3D Geometry

Q1 [8]

(a) [5] Let $P$ divide $AB$ internally so that $AP:PB=m:n$. Since $A,P,B$ are collinear and $P$ is between them, $\vec{AP}=\dfrac{m}{n}\vec{PB}$. Writing componentwise (position vectors), $P-A=\dfrac{m}{n}(B-P)$ [1]. So $n(P-A)=m(B-P)$ [1] → $nP+mP=mB+nA$ → $P=\dfrac{mB+nA}{m+n}$ [1]. Thus $P=\left(\frac{mx_2+nx_1}{m+n},\ \frac{my_2+ny_1}{m+n},\ \frac{mz_2+nz_1}{m+n}\right)$ [1]. The distance-formula consistency check (equal division of directed length) justifies internal division. [1]

(b) [3] With $m:n=1:2$: $x=\frac{1\cdot8+2\cdot2}{3}=\frac{12}{3}=4,\quad y=\frac{1\cdot0+2(-3)}{3}=-2,\quad z=\frac{1\cdot10+2\cdot4}{3}=\frac{18}{3}=6.$ Point $=(4,-2,6)$. [3] (1 each coordinate)

Q2 [9]

(a) [5] Take the line through origin $O$ with a point $P(x,y,z)$ at distance $r=OP$. The direction cosines are $l=\cos\alpha,\ m=\cos\beta,\ n=\cos\gamma$, angles with the axes [1]. By projection, $x=lr,\ y=mr,\ z=nr$ [2]. Then $x^2+y^2+z^2=r^2$ (distance formula) gives $r^2(l^2+m^2+n^2)=r^2$ [1], hence $l^2+m^2+n^2=1$ [1].

(b) [4] $\alpha=60^\circ\Rightarrow l=\tfrac12$; $\beta=45^\circ\Rightarrow m=\tfrac{1}{\sqrt2}$ [1]. $n^2=1-\tfrac14-\tfrac12=\tfrac14$ [1] → $n=\pm\tfrac12$, so $\gamma=60^\circ$ or $120^\circ$ [1]. One set: $\left(\tfrac12,\tfrac{1}{\sqrt2},\tfrac12\right)$ [1].

Q3 [8]

(a) [4]

• Vector: $\vec r=(\hat i+2\hat j+3\hat k)+t(2\hat i-\hat j+2\hat k)$ [1]
• Parametric: $x=1+2t,\ y=2-t,\ z=3+2t$ [1.5]
• Symmetric: $\dfrac{x-1}{2}=\dfrac{y-2}{-1}=\dfrac{z-3}{2}$ [1.5]

(b) [4] $|\vec b|=\sqrt{4+1+4}=3$ [1]. Distance from $A$ = $|t|\,|\vec b|=3|t|=9$ → $t=\pm3$ [1]. $t=3$: $(7,-1,9)$; $t=-3$: $(-5,5,-3)$ [2].

Q4 [8]

(a) [5] $\vec{PQ}=(-3,-3,3)$, $\vec{PR}=(0,-2,3)$ [1]. Normal $\vec n=\vec{PQ}\times\vec{PR}=\begin{vmatrix}\hat i&\hat j&\hat k\\-3&-3&3\\0&-2&3\end{vmatrix}$ $=\hat i(-9+6)-\hat j(-9-0)+\hat k(6-0)=(-3,9,6)$ [2], simplify $\div3\to(-1,3,2)$ [1]. Plane through $P(1,1,-1)$: $-1(x-1)+3(y-1)+2(z+1)=0$ → $-x+3y+2z=0$ [1].

(b) [3] Plane: $-x+3y+2z=0$, i.e. $D=0$, plane passes through origin [2]. Distance $=0$ [1].

Q5 [10]

(a) [4] The common perpendicular to both lines has direction $\vec b_1\times\vec b_2$ [1]. The shortest distance is the length of the projection of the connecting vector $\vec a_2-\vec a_1$ onto this common perpendicular direction [2]: $d=\frac{|(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)|}{|\vec b_1\times\vec b_2|}.$ [1]

(b) [6] $\vec a_1=(1,2,1),\ \vec b_1=(1,-1,1)$; $\vec a_2=(2,-1,-1),\ \vec b_2=(2,1,2)$. $\vec a_2-\vec a_1=(1,-3,-2)$ [1]. $\vec b_1\times\vec b_2=\begin{vmatrix}\hat i&\hat j&\hat k\\1&-1&1\\2&1&2\end{vmatrix}=\hat i(-2-1)-\hat j(2-2)+\hat k(1+2)=(-3,0,3)$ [2]. Numerator: $(1)(-3)+(-3)(0)+(-2)(3)=-3-6=-9\Rightarrow|{-9}|=9$ [1]. $|\vec b_1\times\vec b_2|=\sqrt{9+0+9}=3\sqrt2$ [1]. $d=\dfrac{9}{3\sqrt2}=\dfrac{3}{\sqrt2}=\dfrac{3\sqrt2}{2}$. Nonzero ⇒ skew. [1]

Q6 [7]

(a) [3] The angle $\phi$ between the line's direction $\vec d$ and the plane's normal $\vec n$ satisfies $\cos\phi=\dfrac{|\vec d\cdot\vec n|}{|\vec d||\vec n|}$ [1]. The line–plane angle $\theta$ is the complement, $\theta=90^\circ-\phi$ [1], so $\sin\theta=\cos\phi=\frac{|aA+bB+cC|}{\sqrt{a^2+b^2+c^2}\sqrt{A^2+B^2+C^2}}.$ Sine appears because $\theta$ is measured from the plane, not the normal. [1]

(b) [4] $\vec d=(2,3,6),\ \vec n=(2,-1,2)$ [1]. $\vec d\cdot\vec n=4-3+12=13$ [1]. $|\vec d|=\sqrt{4+9+36}=7$, $|\vec n|=3$ [1]. $\sin\theta=\dfrac{13}{21}$ → $\theta=\sin^{-1}\!\left(\dfrac{13}{21}\right)\approx38.3^\circ$ [1].

[
  {"claim":"Q1b section formula point is (4,-2,6)","code":"m,n=1,2;A=Matrix([2,-3,4]);B=Matrix([8,0,10]);P=(m*B+n*A)/(m+n);result=(P==Matrix([4,-2,6]))"},
  {"claim":"Q5 shortest distance = 3*sqrt(2)/2","code":"a1=Matrix([1,2,1]);a2=Matrix([2,-1,-1]);b1=Matrix([1,-1,1]);b2=Matrix([2,1,2]);cr=b1.cross(b2);d=abs((a2-a1).dot(cr))/cr.norm();result=(simplify(d-3*sqrt(2)/2)==0)"},
  {"claim":"Q4 normal simplifies to (-1,3,2) and plane passes through origin","code":"P=Matrix([1,1,-1]);Q=Matrix([-2,-2,2]);R=Matrix([1,-1,2]);nrm=(Q-P).cross(R-P);result=(nrm==Matrix([-3,9,6]) and nrm.dot(Matrix([1,1,-1]))==0)"},
  {"claim":"Q6b sin(theta)=13/21","code":"d=Matrix([2,3,6]);n=Matrix([2,-1,2]);s=abs(d.dot(n))/(d.norm()*n.norm());result=(simplify(s-Rational(13,21))==0)"}
]