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?

$P_1: 2x+y-z=3$

$P_2: x+2y+z=2$

$\text{To find: which of statements } (1)\text{--}(4) \text{ are TRUE}$

$\vec{n_1}=(2,1,-1),\quad \vec{n_2}=(1,2,1)$

d⃗ = n₁⃗ × n₂⃗ = î ĵ k̂ 2 1 -1 1 2 1

$\vec{d} = (3,-3,3) = 3(1,-1,1)$

$\text{Plan: check (1) ratios, (2) }\perp,\ (3)\ \angle,\ (4)\ P_3,\ d$

$(1,2,-1) \not\propto (1,-1,1) \Rightarrow (1)\ \text{FALSE}$

$\dfrac{3x-4}{9}=\dfrac{x-\frac{4}{3}}{3}$

$\Rightarrow \text{direction } (3,-3,3) \propto (1,-1,1)$

$(1,-1,1)\cdot(1,-1,1)=3 \neq 0 \Rightarrow (2)\ \text{FALSE}$

$\cos\theta=\dfrac{|\vec{n_1}\cdot\vec{n_2}|}{|\vec{n_1}|\,|\vec{n_2}|}$

$\cos\theta=\dfrac{|3|}{\sqrt{6}\cdot\sqrt{6}}=\dfrac{3}{6}=\dfrac{1}{2}$

$\Rightarrow \theta=60^\circ \Rightarrow (3)\ \text{TRUE}$

$\vec{n_3}=(1,-1,1) \parallel \vec{d}$

$P_3: (1)(x-4) - (1)(y-2) + (1)(z+2) = 0$

$\Rightarrow x - y + z = 0$

$d=\dfrac{|Ax_0+By_0+Cz_0-D|}{\sqrt{A^2+B^2+C^2}}$

$d=\dfrac{|2-1+1-0|}{\sqrt{3}}=\dfrac{2}{\sqrt{3}}$

$\Rightarrow (4)\ \text{TRUE}$

$\text{TRUE: } (3)\ \text{and}\ (4)$