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>
\(5\)
\(10\)
\(11\)
\(\sqrt{155}\)
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