JEE Advanced 1994

LEVELJEE Main

Animated Solution for Mathematics - Vector Algebra: Let $p$ and $q$ be the position vectors of $P$ and $Q$ respectively, with respect to $O$ and $∣ p ∣ = p, ∣ q ∣ = q$ . The points $R$ and $S$ divide $P Q$ internally and externally in the ratio $2 : 3$ respectively. If $O R$ and $O S$ are perpendicular then

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Progressive Hints

Hint 1

Visualize the points R and S on the line segment PQ. R is the internal divider, sitting comfortably between P and Q, while S is the external divider, sitting outside on the extended line. The key is to translate the geometric condition of 'perpendicularity' into the algebraic language of the dot product.

Hint 2

Recall your section formula toolkit. For a point dividing a segment in ratio m:n, the position vector is \frac{m\vec{b} + n\vec{a}}{m+n} for internal division and \frac{m\vec{b} - n\vec{a}}{m-n} for external division. Keep these close; they are your primary weapons for this problem.

Hint 3

Write down the expressions for \vec{OR} and \vec{OS} explicitly. Once you have them, set their dot product to zero: \vec{OR} \cdot \vec{OS} = 0. Remember that \vec{v} \cdot \vec{v} = |\vec{v}|^2 and that the dot product is commutative. This will lead you directly to the relationship between p and q.

Visualized Solution

Position Vectors $p$ and $q$

Let the origin be $O$ .
Position vector of $P$ is $O P = p$ .
Position vector of $Q$ is $O Q = q$ .
Given magnitudes: $∣ p ∣ = p$ and $∣ q ∣ = q$ .

Internal Division at $R$

Point $R$ lies on the line segment $P Q$ .
It divides $P Q$ internally in the ratio $2 : 3$ .

Section Formula for $O R$

Using the internal section formula:
$O R = \frac{m q + n p}{m + n}$
$O R = \frac{2 q + 3 p}{2 + 3}$
$O R = \frac{3 p + 2 q}{5}$

External Division at $S$

Point $S$ lies on the extended line $P Q$ .
It divides $P Q$ externally in the ratio $2 : 3$ .

Section Formula for $O S$

Using the external section formula:
$O S = \frac{m q - n p}{m - n}$
$O S = \frac{2 q - 3 p}{2 - 3}$

Simplifying $O S$

The denominator is $2 - 3 = - 1$ .
$O S = \frac{2 q - 3 p}{- 1}$
$O S = 3 p - 2 q$

The Perpendicularity Condition

We are given that $O R$ and $O S$ are perpendicular.
Therefore, the angle between them is $9 0^{\circ}$ .
Mathematically, their dot product must be zero: $O R \cdot O S = 0$ .

Setting Up the Dot Product

Substitute the expressions for $O R$ and $O S$ :
$(\frac{3 p + 2 q}{5}) \cdot (3 p - 2 q) = 0$

Eliminating the Scalar

Multiply both sides by $5$ to remove the denominator.
$(3 p + 2 q) \cdot (3 p - 2 q) = 0$

Expanding the Dot Product

Use the identity $(a + b) \cdot (a - b) = ∣ a ∣^{2} - ∣ b ∣^{2}$ .
Here, $a = 3 p$ and $b = 2 q$ .
$(3 p) \cdot (3 p) - (2 q) \cdot (2 q) = 0$

Applying Vector Magnitudes

Recall that $v \cdot v = ∣ v ∣^{2}$ .
$9∣ p ∣^{2} - 4∣ q ∣^{2} = 0$
Substitute the given magnitudes $∣ p ∣ = p$ and $∣ q ∣ = q$ .
$9 p^{2} - 4 q^{2} = 0$

Final Relation

Rearrange the equation to match the options:
$9 p^{2} = 4 q^{2}$
Key Takeaway: For perpendicular vectors derived from section formulas, the dot product leads to a quadratic relation between the magnitudes.

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The Sigma Insight: Scalar (Dot) Product

Solution Diagram

Analyzing the Setup

We are given two vectors,

p

and

q

, originating from the origin

O

and pointing to points

P

and

Q

, respectively. The magnitudes of these vectors are given as

∣ p ∣ = p

and

∣ q ∣ = q

Point

R

divides the segment

P Q

internally in a ratio of

2 : 3

. Using the internal section formula, the position vector

O R

is:

O R = \frac{2 q + 3 p}{2 + 3} = \frac{3 p + 2 q}{5}

The Adventurer Point

Point

S

divides the segment

P Q

externally in the same ratio of

2 : 3

. Applying the external section formula, we calculate the position vector

O S

as follows:

O S = \frac{2 q - 3 p}{2 - 3}

Simplifying the denominator, we obtain:

O S = \frac{2 q - 3 p}{- 1} = 3 p - 2 q

The Perpendicularity Condition

We are given that the lines

O R

and

O S

are perpendicular. In vector algebra, this implies that their dot product must be zero:

O R \cdot O S = 0

Substituting our derived expressions for

O R

and

O S

into this equation, we get:

(\frac{3 p + 2 q}{5}) \cdot (3 p - 2 q) = 0

Final Calculation

Multiplying both sides by

5

to clear the fraction, we recognize the expression as a difference of squares:

(3 p + 2 q) \cdot (3 p - 2 q) = 0

Expanding the dot product, we have:

(3 p) \cdot (3 p) - (2 q) \cdot (2 q) = 0

Since the dot product of a vector with itself is the square of its magnitude, this simplifies to:

9∣ p ∣^{2} - 4∣ q ∣^{2} = 0

Substituting the given magnitudes

p

and

q

, we arrive at the final relationship:

$9 p^{2} = 4 q^{2}$

Common Traps & FAQs

Why does the external division formula have a minus sign? It feels counter-intuitive.

Think of it as a vector subtraction. When we divide externally, we are essentially moving 'backwards' relative to the segment. The negative sign in the denominator m-n accounts for this direction reversal. It is not just a formula; it is a directional correction that ensures the point S lands exactly where it should on the extended line.

Can I just use the internal formula with a negative ratio for external division?

Absolutely! If you replace n with -n in the internal formula, you get the external one. It is a brilliant shortcut, but be careful with your signs. If you are comfortable with it, use it, but always double-check the denominator to ensure you haven't flipped a sign incorrectly.

Why does the dot product of perpendicular vectors equal zero?

Because the dot product is defined as \vec{A} \cdot \vec{B} = |\vec{A}||\vec{B}| \cos(\theta). When the vectors are perpendicular, the angle \theta is 90 degrees. Since \cos(90\circ) = 0, the entire product vanishes. It is the most elegant way to say 'these two lines have nothing in common' in the language of vectors.

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Visualized Solution

Position Vectors $p$ and $q$

Internal Division at $R$

Section Formula for $O R$

External Division at $S$

Section Formula for $O S$

Simplifying $O S$

The Perpendicularity Condition

Setting Up the Dot Product

Eliminating the Scalar

Expanding the Dot Product

Applying Vector Magnitudes

Final Relation

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The Adventurer Point

The Perpendicularity Condition

Final Calculation

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Comprehension Passage

$∣ O X \times O Y ∣ =$

If the triangle PQR varies, then the minimum value of $cos (P + Q) + cos (Q + R) + cos (R + P)$ is

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Visualized Solution

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

Internal Division at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R

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

External Division at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

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

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

The Perpendicularity Condition

Setting Up the Dot Product

Eliminating the Scalar

Expanding the Dot Product

Applying Vector Magnitudes

Final Relation

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The Adventurer Point

The Perpendicularity Condition

Final Calculation

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Internal Division at $R$

Section Formula for $O R$

External Division at $S$

Section Formula for $O S$

Simplifying $O S$

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