Projection Vectors and Area of Parallelogram

To determine which of the given options are TRUE, we analyze the given vectors and conditions step-by-step.

Step 1: Calculate the Area of Parallelogram $PQRS$

The adjacent side vectors of the parallelogram $PQRS$ are given as: $\vec{PQ} = a\hat{i} + b\hat{j}$ $\vec{PS} = a\hat{i} - b\hat{j}$

The area of the parallelogram is given by the magnitude of the cross product of its adjacent side vectors: $\text{Area} = |\vec{PQ} \times \vec{PS}|$

Computing the cross product: $\vec{PQ} \times \vec{PS} = (a\hat{i} + b\hat{j}) \times (a\hat{i} - b\hat{j}) = -ab\hat{k} - ab\hat{k} = -2ab\hat{k}$

Since the total area of $PQRS$ is equal to $8$: $\text{Area} = |-2ab\hat{k}| = 2ab = 8 \implies ab = 4 \quad \text{--- (Equation 1)}$

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Step 2: Calculate the Projections $|\vec{u}|$ and $|\vec{v}|$, and solve for $a$ and $b$

The vector $\vec{w} = \hat{i} + \hat{j}$ has magnitude: $|\vec{w}| = \sqrt{1^2 + 1^2} = \sqrt{2}$

1. The projection vector $\vec{u}$ of $\vec{w}$ along $\vec{PQ}$: $|\vec{u}| = \frac{|\vec{w} \cdot \vec{PQ}|}{|\vec{PQ}|} = \frac{|(\hat{i} + \hat{j}) \cdot (a\hat{i} + b\hat{j})|}{\sqrt{a^2 + b^2}} = \frac{a + b}{\sqrt{a^2 + b^2}} \quad (\text{since } a, b > 0)$

2. The projection vector $\vec{v}$ of $\vec{w}$ along $\vec{PS}$: $|\vec{v}| = \frac{|\vec{w} \cdot \vec{PS}|}{|\vec{PS}|} = \frac{|(\hat{i} + \hat{j}) \cdot (a\hat{i} - b\hat{j})|}{\sqrt{a^2 + b^2}} = \frac{|a - b|}{\sqrt{a^2 + b^2}}$

Given that $|\vec{u}| + |\vec{v}| = |\vec{w}|$: $\frac{a + b}{\sqrt{a^2 + b^2}} + \frac{|a - b|}{\sqrt{a^2 + b^2}} = \sqrt{2}$ $a + b + |a - b| = \sqrt{2(a^2 + b^2)}$

Now, we consider two cases for $|a - b|$:

• Case 1: $a \ge b \implies |a - b| = a - b$ $a + b + a - b = \sqrt{2(a^2 + b^2)}$ $2a = \sqrt{2(a^2 + b^2)}$ Squaring both sides: $4a^2 = 2(a^2 + b^2) \implies 2a^2 = a^2 + b^2 \implies a^2 = b^2$ Since $a, b > 0$, we get $a = b$.

• Case 2: $a < b \implies |a - b| = b - a$ $a + b + b - a = \sqrt{2(a^2 + b^2)}$ $2b = \sqrt{2(a^2 + b^2)}$ Squaring both sides: $4b^2 = 2(a^2 + b^2) \implies 2b^2 = a^2 + b^2 \implies b^2 = a^2$ Since $a, b > 0$, we get $a = b$.

In both cases, we find that: $a = b$

Using Equation 1 ($ab = 4$): $a^2 = 4 \implies a = 2 \quad \text{and} \quad b = 2$

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Step 3: Evaluate the Options

• Option A: $a + b = 2 + 2 = 4$ Therefore, Option A is TRUE.

• Option B: $a - b = 2 - 2 = 0 \neq 2$ Therefore, Option B is FALSE.

• Option C: The diagonal vector $\vec{PR}$ is given by: $\vec{PR} = \vec{PQ} + \vec{PS} = (2\hat{i} + 2\hat{j}) + (2\hat{i} - 2\hat{j}) = 4\hat{i}$ The length of diagonal $PR$ is: $|\vec{PR}| = |4\hat{i}| = 4$ Therefore, Option C is TRUE.

• Option D: The unit vectors along $\vec{PQ}$ and $\vec{PS}$ are: $\hat{e}_1 = \frac{\vec{PQ}}{|\vec{PQ}|} = \frac{2\hat{i} + 2\hat{j}}{\sqrt{8}} = \frac{\hat{i} + \hat{j}}{\sqrt{2}}$ $\hat{e}_2 = \frac{\vec{PS}}{|\vec{PS}|} = \frac{2\hat{i} - 2\hat{j}}{\sqrt{8}} = \frac{\hat{i} - \hat{j}}{\sqrt{2}}$ The internal angle bisector of $\vec{PQ}$ and $\vec{PS}$ is in the direction of: $\hat{e}_1 + \hat{e}_2 = \frac{2\hat{i}}{\sqrt{2}} = \sqrt{2}\hat{i}$ This is parallel to the x-axis ($\hat{i}$). However, $\vec{w} = \hat{i} + \hat{j}$ points along $\vec{PQ}$ and is not parallel to $\hat{i}$. Therefore, Option D is FALSE.

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Conclusion

The correct options are A and C.