Vectors & 3D Geometry
Plane parallel to a line at given distance — max distance from point
MJAT_TS8_P1
Grade 12
Question:
The equation of the plane $P_1$ parallel to line $L: \frac{x-1}{2}=\frac{y-0}{3}=\frac{z-1}{1}$, at distance 6 from the intersection of $L$ with plane $P: 2x+3y+z-18=0$, and having maximum distance from $Q(5,4,-4)$, is $ax+by+cz+d=0$. The value of $\dfrac{bd}{ac}$ is:
A) $-\dfrac{7}{3}$
B) $\dfrac{7}{3}$
C) $\dfrac{15}{7}$
D) $-\dfrac{15}{7}$
Step-by-Step Solution
Key Concept: Find intersection $P$ of $L$ with plane $P$: parametrise $L=(1+2\lambda,3\lambda,1+\lambda)$ and substitute into $P$. The plane $P_1$ has normal $\vec{PQ}$ direction (for max distance from $Q$). Distance condition selects the correct parallel plane.
$\frac{bd}{ac}=-\frac{15}{7}$. Answer: **D**.
Correct Answer: D