AP Statistics

Consider three planes P₁ : x-y+z=1;, P₂ : x+y-z=-1;, P₃ : x-3y+3z=2 Let L₁, L₂, L₃ be the lines of intersection of the planes P₂ and P₃, P₃ and P₁, P₁ and P₂, respectively. Statement-I : At least two of the lines L₁, L₂ and L₃ are non-parallel and Statement-II : The three planes do not have a common point.

$P_1: x-y+z=1$

$P_2: x+y-z=-1$

$P_3: x-3y+3z=2$

$L_1=P_2\cap P_3,\quad L_2=P_3\cap P_1,\quad L_3=P_1\cap P_2$

$\text{Check Statement-I and Statement-II, then choose the option}$

$\vec{n}_1=(1,-1,1),\quad \vec{n}_2=(1,1,-1),\quad \vec{n}_3=(1,-3,3)$

$L_1=P_2\cap P_3,\quad L_2=P_3\cap P_1,\quad L_3=P_1\cap P_2$

$\text{Plan: } 1.\ \vec{d}_{L_i}=\vec{n}\times\vec{n}\ (\text{the two planes' normals})$

$2.\ \text{compare the three directions} \Rightarrow \text{Statement-I}$

$3.\ \text{solve the system} \Rightarrow \text{Statement-II}$

d⃗_(L₁)=n⃗₂×n⃗₃= î ĵ k̂ 1 1 -1 1 -3 3

$\vec{d}_{L_1}=(0,-4,-4)\parallel(0,1,1)$

$\vec{d}_{L_2}=\vec{n}_3\times\vec{n}_1=(0,2,2)\parallel(0,1,1)$

$\vec{d}_{L_3}=\vec{n}_1\times\vec{n}_2=(0,2,2)\parallel(0,1,1)$

$\vec{d}_{L_1}\parallel\vec{d}_{L_2}\parallel\vec{d}_{L_3}=(0,1,1)$

$\Rightarrow\ \text{all three lines are parallel}\ \Rightarrow\ \text{Statement-I is FALSE}$

$x-y+z=1$

$x+y-z=-1$

$x-3y+3z=2$

$(x-y+z)+(x+y-z)=1+(-1)\ \Rightarrow\ 2x=0$

$2x=0 \Rightarrow x=0$

$z=1+y$

$-3y+3(1+y)=2$

$3=2\ \Rightarrow\ \text{contradiction}$

$\Rightarrow\ \text{no common point}\ \Rightarrow\ \text{Statement-II is TRUE}$

$\text{I: False, II: True}\ \Rightarrow\ \text{option }(4)$

$\boxed{(4)}$