3D Geometry

Question 1

(a) Direction ratios $\langle 2,-1,2\rangle$ and $\langle 1,3,-1\rangle$ are not proportional ($2/1 \ne -1/3$), so not parallel. (1)
$\vec{PQ}=(1,2,2)$. Compute scalar triple product $[\vec{PQ}\ \vec{dA}\ \vec{dB}]$: $\begin{vmatrix} 1&2&2\\ 2&-1&2\\ 1&3&-1\end{vmatrix} = 1(1-6)-2(-2-2)+2(6+1)=-5+8+14=17\ne0.$ (1)
Nonzero ⇒ lines do not intersect ⇒ skew. (1)

(b) $\vec{dA}\times\vec{dB} = \begin{vmatrix} i&j&k\\2&-1&2\\1&3&-1\end{vmatrix} = i(1-6)-j(-2-2)+k(6+1)=(-5,4,7)$. (2)
$|\vec{dA}\times\vec{dB}| = \sqrt{25+16+49}=\sqrt{90}=3\sqrt{10}$. (1)
$d=\frac{|[\vec{PQ}\ \vec{dA}\ \vec{dB}]|}{|\vec{dA}\times\vec{dB}|}=\frac{17}{3\sqrt{10}}=\frac{17\sqrt{10}}{30}\approx1.792.$ (2)

(c) Points: $R_A=P+s\vec{dA}=(1+2s,2-s,3+2s)$, $R_B=Q+t\vec{dB}=(2+t,4+3t,5-t)$. (1)
$\vec{R_AR_B}=(1+t-2s,\;2+3t+s,\;2-t-2s)$ must be $\perp$ both directions.
$\vec{R_AR_B}\cdot\vec{dA}=0:\;2(1+t-2s)-(2+3t+s)+2(2-t-2s)=0 \Rightarrow 4-9s-3t=0$. (1)
$\vec{R_AR_B}\cdot\vec{dB}=0:\;(1+t-2s)+3(2+3t+s)-(2-t-2s)=0 \Rightarrow 5+3s+11t=0$. (1)
Solve: from first $3t=4-9s$. Substitute into second: $5+3s+11t=0 \Rightarrow$ multiply first-derived... Solve system $9s+3t=4$, $3s+11t=-5$.
From these: $9s+3t=4$; $3s+11t=-5 \Rightarrow 9s+33t=-15$. Subtract: $30t=-19\Rightarrow t=-19/30$. Then $9s=4-3t=4+57/30=177/30\Rightarrow s=177/270=59/90$. (1)
$R_A=(1+2(59/90),\,2-59/90,\,3+2(59/90))=(104/45,\,121/90,\,194/45)$.
$R_B=(2-19/30,\,4-57/30,\,5+19/30)=(41/30,\,63/30,\,169/30)$. (1)
Midpoint (detector) $=\tfrac12(R_A+R_B)$. Numerically $R_A\approx(2.311,1.344,4.311)$, $R_B\approx(1.367,2.100,5.633)$, midpoint $\approx(1.839,1.722,4.972)$. (1)

(d) (4)

import numpy as np
def shortest_dist(P, dA, Q, dB):
    P,dA,Q,dB = map(np.array,(P,dA,Q,dB))
    n = np.cross(dA,dB)
    nn = np.linalg.norm(n)
    if nn < 1e-12:      # parallel: cross product ~ 0
        # fall back to point-line distance
        return np.linalg.norm(np.cross(Q-P,dA))/np.linalg.norm(dA)
    return abs(np.dot(Q-P,n))/nn

Formula divides by $|\vec{dA}\times\vec{dB}|$, which is $0$ for parallel lines (division fails); detect via nn < tol. (1 for detection sentence)

Question 2

(a) Line dir $\vec{b}=\langle3,-2,4\rangle$, normal $\vec{n}=\langle2,-2,1\rangle$. (1)
$\sin\theta = \dfrac{|\vec{b}\cdot\vec{n}|}{|\vec{b}||\vec{n}|}$. $\vec{b}\cdot\vec{n}=6+4+4=14$. (1)
$|\vec{b}|=\sqrt{29}$, $|\vec{n}|=3$. (1)
$\theta=\sin^{-1}\frac{14}{3\sqrt{29}}\approx\sin^{-1}(0.8665)\approx60.06^\circ.$ (1)

(b) Proof: Line is parallel to plane iff its direction $\perp$ normal (so it never turns toward/away from plane) i.e. $aA+bB+cC=0$, AND the line is not contained in the plane, requiring at least one line point off the plane. If a point lies on the plane too, the line lies in the plane (not "parallel and separate"). (3)
Here $\vec{b}\cdot\vec{n}=14\ne0$, so $L$ is not parallel to $\pi$ — it intersects. (2)

(c) Parametrize $L$: $(1+3\lambda,\,-2-2\lambda,\,4\lambda)$. Substitute:
$2(1+3\lambda)-2(-2-2\lambda)+4\lambda=5 \Rightarrow 2+6\lambda+4+4\lambda+4\lambda=5 \Rightarrow 6+14\lambda=5\Rightarrow\lambda=-\tfrac{1}{14}$. (2)
Point $=(1-3/14,\,-2+2/14,\,-4/14)=\left(\tfrac{11}{14},\,-\tfrac{13}{7},\,-\tfrac{2}{7}\right)$. (2)

(d) Let $P_1(x_1,y_1,z_1)$ be any point on the plane, so $Ax_1+By_1+Cz_1+D=0$. (1)
Distance $=$ projection of $\vec{P_1P_0}$ onto unit normal $\hat n=\frac{(A,B,C)}{\sqrt{A^2+B^2+C^2}}$: $d=|\vec{P_1P_0}\cdot\hat n|=\frac{|A(x_0-x_1)+B(y_0-y_1)+C(z_0-z_1)|}{\sqrt{A^2+B^2+C^2}}.$ (2)
Using $Ax_1+By_1+Cz_1=-D$: $d=\frac{|Ax_0+By_0+Cz_0+D|}{\sqrt{A^2+B^2+C^2}}.$ (1)

Question 3

(a) $\vec{AB}=(3,4,-4)$, $|\vec{AB}|=\sqrt{9+16+16}=\sqrt{41}$. (1)
DCs: $l=\tfrac{3}{\sqrt{41}},\,m=\tfrac{4}{\sqrt{41}},\,n=\tfrac{-4}{\sqrt{41}}$. (2)
Check: $\frac{9+16+16}{41}=\frac{41}{41}=1$. ✓ (1)

(b) $\vec F = 10\hat u = \frac{10}{\sqrt{41}}(3,4,-4) = \left(\tfrac{30}{\sqrt{41}},\tfrac{40}{\sqrt{41}},\tfrac{-40}{\sqrt{41}}\right)\approx(4.685,6.247,-6.247)\,$N. (3)

(c) $\cos\alpha=l,\cos\beta=m,\cos\gamma=n$ and by definition $l^2+m^2+n^2=1$, hence $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$. (2)
Then $\sin^2\alpha+\sin^2\beta+\sin^2\gamma=(1-\cos^2\alpha)+(1-\cos^2\beta)+(1-\cos^2\gamma)=3-1=2$. (2)

(d) Rod1 dir $\langle3,4,-4\rangle$, Rod2 dir $\langle1,-2,2\rangle$. (1)
$\cos\theta=\dfrac{|3-8-8|}{\sqrt{41}\sqrt{9}}=\dfrac{13}{3\sqrt{41}}\approx0.6768$. (2)
$\theta\approx47.4^\circ.$ (1)

[
  {"claim":"Q1 scalar triple product = 17 (skew) and shortest distance = 17/(3*sqrt(10))",
   "code":"PQ=Matrix([1,2,2]);dA=Matrix([2,-1,2]);dB=Matrix([1,3,-1]);stp=PQ.dot(dA.cross(dB));cr=dA.cross(dB);d=Abs(stp)/cr.norm();result=(stp==17) and (simplify(d-Rational(17,1)/(3*sqrt(10)))==0)"},
  {"claim":"Q2a sin(theta)=14/(3*sqrt(29))",
   "code":"b=Matrix([3,-2,4]);n=Matrix([2,-2,1]);s=Abs(b.dot(n))/(b.norm()*n.norm());result=simplify(s-Rational(14,1)/(3*sqrt(29)))==0"},
  {"claim":"Q2c intersection param lambda=-1/14",
   "code":"lam=symbols('lam');expr=2*(1+3*lam)-2*(-2-2*lam)+4*lam-5;sol=solve(expr,lam)[0];result=sol==Rational(-1,14)"},
  {"claim":"Q3a direction cosines squared sum to 1",
   "code":"result=simplify((Rational(3)**2+Rational(4)**2+Rational(4)**2)/41-1)==0"},
  {"claim":"Q3d cos theta between rods = 13/(3*sqrt(41))",
   "code":"u=Matrix([3,4,-4]);v=Matrix([1,-2,2]);c=Abs(u.dot(v))/(u.norm()*v.norm());result=simplify(c-Rational(13,1)/(3*sqrt(41)))==0"}
]