Vector Algebra
Parallelogram with Diagonal Conditions
Grade 12
Question:
<p>Let \(\overrightarrow{PQ}=-2\hat{a}+\hat{b}\) and
\(\overrightarrow{PS}=3\hat{a}-4\hat{b}\), where \(\hat{a}\) and \(\hat{b}\)
are mutually perpendicular unit vectors. The area of parallelogram PQRS is:</p>
<li>\(5\)</li>
<li>\(10\)</li>
<li>\(11\)</li>
<li>\(\sqrt{155}\)</li>
Step-by-Step Solution
Key Concept: Area = |PQ \times PS|. For perpendicular unit vectors â,b̂, |â \times b̂| = 1.
\(\overrightarrow{PQ}\times\overrightarrow{PS}
=(-2\hat{a}+\hat{b})\times(3\hat{a}-4\hat{b})\)
\(=-6(\hat{a}\times\hat{a})+8(\hat{a}\times\hat{b})+3(\hat{b}\times\hat{a})-4(\hat{b}\times\hat{b})\)
\(=8(\hat{a}\times\hat{b})-3(\hat{a}\times\hat{b})=5(\hat{a}\times\hat{b})\).
\(|\overrightarrow{PQ}\times\overrightarrow{PS}|=5|\hat{a}\times\hat{b}|=5\cdot1=\boxed{5}\).
Correct Answer: A