Level 1 — Recognition3D Geometry

3D Geometry

⏱ 20 minutes30 marksprintable — key stays hidden on paper

Time Limit: 20 minutes
Total Marks: 30
Instructions: Attempt all questions. For True/False items, a brief justification is required for full marks.

Section A — Multiple Choice (1 mark each)

Q1. The number of octants into which 3D space is divided by the coordinate planes is: (a) 4 (b) 6 (c) 8 (d) 12

Q2. The point $(2, -3, 5)$ lies in the octant where the signs of $(x, y, z)$ are: (a) $(+,+,+)$ (b) $(+,-,+)$ (c) $(-,-,+)$ (d) $(+,-,-)$

Q3. The distance of the point $(3, 4, 12)$ from the origin is: (a) 12 (b) 13 (c) $\sqrt{19}$ (d) 19

Q4. The midpoint of the segment joining $(1, 2, 3)$ and $(5, 6, 7)$ is: (a) $(3, 4, 5)$ (b) $(2, 3, 4)$ (c) $(6, 8, 10)$ (d) $(4, 4, 4)$

Q5. If the direction ratios of a line are $(2, -1, 2)$ , its direction cosines are: (a) $\left(\tfrac{2}{3}, -\tfrac{1}{3}, \tfrac{2}{3}\right)$ (b) $(2,-1,2)$ (c) $\left(\tfrac{1}{2}, -1, \tfrac{1}{2}\right)$ (d) $\left(\tfrac{2}{9}, -\tfrac{1}{9}, \tfrac{2}{9}\right)$

Q6. For direction cosines $l, m, n$ , the value of $l^2 + m^2 + n^2$ is: (a) 0 (b) $\tfrac{1}{2}$ (c) 1 (d) 3

Q7. The symmetric form of a line through $(x_1,y_1,z_1)$ with direction ratios $(a,b,c)$ is: (a) $\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{z-z_1}{c}$
(b) $ax_1+by_1+cz_1=0$
(c) $\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$
(d) $(x-x_1)a=(y-y_1)b=(z-z_1)c$

Q8. The general equation of a plane is: (a) $ax+by+cz=0$ only (b) $Ax+By+Cz+D=0$ (c) $lx+my+nz=1$ (d) $x^2+y^2+z^2=r^2$

Q9. A normal vector to the plane $2x - 3y + z = 7$ is: (a) $(7,7,7)$ (b) $(2,-3,1)$ (c) $(2,3,1)$ (d) $(-7,-7,-7)$

Q10. The distance from point $(0,0,0)$ to the plane $x + 2y + 2z = 9$ is: (a) 9 (b) 3 (c) $\sqrt{9}$ (d) 1

Section B — Matching (5 marks)

Q11. Match Column I with Column II. (1 mark each correct pair)

Column I		Column II
(i) Angle between two lines (dir. ratios)		(P) $\dfrac{
(ii) Distance point to plane		(Q) $\cos\theta=\dfrac{
(iii) Angle between two planes		(R) $\sin\theta=\dfrac{
(iv) Angle between line and plane		(S) $\cos\theta=\dfrac{
(v) Shortest distance (skew lines)		(T) $\dfrac{

Section C — True / False with Justification (2 marks each)

Q12. Two lines with direction ratios $(1,2,3)$ and $(2,4,6)$ are parallel.

Q13. The plane $x - y + z = 4$ and the line with direction ratios $(1,1,0)$ are perpendicular.

Q14. The point $(2, 5, 0)$ lies on the $z$ -axis.

Q15. Intersecting lines always have a nonzero shortest distance.

Q16. The intercept form $\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$ meets the $x$ -axis at $(a,0,0)$ .

Q17. Direction cosines of the positive $y$ -axis are $(0,1,0)$ .

Answer keyMark scheme & solutions

Section A (1 mark each)

Q1. (c) 8 — The three mutually perpendicular coordinate planes divide space into $2^3 = 8$ octants.

Q2. (b) $(+,-,+)$ — $x=2>0$ , $y=-3<0$ , $z=5>0$ .

Q3. (b) 13 — $\sqrt{3^2+4^2+12^2}=\sqrt{9+16+144}=\sqrt{169}=13$ .

Q4. (a) $(3,4,5)$ — Midpoint $=\left(\tfrac{1+5}{2},\tfrac{2+6}{2},\tfrac{3+7}{2}\right)=(3,4,5)$ .

Q5. (a) $\left(\tfrac{2}{3},-\tfrac{1}{3},\tfrac{2}{3}\right)$ — Magnitude $=\sqrt{4+1+4}=3$ ; divide each ratio by 3.

Q6. (c) 1 — Fundamental relation $l^2+m^2+n^2=1$ .

Q7. (a) — Standard symmetric form of a line.

Q8. (b) $Ax+By+Cz+D=0$ — General (Cartesian) equation of a plane.

Q9. (b) $(2,-3,1)$ — Coefficients of $x,y,z$ give the normal.

Q10. (b) 3 — $\dfrac{|0+0+0-9|}{\sqrt{1+4+4}}=\dfrac{9}{3}=3$ .

Section B

Q11. Correct matches (1 mark each):

(i) → (Q) (angle between lines uses direction vectors)
(ii) → (P) (point-to-plane distance formula)
(iii) → (S) (angle between planes uses normals)
(iv) → (R) (line-plane uses sine with line-dir & normal)
(v) → (T) (skew-line shortest distance)

Section C (2 marks each: 1 for correct T/F, 1 for justification)

Q12. TRUE. $(2,4,6)=2(1,2,3)$ , so the direction ratios are proportional ⇒ lines parallel.

Q13. FALSE. For line ⊥ plane, the line's direction must be parallel to the normal $(1,-1,1)$ . Here $(1,1,0)$ is not proportional to $(1,-1,1)$ . (Also $\vec b\cdot\vec n = 1-1+0=0$ , so the line is actually parallel to the plane, not perpendicular.)

Q14. FALSE. Points on the $z$ -axis have $x=0,y=0$ . Here $x=2,y=5\neq0$ ; the point lies in the $xy$ -plane.

Q15. FALSE. Intersecting lines share a common point, so their shortest distance is $0$ . (Nonzero shortest distance characterises skew lines.)

Q16. TRUE. Setting $y=0,z=0$ gives $\tfrac{x}{a}=1\Rightarrow x=a$ , i.e. intercept point $(a,0,0)$ .

Q17. TRUE. The positive $y$ -axis makes angles $90°,0°,90°$ with $x,y,z$ axes, giving cosines $(\cos90°,\cos0°,\cos90°)=(0,1,0)$ ; and $0^2+1^2+0^2=1$ .

[
  {"claim":"Q3 distance of (3,4,12) from origin is 13","code":"result = sqrt(3**2+4**2+12**2)==13"},
  {"claim":"Q5 direction cosines of (2,-1,2) sum of squares equals 1","code":"mag=sqrt(2**2+(-1)**2+2**2); l=Rational(2,3); m=Rational(-1,3); n=Rational(2,3); result = simplify(l**2+m**2+n**2)==1 and mag==3"},
  {"claim":"Q10 distance origin to x+2y+2z=9 is 3","code":"result = Rational(abs(-9),sqrt(1+4+4))==3"},
  {"claim":"Q12 (1,2,3) and (2,4,6) are parallel (cross product zero)","code":"result = Matrix([1,2,3]).cross(Matrix([2,4,6]))==Matrix([0,0,0])"},
  {"claim":"Q13 line dir (1,1,0) dot plane normal (1,-1,1) is 0 (parallel not perpendicular)","code":"result = (Matrix([1,1,0]).dot(Matrix([1,-1,1])))==0"}
]