3D Geometry

Q1

(a) Direction vector $\vec{AB} = (3, -3, 2)$, magnitude $\sqrt{9+9+4} = \sqrt{22}$. Direction cosines: $\left(\dfrac{3}{\sqrt{22}}, \dfrac{-3}{\sqrt{22}}, \dfrac{2}{\sqrt{22}}\right)$. [3] (1 for direction vector, 1 for magnitude, 1 for DCs.)

(b) Parametrize: $(1+3t,\ 2-3t,\ 3+2t)$. Substitute into $2x - y + z = 12$: $2(1+3t) - (2-3t) + (3+2t) = 12$ $2 + 6t - 2 + 3t + 3 + 2t = 12 \Rightarrow 11t + 3 = 12 \Rightarrow t = \tfrac{9}{11}$. Point $= \left(1 + \tfrac{27}{11},\ 2 - \tfrac{27}{11},\ 3 + \tfrac{18}{11}\right) = \left(\tfrac{38}{11}, -\tfrac{5}{11}, \tfrac{51}{11}\right)$. [4] (2 for substitution & solving $t$, 2 for point.)

(c) Distance from origin: $\sqrt{\left(\tfrac{38}{11}\right)^2 + \left(\tfrac{5}{11}\right)^2 + \left(\tfrac{51}{11}\right)^2} = \frac{\sqrt{1444 + 25 + 2601}}{11} = \frac{\sqrt{4070}}{11} \approx \frac{63.8}{11} \approx 5.80$. Since $\sqrt5 \approx 2.24 < 5.80$, the requirement is NOT met. [3] (2 for distance, 1 for conclusion.)

Q2

(a) $\vec{d_1}=(2,-1,3)$, $\vec{d_2}=(1,4,2)$. Not parallel (ratios differ). Point difference $\vec{a_2}-\vec{a_1} = (1,4,-3)$. Scalar triple product: $\begin{vmatrix}1&4&-3\\2&-1&3\\1&4&2\end{vmatrix}$ $= 1(-1\cdot2 - 3\cdot4) - 4(2\cdot2 - 3\cdot1) + (-3)(2\cdot4 - (-1)\cdot1)$ $= 1(-2-12) - 4(4-3) -3(8+1) = -14 - 4 - 27 = -45 \neq 0$. Non-zero ⇒ not coplanar and not parallel ⇒ skew. [3] (1 not parallel, 2 triple product ≠ 0.)

(b) $\vec{d_1}\times\vec{d_2} = \begin{vmatrix}i&j&k\\2&-1&3\\1&4&2\end{vmatrix} = i(-2-12) - j(4-3) + k(8+1) = (-14, -1, 9)$. $|\vec{d_1}\times\vec{d_2}| = \sqrt{196+1+81} = \sqrt{278}$. Shortest distance $= \dfrac{|(\vec{a_2}-\vec{a_1})\cdot(\vec{d_1}\times\vec{d_2})|}{|\vec{d_1}\times\vec{d_2}|} = \dfrac{|-45|}{\sqrt{278}} = \dfrac{45}{\sqrt{278}} \approx 2.70$. [4] (2 cross product, 2 distance.)

(c) $\cos\theta = \dfrac{|\vec{d_1}\cdot\vec{d_2}|}{|\vec{d_1}||\vec{d_2}|} = \dfrac{|2 -4 +6|}{\sqrt{14}\sqrt{21}} = \dfrac{4}{\sqrt{294}} = \dfrac{4}{17.146} \approx 0.2333$. $\theta \approx 76.5° \approx \mathbf{77°}$. [3] (1 dot, 1 magnitudes, 1 angle.)

Q3

(a) $\vec{PQ} = (-1,3,1)$, $\vec{PR} = (2,1,-2)$. Normal $\vec n = \vec{PQ}\times\vec{PR} = \begin{vmatrix}i&j&k\\-1&3&1\\2&1&-2\end{vmatrix}$ $= i(3\cdot(-2) - 1\cdot1) - j((-1)(-2) - 1\cdot2) + k((-1)(1) - 3\cdot2)$ $= i(-6-1) - j(2-2) + k(-1-6) = (-7, 0, -7)$. Simplify $\vec n = (1, 0, 1)$. Plane through $P(2,0,1)$: $x + z = 3$. So $\mathbf{x + z = 3}$. [4] (2 normal, 2 equation.)

(b) Ground plane $z=0$ has normal $(0,0,1)$. $\cos\theta = \dfrac{|(1,0,1)\cdot(0,0,1)|}{\sqrt2 \cdot 1} = \dfrac{1}{\sqrt2}$ ⇒ $\theta = \mathbf{45°}$. [3]

(c) Distance from $T(6,5,4)$ to $x+z-3=0$: $\dfrac{|6 + 4 - 3|}{\sqrt{1+1}} = \dfrac{7}{\sqrt2} = \dfrac{7\sqrt2}{2} \approx 4.95$. [3]

Q4

(a) Section formula (ratio $k:1$, $C=(3,-2,4)$, $D=(-1,6,-2)$): $M = \left(\frac{k(-1)+3}{k+1},\ \frac{k(6)+(-2)}{k+1},\ \frac{k(-2)+4}{k+1}\right) = \left(\frac{3-k}{k+1}, \frac{6k-2}{k+1}, \frac{4-2k}{k+1}\right).$ [3]

(b) $x+y+z = 5$: $\dfrac{(3-k) + (6k-2) + (4-2k)}{k+1} = 5$ $\Rightarrow \dfrac{3k + 5}{k+1} = 5 \Rightarrow 3k+5 = 5k+5 \Rightarrow 2k = 0 \Rightarrow k = 0$. But $k=0$ gives $M=C=(3,-2,4)$; check: $3-2+4=5$ ✓. So $\mathbf{k=0}$ (M coincides with C). [4] (2 setup, 2 solve. Accept noting M=C.)

(c) $M = (3, -2, 4)$: $x>0, y<0, z>0$ ⇒ octant IV (signs $+,-,+$). [3] (2 coordinates check, 1 octant.)

Q5

(a) Line direction $\vec d=(1,2,2)$, plane normal $\vec n=(1,-2,2)$. $\sin\theta = \dfrac{|\vec d\cdot\vec n|}{|\vec d||\vec n|} = \dfrac{|1 -4 +4|}{3\cdot3} = \dfrac{1}{9}$. $\theta = \arcsin(1/9) \approx \mathbf{6.4°}$. [4] (2 sin formula, 2 angle.)

(b) Foot of perpendicular from $A(0,1,-1)$ to $\Pi: x-2y+2z=6$. Line: $(t, 1-2t, -1+2t)$. Substitute: $t - 2(1-2t) + 2(-1+2t) = 6 \Rightarrow t - 2 + 4t - 2 + 4t = 6 \Rightarrow 9t - 4 = 6 \Rightarrow t = \tfrac{10}{9}$. Foot $F = \left(\tfrac{10}{9},\ 1 - \tfrac{20}{9},\ -1 + \tfrac{20}{9}\right) = \left(\tfrac{10}{9}, -\tfrac{11}{9}, \tfrac{11}{9}\right)$. [4]

(c) Image $A' = 2F - A = \left(\tfrac{20}{9},\ -\tfrac{22}{9} - 1,\ \tfrac{22}{9} + 1\right) = \left(\tfrac{20}{9}, -\tfrac{31}{9}, \tfrac{31}{9}\right)$. [2]

[
  {"claim":"Q1(b) line meets plane at t=9/11 giving point (38/11,-5/11,51/11)","code":"t=Rational(9,11); P=(1+3*t, 2-3*t, 3+2*t); result = (2*P[0]-P[1]+P[2]==12) and P==(Rational(38,11),Rational(-5,11),Rational(51,11))"},
  {"claim":"Q2(b) shortest distance between skew lines = 45/sqrt(278)","code":"d1=Matrix([2,-1,3]); d2=Matrix([1,4,2]); diff=Matrix([1,4,-3]); c=d1.cross(d2); dist=abs(diff.dot(c))/sqrt(c.dot(c)); result = simplify(dist - Rational(45,1)/sqrt(278))==0"},
  {"claim":"Q3(c) distance from T(6,5,4) to plane x+z=3 is 7/sqrt(2)","code":"dist=Abs(6+4-3)/sqrt(2); result = simplify(dist - 7/sqrt(2))==0"},
  {"claim":"Q5(b) foot of perpendicular is (10/9,-11/9,11/9) and lies on plane","code":"F=(Rational(10,9),Rational(-11,9),Rational(11,9)); result = (F[0]-2*F[1]+2*F[2]==6)"}
]