Noesis
Level 2 — Recall3D Geometry

3D Geometry

40 marksprintable — key stays hidden on paper

Level: 2 (Recall / Standard textbook problems) Time: 30 minutes Total Marks: 40

Q1. Find the distance between the points A(1,−2,3)A(1, -2, 3) and B(4,1,−1)B(4, 1, -1). (3 marks)

Q2. Find the coordinates of the point which divides the line joining P(2,−1,4)P(2, -1, 4) and Q(4,3,−2)Q(4, 3, -2) internally in the ratio 2:12:1. (4 marks)

Q3. State the relation between the direction cosines l,m,nl, m, n of a line. If a line has direction ratios 2,−3,62, -3, 6, find its direction cosines. (4 marks)

Q4. In which octant do the following points lie? (a) (−1,2,3)(-1, 2, 3) (b) (4,−2,−5)(4, -2, -5) (c) (−3,−1,2)(-3, -1, 2) (d) (5,6,7)(5, 6, 7) (4 marks)

Q5. Write the vector and Cartesian (symmetric) equations of the line passing through the point (1,2,3)(1, 2, 3) and parallel to the vector 2i^−j^+4k^2\hat{i} - \hat{j} + 4\hat{k}. (4 marks)

Q6. Find the angle between the two lines whose direction ratios are (1,2,2)(1, 2, 2) and (2,−2,1)(2, -2, 1). (5 marks)

Q7. Find the equation of the plane passing through the point (2,1,−1)(2, 1, -1) with normal vector 3i^−2j^+k^3\hat{i} - 2\hat{j} + \hat{k}. Also write it in general form. (4 marks)

Q8. Find the distance of the point (3,−2,1)(3, -2, 1) from the plane 2x−y+2z+3=02x - y + 2z + 3 = 0. (4 marks)

Q9. Find the angle between the two planes x+2y+2z=5x + 2y + 2z = 5 and 3x+4y=103x + 4y = 10. (4 marks)

Q10. Find the shortest distance between the skew lines r⃗=(i^+2j^+k^)+λ(i^−j^+k^)andr⃗=(2i^−j^−k^)+μ(2i^+j^+2k^).\vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k}) \quad\text{and}\quad \vec{r} = (2\hat{i} - \hat{j} - \hat{k}) + \mu(2\hat{i} + \hat{j} + 2\hat{k}). (4 marks)

Answer keyMark scheme & solutions

Q1. (3 marks) Distance formula: d=(x2−x1)2+(y2−y1)2+(z2−z1)2d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2} (1) =(4−1)2+(1+2)2+(−1−3)2=9+9+16= \sqrt{(4-1)^2+(1+2)^2+(-1-3)^2} = \sqrt{9+9+16} (1) =34= \sqrt{34} (1)

Q2. (4 marks) Section formula (internal, ratio m:n=2:1m:n = 2:1): (mx2+nx1m+n,my2+ny1m+n,mz2+nz1m+n)\left(\dfrac{mx_2+nx_1}{m+n}, \dfrac{my_2+ny_1}{m+n}, \dfrac{mz_2+nz_1}{m+n}\right) (1) x=2(4)+1(2)3=103x = \dfrac{2(4)+1(2)}{3} = \dfrac{10}{3} (1) y=2(3)+1(−1)3=53y = \dfrac{2(3)+1(-1)}{3} = \dfrac{5}{3} (1) z=2(−2)+1(4)3=03=0z = \dfrac{2(-2)+1(4)}{3} = \dfrac{0}{3}=0; Point =(103,53,0)= \left(\dfrac{10}{3}, \dfrac{5}{3}, 0\right) (1)

Q3. (4 marks) Relation: l2+m2+n2=1l^2 + m^2 + n^2 = 1 (1) Magnitude of direction ratios: 22+(−3)2+62=4+9+36=49=7\sqrt{2^2+(-3)^2+6^2} = \sqrt{4+9+36} = \sqrt{49} = 7 (1) Direction cosines =aΣa2= \dfrac{a}{\sqrt{\Sigma a^2}}: (27,−37,67)\left(\dfrac{2}{7}, \dfrac{-3}{7}, \dfrac{6}{7}\right) (2)

Q4. (4 marks) — 1 mark each Sign convention determines octant: (a) (−1,2,3)(-1,2,3): (−,+,+)(-,+,+) → Octant II (1) (b) (4,−2,−5)(4,-2,-5): (+,−,−)(+,-,-) → Octant VIII (1) (c) (−3,−1,2)(-3,-1,2): (−,−,+)(-,-,+) → Octant III (1) (d) (5,6,7)(5,6,7): (+,+,+)(+,+,+) → Octant I (1)

Q5. (4 marks) Vector form: r⃗=a⃗+λb⃗\vec{r} = \vec{a} + \lambda\vec{b} r⃗=(i^+2j^+3k^)+λ(2i^−j^+4k^)\vec{r} = (\hat{i}+2\hat{j}+3\hat{k}) + \lambda(2\hat{i}-\hat{j}+4\hat{k}) (2) Cartesian (symmetric) form: x−12=y−2−1=z−34\dfrac{x-1}{2} = \dfrac{y-2}{-1} = \dfrac{z-3}{4} (2)

Q6. (5 marks) cos⁡θ=∣a1a2+b1b2+c1c2∣a12+b12+c12 a22+b22+c22\cos\theta = \dfrac{|a_1a_2+b_1b_2+c_1c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\,\sqrt{a_2^2+b_2^2+c_2^2}} (1) Numerator: ∣1(2)+2(−2)+2(1)∣=∣2−4+2∣=0|1(2)+2(-2)+2(1)| = |2-4+2| = 0 (2) Denominators: 9=3\sqrt{9}=3 and 9=3\sqrt{9}=3 (1) cos⁡θ=0⇒θ=90°\cos\theta = 0 \Rightarrow \theta = 90° (lines are perpendicular) (1)

Q7. (4 marks) Plane through a⃗\vec{a} with normal n⃗\vec{n}: n⃗⋅(r⃗−a⃗)=0\vec{n}\cdot(\vec{r}-\vec{a})=0 (1) 3(x−2)−2(y−1)+1(z+1)=03(x-2) - 2(y-1) + 1(z+1) = 0 (2) 3x−6−2y+2+z+1=0⇒3x−2y+z−3=03x - 6 - 2y + 2 + z + 1 = 0 \Rightarrow 3x - 2y + z - 3 = 0 (1)

Q8. (4 marks) Distance =∣ax1+by1+cz1+d∣a2+b2+c2= \dfrac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}} (1) Numerator: ∣2(3)−(−2)+2(1)+3∣=∣6+2+2+3∣=13|2(3) - (-2) + 2(1) + 3| = |6+2+2+3| = 13 (2) Denominator: 4+1+4=3\sqrt{4+1+4} = 3; Distance =133= \dfrac{13}{3} (1)

Q9. (4 marks) Normals: n1⃗=(1,2,2)\vec{n_1}=(1,2,2), n2⃗=(3,4,0)\vec{n_2}=(3,4,0) (1) cos⁡θ=∣1(3)+2(4)+2(0)∣1+4+4 9+16=∣11∣3⋅5=1115\cos\theta = \dfrac{|1(3)+2(4)+2(0)|}{\sqrt{1+4+4}\,\sqrt{9+16}} = \dfrac{|11|}{3\cdot 5} = \dfrac{11}{15} (2) θ=cos⁡−11115≈42.8°\theta = \cos^{-1}\dfrac{11}{15} \approx 42.8° (1)

Q10. (4 marks) a1⃗=(1,2,1), a2⃗=(2,−1,−1)\vec{a_1}=(1,2,1),\ \vec{a_2}=(2,-1,-1); b1⃗=(1,−1,1), b2⃗=(2,1,2)\vec{b_1}=(1,-1,1),\ \vec{b_2}=(2,1,2) b1⃗×b2⃗=∣i^j^k^1−11212∣=(−2−1)i^−(2−2)j^+(1+2)k^=(−3,0,3)\vec{b_1}\times\vec{b_2} = \begin{vmatrix}\hat i&\hat j&\hat k\\1&-1&1\\2&1&2\end{vmatrix} = (-2-1)\hat i-(2-2)\hat j+(1+2)\hat k = (-3,0,3) (1) ∣b1⃗×b2⃗∣=9+0+9=18=32|\vec{b_1}\times\vec{b_2}| = \sqrt{9+0+9}=\sqrt{18}=3\sqrt2 a2⃗−a1⃗=(1,−3,−2)\vec{a_2}-\vec{a_1} = (1,-3,-2); dot with (−3,0,3)(-3,0,3): −3+0−6=−9-3+0-6 = -9 (2) d=∣−9∣32=932=32=322d = \dfrac{|-9|}{3\sqrt2} = \dfrac{9}{3\sqrt2} = \dfrac{3}{\sqrt2} = \dfrac{3\sqrt2}{2} (1)

[
  {"claim":"Q1 distance is sqrt(34)","code":"d=sqrt((4-1)**2+(1+2)**2+(-1-3)**2); result = simplify(d - sqrt(34))==0"},
  {"claim":"Q2 section point is (10/3,5/3,0)","code":"x=(2*4+1*2)/3; y=(2*3+1*(-1))/3; z=(2*(-2)+1*4)/3; result = (x==Rational(10,3)) and (y==Rational(5,3)) and (z==0)"},
  {"claim":"Q3 direction cosines are (2/7,-3/7,6/7)","code":"mag=sqrt(2**2+(-3)**2+6**2); result = (mag==7) and (Rational(2,7)**2+Rational(-3,7)**2+Rational(6,7)**2==1)"},
  {"claim":"Q8 distance is 13/3","code":"num=abs(2*3-(-2)+2*1+3); den=sqrt(4+1+4); result = simplify(num/den - Rational(13,3))==0"},
  {"claim":"Q10 shortest distance is 3/sqrt(2)","code":"b1=Matrix([1,-1,1]); b2=Matrix([2,1,2]); c=b1.cross(b2); a=Matrix([1,-3,-2]); d=abs(a.dot(c))/c.norm(); result = simplify(d - 3/sqrt(2))==0"}
]