3D Geometry
Plane normal to a vector; equation of plane
nta_pyq_2023_jan
Grade 12
Question:
A vector $\vec{v}$ in the first octant is inclined to the x-axis at $60°$, to the y-axis at $45°$ and to the z-axis at an acute angle. If a plane passing through the points $(\sqrt{2}, -1, 1)$ and $(a, b, c)$, is normal to $\vec{v}$, then
\sqrt{2}a + b + c = 1
a + b + \sqrt{2}c = 1
a + \sqrt{2}b + c = 1
\sqrt{2}a - b + c = 1
Step-by-Step Solution
Key Concept: Find $\vec{v}$ from direction cosines, write plane equation through $(\sqrt{2},-1,1)$ with normal $\vec{v}$.
$\vec{v} \propto (1, \sqrt{2}, 1)$. Plane through $(\sqrt{2},-1,1)$: $(x-\sqrt{2})+\sqrt{2}(y+1)+(z-1)=0 \Rightarrow x+\sqrt{2}y+z=1$. So $(a,b,c)$ satisfies $a+\sqrt{2}b+c=1$. Answer: (3)
Correct Answer: $a + \sqrt{2}b + c = 1$