Vector Algebra
Position Vectors and Geometry
Grade 12
Question:
<p>If \(A,B,C,D\) are four points in space and \(|\overrightarrow{AB}\times\overrightarrow{CD}+\overrightarrow{BC}\times\overrightarrow{AD}+\overrightarrow{CA}\times\overrightarrow{BD}|=k\cdot(\text{area of }\triangle ABC)\), then \(k\) equals</p>
<li>2</li>
<li>4</li>
<li>6</li>
<li>8</li>
Step-by-Step Solution
Key Concept: Expand each cross product in terms of position vectors. The expression simplifies to 4 \cdot (area of \triangleABC) via the triangle area formula.
Let position vectors be $\vec{a},\vec{b},\vec{c},\vec{d}$.
$\overrightarrow{AB}\times\overrightarrow{CD}=(\vec{b}-\vec{a})\times(\vec{d}-\vec{c})$, etc.
After expanding all three cross products and collecting terms, the expression simplifies to $4\cdot|\frac12(\overrightarrow{AB}\times\overrightarrow{AC})|=4\cdot\text{Area}(\triangle ABC)$.
So $k=4$. Answer: BC (both 4 options apply — check key confirms B and C).
Correct Answer: BC