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

$O = (0,0,0);\quad OP \text{ along } x\text{-axis},\ OR \text{ along } y\text{-axis}$ $OPQR \text{ is a square}, OP = 3;\quad S \text{ above } T, TS = 3$ $\text{Find: which of (1)--(4) is/are true}$ $1.\ O,P,Q,R,S,T$ $2.\ \angle(OQ, OS)$ $3.\ \text{plane } OQS,\ \text{dist}(P,\Pi),\ \text{dist}(O, RS)$ $P(3,0,0),\quad R(0,3,0)$ $Q = P + R = (3,3,0)$ $T = \left(\dfrac32,\dfrac32,0\right),\quad S = \left(\dfrac32,\dfrac32,3\right)$ $\vec{OQ} = (3,3,0),\quad \vec{OS} = \left(\dfrac32,\dfrac32,3\right)$ $\vec{OQ}\cdot\vec{OS} = \dfrac92 + \dfrac92 + 0 = 9$ $|OQ| = 3\sqrt2,\quad |OS| = \dfrac{3\sqrt6}{2}$ $\cos\theta = \dfrac{9}{3\sqrt2 \cdot \frac{3\sqrt6}{2}} = \dfrac{1}{\sqrt3} \neq \dfrac12 = \cos\dfrac{\pi}{3}$ $\vec{OQ}\times\vec{OS} = \begin{vmatrix} \hat i & \hat j & \hat k \\ 3 & 3 & 0 \\ \frac32 & \frac32 & 3 \end{vmatrix}$ $= 9\hat i - 9\hat j + 0\hat k = 9(\hat i - \hat j)$ $\vec n = (1,-1,0)\ \text{through } O \Rightarrow x - y = 0$ $\Rightarrow (2) \text{ TRUE}$ $\text{dist}(P,\Pi) = \dfrac{|3-0|}{\sqrt{1^2+1^2}} = \dfrac{3}{\sqrt2}$ $\Rightarrow (3) \text{ TRUE}$ $RS: R(0,3,0),\quad \vec d = S - R = \left(\dfrac32,-\dfrac32,3\right) \parallel (1,-1,2)$ $d = \dfrac{|\vec{RO} \times \vec d|}{|\vec d|} = \dfrac{|(0,-3,0)\times(1,-1,2)|}{\sqrt6}$ $= \dfrac{\sqrt{45}}{\sqrt6} = \dfrac{3\sqrt5}{\sqrt6} = \dfrac{\sqrt{30}}{2} \neq \dfrac{\sqrt{15}}{2}$ $\boxed{(2)\ \text{and}\ (3)\ \text{are TRUE}}$