Vectors & 3D Geometry
Area of region in plane with inequality constraints
MJAT_TS5_P1
Grade 12
Question:
Let $\alpha$ be the region in the plane $x+y+z=1$ with $x,y,z\geq 0$. Let $\beta$ be the set of points $(a,b,c)\in\alpha$ such that only two of the three inequalities $a\leq\frac{1}{2}$, $b\leq\frac{1}{3}$, $c\leq\frac{1}{6}$ hold simultaneously. If $A(R)$ denotes area of region $R$, then:
A) $A(\alpha)=\dfrac{\sqrt{3}}{4}$
B) $A(\alpha)=\dfrac{\sqrt{3}}{2}$
C) $A(\beta)=\dfrac{7\sqrt{3}}{36}$
D) $A(\beta)=\dfrac{11\sqrt{3}}{36}$
Step-by-Step Solution
Key Concept: $\alpha$ is the equilateral triangle in plane $x+y+z=1$ with vertices $(1,0,0)$, $(0,1,0)$, $(0,0,1)$. Side length $=\sqrt{2}$. $A(\alpha)=\frac{\sqrt{3}}{4}(\sqrt{2})^2=\frac{\sqrt{3}}{2}$ (B ✓). For $\beta$: compute area where exactly 2 inequalities hold using inclusion-exclusion.
B ✓ ($A(\alpha)=\sqrt{3}/2$), D ✓ ($A(\beta)=11\sqrt{3}/36$). Answer: B, D.
Correct Answer: BD