Cut a cuboid and check where a perpendicular drill exits

From Figure 1, $O=(0,0,0)$, $D=(0,0,30)$, $E=(0,10,30)$, $A=(0,10,0)$ and $F=(-20,10,30)$. Moving from A towards B decreases x. The Teacher completions give

$\begin{aligned}\overrightarrow{OP}&=(0,5,30),\\\overrightarrow{OQ}&=(0,10,15),\\\overrightarrow{OR}&=(-10,10,30).\end{aligned}$

Subtract the base point P:

$\begin{aligned}\overrightarrow{PQ}&=(0,10,15)-(0,5,30)\\&=(0,5,-15),\\\overrightarrow{PR}&=(-10,10,30)-(0,5,30)\\&=(-10,5,0).\end{aligned}$

These are displacement vectors in centimetres, not the position vectors of Q and R.

Scale the two directions from (i) by $1/5$ before taking their cross product:

$\begin{aligned}&\mathbf n\\ & =(0,1,-3)\times(-2,1,0)\\ & =(3,6,2).\end{aligned}$

The component calculation is $n_x=1(0)-(-3)(1)=3$, $n_y=(-3)(-2)-0(0)=6$, $n_z=0(1)-1(-2)=2$. Use Q to find the constant, as in the source:

$\begin{aligned}\mathbf r\cdot(3,6,2)&=(0,10,15)\cdot(3,6,2)\\&=90.\end{aligned}$

Hence $3x+6y+2z=90$. P and R also satisfy this equation, which checks that the plane contains the whole triangle.

The plane DEFG is the horizontal top face $z=30$, with normal $(0,0,1)$, completing the Teacher blank. For the acute plane angle,

$\begin{aligned}\cos\theta&=\frac{|(0,0,1)\cdot(3,6,2)|}{\sqrt{9+36+4}}\\&=\frac{|2|}{\sqrt{49}}=\frac27,\\\theta&=\cos^{-1}(2/7)=73.4^\circ\ \text{(to 1 d.p.)}.\end{aligned}$

Use cosine for the angle between two planes.

First average the endpoint position vectors of QR:

$\begin{aligned}\overrightarrow{OS}&=\tfrac12[(0,10,15)+(-10,10,30)]\\&=(-5,10,45/2).\end{aligned}$

Since $PT:TS=4:1$, T is four fifths of the way from P to S. The source section formula is

$\begin{aligned}\overrightarrow{OT}&=\frac{4\overrightarrow{OS}+\overrightarrow{OP}}{4+1}\\&=\frac15(-20,45,120)\\&=(-4,9,24).\end{aligned}$

The larger weight goes on S, since T lies closer to S. As a check, $3(-4)+6(9)+2(24)=90$, so T lies in the cut plane.

The drill passes through T and is perpendicular to triangle PQR, so its direction is the plane normal $(3,6,2)$ from (ii). This gives the printed vector equation. The line allows every real parameter; the physical drill travels from T into the remaining cuboid in the negative-$\lambda$ direction.

A point on the drill line is $(-4+3\lambda,9+6\lambda,24+2\lambda)$. The cuboid occupies

$-20\le x\le0,\quad0\le y\le10,\quad0\le z\le30.$

Case 1: OABC. Its supporting plane is $z=0$, or $\mathbf r\cdot(0,0,1)=0$. Substitution gives

$\begin{aligned}24+2\lambda&=0,\\\lambda&=-12,\\\mathbf r&=(-40,-63,0).\end{aligned}$

This point lies outside the finite bottom face because its x and y coordinates fail the cuboid bounds. Thus the drill does not exit through OABC, even though its infinite line intersects the infinite plane $z=0$.

Case 2: ODGC. Its supporting plane is $y=0$, or $\mathbf r\cdot(0,1,0)=0$:

$\begin{aligned}9+6\lambda&=0,\\\lambda&=-\frac32,\\\mathbf r&=(-\tfrac{17}{2},0,21).\end{aligned}$

Here $-20\le-17/2\le0$ and $0\le21\le30$, so the point lies on the finite face ODGC. It is reached from T in the inward direction before $\lambda=-12$. Along $-3/2\le\lambda\le0$, all cuboid bounds hold; also $3x+6y+2z=90+49\lambda\le90$, so the segment stays in the remaining solid after E is cut away. Hence ODGC is the exit face.

Source correction. In the first case the source calls an intermediate intersection “plane ODGC”, although the case heading, normal $(0,0,1)$ and equation are for OABC. The calculation above uses OABC consistently; the original wording remains in the native transcription.