Consider a pyramid OPQRS located in the first octant ($x \geq 0$, $y \geq 0$, $z \geq 0$) with O as origin, and OP and OR along the x–axis and the y–axis, respectively. The base OPQR of the pyramid is a square with $OP = 3$. The point S is directly above the mid-point, T of diagonal OQ such that $TS = 3$. Then

$\text{Given: pyramid OPQRS in first octant, } O = \text{origin}$

OP ∥ x-axis, OR ∥ y-axis, OPQR square

$OP = 3;\quad S \text{ above } T \text{ (midpoint of } OQ),\quad TS = 3$

$\text{To check: statements (1), (2), (3), (4)}$

(angle OQ-OS, plane OQS, dist(P, OQS), dist(O, RS)) $O(0,0,0),\quad P(3,0,0),\quad R(0,3,0)$ $Q = P + R = (3,3,0),\quad T = \text{mid}(OQ) = \left(\tfrac32,\tfrac32,0\right)$ $S = \left(\tfrac32,\tfrac32,3\right)$ $(1)\ \cos\theta = \dfrac{\vec{OQ}\cdot\vec{OS}}{|\vec{OQ}|\,|\vec{OS}|}$ $(2)\ \text{normal of plane } OQS = \vec{OQ}\times\vec{OS}$ $(3)\ \text{dist}(P,\text{plane }OQS) = \dfrac{|\vec{OP}\cdot\vec{n}|}{|\vec{n}|},\quad (4)\ \text{dist}(O,RS) = \dfrac{|\vec{RO}\times\vec{RS}|}{|\vec{RS}|}$ $\vec{OQ} = (3,3,0),\qquad \vec{OS} = \left(\tfrac32,\tfrac32,3\right)$ $\vec{OQ}\cdot\vec{OS} = \tfrac92+\tfrac92+0 = 9$ $|\vec{OQ}| = 3\sqrt2,\qquad |\vec{OS}| = \sqrt{\tfrac{27}{2}} = \dfrac{3\sqrt3}{\sqrt2}$ $\cos\theta = \dfrac{9}{3\sqrt2\cdot\frac{3\sqrt3}{\sqrt2}} = \dfrac{1}{\sqrt3}$ $\theta \approx 54.7^\circ \neq \dfrac{\pi}{3} \Rightarrow (1)\ \text{false}$ $\vec{n} = \vec{OQ}\times\vec{OS} = (9,\ -9,\ 0) \parallel (1,\ -1,\ 0)$ $\vec{n}\cdot(x,y,z) = 0 \Rightarrow x - y = 0 \Rightarrow (2)\ \text{true}$ $\text{dist}(P,\, x-y=0) = \dfrac{|3-0-0|}{\sqrt{1^2+(-1)^2}} = \dfrac{3}{\sqrt2} \Rightarrow (3)\ \text{true}$ $\vec{RS} = \left(\tfrac32,\ -\tfrac32,\ 3\right),\qquad \vec{RO} = O - R = (0,\ -3,\ 0)$ $\text{dist}(O, RS) = \dfrac{|\vec{RO}\times\vec{RS}|}{|\vec{RS}|}$ $\vec{RO}\times\vec{RS} = (-9,\ 0,\ \tfrac92),\qquad |\vec{RO}\times\vec{RS}| = \sqrt{81+\tfrac{81}{4}} = \dfrac{9\sqrt5}{2}$ $|\vec{RS}| = \sqrt{\tfrac94+\tfrac94+9} = \sqrt{\tfrac{27}{2}} = \dfrac{3\sqrt6}{2}$ $\text{dist}(O, RS) = \dfrac{9\sqrt5/2}{3\sqrt6/2} = \dfrac{3\sqrt5}{\sqrt6} = \dfrac{\sqrt{30}}{2} = \sqrt{\dfrac{15}{2}} \Rightarrow (4)\ \text{true}$ $\boxed{\text{Answer: (2), (3), (4)}}$