Vectors & 3D Geometry
Plane through line intersection; minimum cross-section area
MJAT_TS2_P1
Grade 12
Question:
Consider a plane $P_1$ through the intersection of $x-y-z-4=0$ and $x+y+2z-4=0$, parallel to the line of intersection of $2x+3y+z=1$ and $x+3y+2z=2$. Express $P_1$ as $x+\alpha y+\beta z+\gamma=0$.
Four parallel lines $L_1,L_2,L_3,L_4$ (parallel to $\vec{u}=2\hat{i}+\hat{j}+2\hat{k}$) pass through $(3,3,3)$, $(3,3,0)$, $(0,3,0)$, $(0,3,3)$ respectively. A second plane $P_2$ intersects them at $V_1,V_2,V_3,V_4$. Let $S$ be the minimum area of quadrilateral $V_1V_2V_3V_4$.
Which statements are correct?
A) $|\alpha+\beta+\gamma|=11$
B) Minimum area $S=5$
C) Normal to $P_1$ is parallel to $2\hat{i}-\hat{j}-3\hat{k}$
D) $|\alpha+\beta+\gamma|+S=14$
Step-by-Step Solution
Key Concept: From the family of planes through the intersection, use $\lambda=-1/2$ to get $P_1: x-3y-4z-4=0$, so $\alpha=-3,\beta=-4,\gamma=-4$ and $|\alpha+\beta+\gamma|=11$ (A). The four lines form a rectangle in $y=3$ plane with area $9$; projected onto $P_2\perp\hat{u}$: min area $S=9|\hat{n}\cdot\hat{u}|=9\cdot|\frac{1}{3}|=3$.
$\lambda=-1/2$: $P_1: x-3y-4z-4=0$. $|\alpha+\beta+\gamma|=|-3-4-4|=11$ (A ✓). Projection: $S=|\vec{A}\cdot\hat{u}|=|9\hat{j}\cdot\frac{1}{3}(2,1,2)|=3$ (B ✗). C: Normal $(1,-3,-4)$ not parallel to $(2,-1,-3)$ (C ✗). D: $11+3=14$ ✓.
Correct Answer: AD