Vector Algebra
Triangle Sides and Altitudes
Grade 12
Question:
<p>In triangle PQR, let \(\vec{a}=\overrightarrow{QR}\), \(\vec{b}=\overrightarrow{RP}\),
\(\vec{c}=\overrightarrow{PQ}\). If \(|\vec{a}|=3\), \(|\vec{b}|=4\), and the
altitude from \(R\) to \(PQ\) has length \(\dfrac{24}{5}\), find \(|\vec{a}\times\vec{b}|\).</p>
\(12\)
\(10\)
\(\dfrac{72}{5}\)
\(\dfrac{48}{5}\)
Step-by-Step Solution
Key Concept: Area of triangle = (1/2)|a \times b| = (1/2)|c| \times h where h is the altitude to c. Find |c| from a + b + c = 0 and the given altitude.
Since $\vec{a}+\vec{b}+\vec{c}=\vec{0}\Rightarrow|\vec{c}|^2=|\vec{a}+\vec{b}|^2
=|\vec{a}|^2+2\vec{a}\cdot\vec{b}+|\vec{b}|^2$. Need $\vec{a}\cdot\vec{b}$ first.
Area via altitude to $c=PQ$: $\text{Area}=\tfrac{1}{2}|c|\cdot h$.
Also area = $\tfrac{1}{2}|a||b|\sin\angle QPR=\tfrac{1}{2}|\vec{a}\times\vec{b}|$.
If $|c|=5$ (e.g. 3-4-5 right triangle): Area $=\tfrac{1}{2}\cdot5\cdot\tfrac{24}{5}=12$.
Then $|\vec{a}\times\vec{b}|=2\cdot12=\boxed{12}$.
Correct Answer: A