Let $P_1 : 2x + y - z = 3$ and $P_2 : x + 2y + z = 2$ be two planes. Then, which of the following statement(s) is (are) TRUE?

Let P₁ : 2x + y - z = 3 and P₂ : x + 2y + z = 2 be two planes. Then, which of the following statement(s) is (are) TRUE?

$\text{Given: } P_1: 2x + y - z = 3, \quad P_2: x + 2y + z = 2$

$\vec{n}_1 = (2, 1, -1), \quad \vec{n}_2 = (1, 2, 1)$

$\text{To find: which of } (1), (2), (3), (4) \text{ is/are TRUE}$

$\text{Two planes meet in a line } L = P_1 \cap P_2$ $\vec{n}_1 = (2,1,-1), \quad \vec{n}_2 = (1,2,1)$ $\text{Plan: } 1.\ \vec{d}_L = \vec{n}_1 \times \vec{n}_2 \Rightarrow (1), (2)$ $2.\ \text{angle between } P_1, P_2 \Rightarrow (3)$ $3.\ P_3 \text{ and distance} \Rightarrow (4)$ $\vec{d}_L = \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 1 & 2 & 1 \end{vmatrix}$ $= \hat{i}(1+2) - \hat{j}(2+1) + \hat{k}(4-1)$ $\vec{d}_L = (3,-3,3) \propto (1,-1,1)$ $(1,-1,1) \neq k(1,2,-1) \Rightarrow (1) \text{ is FALSE}$ $\dfrac{3x-4}{9}=\dfrac{x-\frac43}{3},\quad \dfrac{1-3y}{9}=\dfrac{y-\frac13}{-3}$ $\vec{d}_2 = (3,-3,3) \propto (1,-1,1)$ $\vec{d}_2 \parallel \vec{d}_L: \quad \vec{d}_2 \cdot \vec{d}_L = 3+3+3 = 9 \neq 0$ $\Rightarrow (2) \text{ is FALSE}$ $\cos\theta = \dfrac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1|\,|\vec{n}_2|} = \dfrac{3}{\sqrt6\cdot\sqrt6} = \dfrac12$ $\theta = 60^\circ \Rightarrow (3) \text{ is TRUE}$ $\vec{n}_3 = \vec{d}_L = (1,-1,1)$ $P_3: 1(x-4) - 1(y-2) + 1(z+2) = 0$ $P_3: x - y + z = 0$ $d = \dfrac{|2 - 1 + 1|}{\sqrt{1+1+1}} = \dfrac{2}{\sqrt3}$ $\Rightarrow (4) \text{ is TRUE}$ $\text{TRUE: } (3) \text{ and } (4)$