Vectors
Projection
MJAT None
Grade 12
Question:
Let $a$ and $b$ be positive real numbers. Suppose $\overrightarrow{PQ} = a\hat{i} + b\hat{j}$ and $\overrightarrow{PS} = a\hat{i} - b\hat{j}$ are adjacent sides of a parallelogram $PQRS$. Let $\overrightarrow{u}$ and $\overrightarrow{v}$ be the projection vectors of $\overrightarrow{w} = \hat{i} + \hat{j}$ along $\overrightarrow{PQ}$ and $\overrightarrow{PS}$, respectively. If $|\overrightarrow{u}| + |\overrightarrow{v}| = |\overrightarrow{w}|$ and if the area of the parallelogram $PQRS$ is 8, then which of the following statements is/are TRUE?
A) $a + b = 4$
B) $a - b = 2$
C) The length of the diagonal $PR$ of the parallelogram $PQRS$ is 4
D) $\overrightarrow{w}$ is an angle bisector of the vectors $\overrightarrow{PQ}$ and $\overrightarrow{PS}$
Step-by-Step Solution
Key Concept: The area condition gives $ ab = 4 $, while the projection condition forces $ a = b $, leading to $ a = b = 2 $.
1. **Area of Parallelogram**: The area is $ |\overrightarrow{PQ} \times \overrightarrow{PS}| = |(a\hat{i} + b\hat{j}) \times (a\hat{i} - b\hat{j})| = | -2ab\hat{k} | = 2ab $. Given area is 8, so $ ab = 4 $.
2. **Projection Magnitudes**:
- Projection of $ \overrightarrow{w} $ along $ \overrightarrow{PQ} $:
$ |\overrightarrow{u}| = \frac{|\overrightarrow{w} \cdot \overrightarrow{PQ}|}{|\overrightarrow{PQ}|} = \frac{|a + b|}{\sqrt{a^2 + b^2}} $.
- Projection of $ \overrightarrow{w} $ along $ \overrightarrow{PS} $:
$ |\overrightarrow{v}| = \frac{|\overrightarrow{w} \cdot \overrightarrow{PS}|}{|\overrightarrow{PS}|} = \frac{|a - b|}{\sqrt{a^2 + b^2}} $.
3. **Equation from Given Condition**:
$ |\overrightarrow{u}| + |\overrightarrow{v}| = \sqrt{2} $. Squaring both sides:
$$
\left( \frac{|a + b| + |a - b|}{\sqrt{a^2 + b^2}} \right)^2 = 2.
$$
Expanding and simplifying leads to $ |a^2 - b^2| = 0 \Rightarrow a = b $.
4. **Solving for $ a $ and $ b $**:
From $ ab = 4 $ and $ a = b $, we get $ a = b = 2 $.
Correct Answer: A, C, D