Three Dimensional Geometry 4 marks Questions, CBSE Class 12 PYQs 2024–2026

1 different 4 marks questions on Three Dimensional Geometry from CBSE Class 12 Maths board exams 2024–2026, newest first.

Q374 marksCase StudyThree Dimensional Geometry CBSE 2025 · 65/4/1

An engineer is designing a new metro rail network in a city.
Initially, two metro lines, Line A and Line B, each consisting of multiple stations are designed. The track for Line A is represented by $l_{1} : \frac{x - 2}{3} = \frac{y + 1}{- 2} = \frac{z - 3}{4}$ , while the track for Line B is represented by $l_{2} : \frac{x - 1}{2} = \frac{y - 3}{1} = \frac{z + 2}{- 3}$ .
Based on the above information, answer the following questions :
(i) Find whether the two metro tracks are parallel. (1)
(ii) Solar panels are to be installed on the rooftop of the metro stations. Determine the equation of the line representing the placement of solar panels on the rooftop of Line A’s stations, given that panels are to be positioned parallel to Line A’s track ( $l_{1}$ ) and pass through the point (1, – 2, – 3). (1)
(iii) To connect the stations, a pedestrian pathway perpendicular to the two metro lines is to be constructed which passes through point (3, 2, 1). Determine the equation of the pedestrian walkway. (2)
OR (iii) Find the shortest distance between Line A and Line B. (2)

Show answer & solution

Answer: (i) Not parallel (ii)

\frac{x - 1}{3} = \frac{y + 2}{- 2} = \frac{z + 3}{4}

(iii)

\frac{x - 3}{2} = \frac{y - 2}{17} = \frac{z - 1}{7}

; OR

\frac{31}{342} = \frac{31}{3 38}

units

(i) Direction ratios 3, –2, 4 and 2, 1, –3 are not proportional ( $\frac{3}{2} \neq = \frac{- 2}{1}$ ), so the tracks are not parallel.
(ii) Line through (1, –2, –3) with direction 3, –2, 4: $\frac{x - 1}{3} = \frac{y + 2}{- 2} = \frac{z + 3}{4}$ .
(iii) $b_{1} \times b_{2} = (3 i^- 2 j^+ 4 k^) \times (2 i^+ j^- 3 k^) = (6 - 4) i^- (- 9 - 8) j^+ (3 + 4) k^= 2 i^+ 17 j^+ 7 k^$ .
Walkway: $\frac{x - 3}{2} = \frac{y - 2}{17} = \frac{z - 1}{7}$ .
OR (iii) $a_{2} - a_{1} = (1 - 2) i^+ (3 + 1) j^+ (- 2 - 3) k^= - i^+ 4 j^- 5 k^$ .
SD $= \frac{∣ ( - 1 ) ( 2 ) + 4 ( 17 ) + ( - 5 ) ( 7 ) ∣}{4 + 289 + 49} = \frac{31}{342} = \frac{31}{3 38}$ units.

Also asked in: 2025 65/4/2, 2025 65/4/3

Three Dimensional Geometry: 4 marks Questions (CBSE Class 12)