Let a and b be positive real numbers. Suppose PQ = ai + bj and PS = ai - bj are adjacent sides of a

Q: Why do we need the modulus in the numerator of the projection formula? Can't we just drop it?

That is a dangerous trap! The dot product $\vec{w} \cdot \vec{PS}$ can be negative if the angle between the vectors is obtuse. Since a length (magnitude) must always be positive, the modulus ensures we are dealing with a physical distance, not a signed scalar. Always respect the modulus!

Q: Can we just assume $a > b$ and ignore the other case? It feels like extra work.

In a competition setting, you might get lucky, but mathematically, you must prove the other case is impossible. By checking $a < b$, we find it leads to a contradiction, which confirms our initial assumption was the only valid path. Rigor is the hallmark of a true JEE topper.

Q: Is there a faster way to solve this without all the algebra?

While intuition helps, the algebra is your safety net. The projection sum condition is a specific constraint that forces the geometry of the parallelogram to be symmetric, which is why $a=b$ emerges so cleanly. Trust the process, and the math will reveal the symmetry.

Question

Let $a$ and $b$ be positive real numbers. Suppose $\vec{PQ} = a\hat{i} + b\hat{j}$ and $\vec{PS} = a\hat{i} - b\hat{j}$ are adjacent sides of a parallelogram $PQRS$. Let $\vec{u}$ and $\vec{v}$ be the projection vectors of $\vec{w} = \hat{i} + \hat{j}$ along $\vec{PQ}$ and $\vec{PS}$, respectively. If $|\vec{u}| + |\vec{v}| = |\vec{w}|$ and if the area of the parallelogram $PQRS$ is 8, then which of the following statements is/are TRUE?

Options:

(A) (A) $a + b = 4$

(B) (B) $a - b = 2$

(C) (C) The length of the diagonal $PR$ of the parallelogram $PQRS$ is 4

(D) (D) $\vec{w}$ is an angle bisector of the vectors $\vec{PQ}$ and $\vec{PS}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a flat plane, looking down at a parallelogram $PQRS$. You have two vectors, $\vec{PQ} = a\hat{i} + b\hat{j}$ and $\vec{PS} = a\hat{i} - b\hat{j}$, defining the shape.

Now, introduce a third vector, $\vec{w} = \hat{i} + \hat{j}$, cutting across the space. The problem asks us to consider the projections of $\vec{w}$ onto these sides.

Think of a projection as a shadow. If you shine a light perpendicular to the side $\vec{PQ}$, the shadow cast by $\vec{w}$ is the vector $\vec{u}$. The magnitude of this shadow is given by the elegant formula:

$$|\vec{u}| = \frac{|\vec{w} \cdot \vec{PQ}|}{|\vec{PQ}|}$$

### The Algebraic Dance

Let us calculate these shadows. For $\vec{u}$, the dot product $\vec{w} \cdot \vec{PQ}$ is $(\hat{i} + \hat{j}) \cdot (a\hat{i} + b\hat{j}) = a + b$.

Since $a$ and $b$ are positive, the magnitude is simply:

$$|\vec{u}| = \frac{a+b}{\sqrt{a^2+b^2}}$$

Now, for $\vec{v}$, the projection onto $\vec{PS}$, the dot product is $(\hat{i} + \hat{j}) \cdot (a\hat{i} - b\hat{j}) = a - b$. Here, we must be careful! We do not know if $a > b$ or $b > a$, so we must keep the modulus:

$$|\vec{v}| = \frac{|a-b|}{\sqrt{a^2+b^2}}$$

The problem gives us a beautiful constraint: $|\vec{u}| + |\vec{v}| = |\vec{w}|$. Since $|\vec{w}| = \sqrt{1^2 + 1^2} = \sqrt{2}$, our equation becomes:

$$\frac{a+b+|a-b|}{\sqrt{a^2+b^2}} = \sqrt{2}$$

### The Area Constraint

We have one equation but two variables. We need more information. The problem provides the area of the parallelogram, which is $8$.

The area of a parallelogram defined by vectors $\vec{A}$ and $\vec{B}$ is the magnitude of their cross product, $|\vec{A} 	imes \vec{B}|$. Calculating $\vec{PQ} 	imes \vec{PS} = (a\hat{i} + b\hat{j}) 	imes (a\hat{i} - b\hat{j})$, we find the result is $-2ab\hat{k}$.

The magnitude is $2ab$. Thus, $2ab = 8$, which simplifies to:

$$ab = 4$$

### Solving the Mystery

Now, we return to our projection equation. We must analyze the modulus $|a-b|$.

If we assume $a \geq b$, the equation simplifies to:

$$\frac{2a}{\sqrt{a^2+b^2}} = \sqrt{2}$$

Squaring both sides yields $4a^2 = 2(a^2+b^2)$, which simplifies to $2a^2 = 2b^2$, or $a=b$. If we assume $a < b$, we get a contradiction. Therefore, **$a=b$ is the only solution**.

With $ab=4$ and $a=b$, we find $a^2=4$, so **$a=2$ and $b=2$**.

### The Final Verification

We have found $a=2$ and $b=2$. Let us check the options:

*   Option (A) $a+b = 2+2 = 4$, which is **TRUE**.
*   Option (B) $a-b = 0 
eq 2$, which is **FALSE**.
*   For Option (C), the diagonal $\vec{PR} = \vec{PQ} + \vec{PS} = (2\hat{i} + 2\hat{j}) + (2\hat{i} - 2\hat{j}) = 4\hat{i}$. The length is $4$, so Option (C) is **TRUE**.
*   Finally, for Option (D), the angle bisector of two vectors of equal magnitude lies along their sum, which is $4\hat{i}$ (the x-axis). Since $\vec{w} = \hat{i} + \hat{j}$ is not on the x-axis, Option (D) is **FALSE**.

You have successfully navigated the vector space! Keep this clarity of thought, and no problem will ever be too complex.

Select Answer:

Visualized Solution

Vectors $P Q$ and $P S$

Vector $w$ and Projections

Magnitude of Projection

Evaluating $∣ u ∣$

Evaluating $∣ v ∣$

The Main Equation

Area of Parallelogram

Calculating Cross Product

Using the Area Value

Handling the Modulus (Case 1)

Finding Values of $a$ and $b$

What about Case 2?

Checking Options (A) and (B)

Checking Option (C): Diagonal

Checking Option (D) & Conclusion

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Algebraic Dance

The Area Constraint

Solving the Mystery

The Final Verification

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