3D Geometry
Planes
MMTS_Full_Test_08
Grade 12

Question:

The equation of the plane containing line $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ and perpendicular to plane $4x+5y-3z-8=0$ is
$7x+5y-z=0$
$5x+y+3z=0$
$7x-5y+z=0$
$3x-5y+z=0$

Step-by-Step Solution

Key Concept: Plane through origin (line passes through origin); normal $\perp$ to direction $(1,2,3)$ and to $(4,5,-3)$
Normal $=(1,2,3)\times(4,5,-3)=(-6-15,12+3,5-8)=(-21,15,-3)$. Plane: $-21x+15y-3z=0\Rightarrow 7x-5y+z=0$.
Correct Answer: 4

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