec{OP}$ term disappears, leaving you with a much simpler expression. It is a beautiful example of how vector properties can do the heavy lifting for you.

Q: What if $r$ was negative? Does that change anything?

Great question! The problem explicitly states $r > 0$, which corresponds to internal division. If $r$ were negative, it would imply external division. In such a case, the formula for the position vector would still hold, but the geometric interpretation would shift. Always check the constraints given in the problem statement; they are there to guide you!

Question

Let the point A divide the line segment joining the points $P(-1,-1,2)$ and $Q(5,5,10)$ internally in the ratio $r:1(r>0)$ If O is the origin and $(\vec{OQ}\cdot\vec{OA})-\frac{1}{5}|\vec{OP}	imes\vec{OA}|^{2}=10$ then the value of r is:

Options:

(A) $\sqrt{7}$

(B) 14

(C) 3

(D) 7

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the origin of a 3D coordinate system. You have two points, $P$ and $Q$, floating in space, anchored to your position by their respective vectors $\vec{p}$ and $\vec{q}$.

A point $A$ lies on the segment connecting them, dividing it in a ratio $r:1$. By using the section formula, we define the position vector of $A$ as:

$$\vec{OA} = \frac{r\vec{q} + \vec{p}}{r+1}$$

This single equation is our master key, unlocking the entire problem.

### The Vector Toolkit

Before we dive into the main equation, we must prepare our tools. We need the magnitude squared of $\vec{q}$ and the dot product of $\vec{p}$ and $\vec{q}$.

Calculating $|\vec{q}|^2$ gives us $5^2 + 5^2 + 10^2 = 150$. The dot product $\vec{p} \cdot \vec{q}$ yields $(-1)(5) + (-1)(5) + (2)(10) = 10$.

Now, let us look at the first term of our given equation: $\vec{OQ} \cdot \vec{OA}$. Substituting our expression for $\vec{OA}$, we get $\vec{q} \cdot (\frac{r\vec{q} + \vec{p}}{r+1})$.

Distributing the dot product, we arrive at:

$$\vec{OQ} \cdot \vec{OA} = \frac{r|\vec{q}|^2 + \vec{p} \cdot \vec{q}}{r+1} = \frac{150r + 10}{r+1}$$

### The Magic of the Cross Product

Now, we face the second term: $\vec{OP} 	imes \vec{OA}$. Applying the distributive property of the cross product, we have:

$$\vec{p} 	imes \left(\frac{r\vec{q} + \vec{p}}{r+1}ight) = \frac{r(\vec{p} 	imes \vec{q}) + (\vec{p} 	imes \vec{p})}{r+1}$$

Since the cross product of any vector with itself is zero, the term $\vec{p} 	imes \vec{p}$ vanishes. We are left with $\frac{r}{r+1} (\vec{p} 	imes \vec{q})$.

Calculating $\vec{p} 	imes \vec{q}$ using the determinant method gives us $-20\hat{i} + 20\hat{j}$, and its magnitude squared is $800$. Thus, the second term of our equation becomes:

$$\frac{1}{5} \left|\frac{r}{r+1} (\vec{p} 	imes \vec{q})ight|^2 = \frac{1}{5} \frac{r^2}{(r+1)^2} (800) = \frac{160r^2}{(r+1)^2}$$

### The Final Algebraic Dance

We have successfully reduced the complex vector equation to a simple algebraic one:

$$\frac{150r + 10}{r+1} - \frac{160r^2}{(r+1)^2} = 10$$

To solve this, we multiply the entire equation by $(r+1)^2$ to clear the denominators. This leads us to:

$$(150r + 10)(r+1) - 160r^2 = 10(r+1)^2$$

Expanding both sides, we get $150r^2 + 160r + 10 - 160r^2 = 10(r^2 + 2r + 1)$, which simplifies to $-10r^2 + 160r + 10 = 10r^2 + 20r + 10$.

The constants cancel out, leaving us with $20r^2 - 140r = 0$. Factoring this, we find $20r(r - 7) = 0$.

Since $r > 0$, we reject $r=0$ and conclude that **$r=7$**.

Animated Solution for Mathematics - Vector Algebra: Let the point A divide the line segment joining the points $P (- 1, - 1, 2)$ and $Q (5, 5, 10)$ internally in the ratio $r : 1 (r > 0)$ If O is the origin and $(O Q \cdot O A) - \frac{1}{5} ∣ O P \times O A ∣^{2} = 10$ then the value of r is:

Select Answer:

Visualized Solution

Visualizing the 3D Setup

Applying the Section Formula

Pre-calculating Vector Properties

Evaluating the First Term: $O Q \cdot O A$

Simplifying the Cross Product Term

Computing $p \times q$

Magnitude Squared of the Cross Product

Setting up the Final Equation

Clearing Denominators

Solving the Quadratic Equation

Final Answer: Finding $r$

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Vector Toolkit

The Magic of the Cross Product

The Final Algebraic Dance

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