Previous year questions: Three Dimensional Geometry

Previous year questions: three dimensional geometry

Six questions here are printed board questions from the 2023, 2024, 2025 and 2026 exams, and four are from official sample papers. Every heading gives the paper and the marks. Read the question, note the point and the direction, then start.

Q1 (CBSE Board 2024, 1 mark)

The vector equation of a line passing through the point (1, -1, 0) and parallel to the Y-axis is: (A) r = i - j + lambda(i - j), (B) r = i - j + lambda j, (C) r = i - j + lambda k, (D) r = lambda j.

Answer: Option (B). The point (1, -1, 0) has position vector i - j, and a line parallel to the Y-axis has direction j. So the equation is r = i - j + lambda j, which is option (B). Option (D) passes through the origin, and options (A) and (C) have the wrong directions.

Q2 (CBSE Board 2024, 1 mark)

The lines (1 - x)/2 = (y - 1)/3 = z/1 and (2x - 3)/(2p) = y/(-1) = (z - 4)/7 are perpendicular to each other for p equal to: (A) -1/2, (B) 1/2, (C) 2, (D) 3.

Answer: Option (C). The first line has direction ratios (-2, 3, 1), because (1 - x)/2 is -(x - 1)/2. The second line has direction ratios (p, -1, 7). Perpendicular lines have zero dot product, so -2p - 3 + 7 = 0, which gives p = 2.

Q3 (CBSE Board 2025, 2 marks)

Find the angle between the lines r = (3 + 2 lambda)i - (2 - 2 lambda)j + (6 + 2 lambda)k and r = (2j - 5k) + mu(6i + 3j + 2k).

Answer: The first line is (3i - 2j + 6k) plus lambda(2i + 2j + 2k), so its direction is (2, 2, 2), which is parallel to (1, 1, 1). The second direction is (6, 3, 2). The dot product is 6 + 3 + 2 = 11, and the lengths are root 3 and 7. So the cosine of the angle is 11 divided by 7 root 3, and the angle is the inverse cosine of that value, about 24.9 degrees.

Q4 (CBSE Board 2025, 5 marks)

Two lines are given by l1: r = i + 2j - 4k + lambda(4i + 6j + 12k) and l2: r = 3i + 3j - 5k + mu(6i + 9j + 18k). Work out the shortest distance between them.

Answer: The directions are (4, 6, 12) and (6, 9, 18). The second is one and a half times the first, so the lines are parallel and the skew formula is not used. Take the shared direction as (2, 3, 6), whose length is 7. The joining vector is (3 - 1, 3 - 2, -5 + 4) = (2, 1, -1). Its cross product with (2, 3, 6) is (9, -14, 4), whose magnitude is root 293. The distance is root 293 divided by 7.

Q5 (CBSE Board 2026, 5 marks)

Find the foot of the perpendicular from the point (0, 2, 3) on the line (x + 3)/5 = (y - 1)/2 = (z + 4)/3, and hence find the length of the perpendicular.

Answer: Write the line in parametric form: x = -3 + 5t, y = 1 + 2t, z = -4 + 3t. A general point on the line is F = (-3 + 5t, 1 + 2t, -4 + 3t). The vector from the given point P(0, 2, 3) to F must be perpendicular to the direction (5, 2, 3). So (-3 + 5t, -1 + 2t, -7 + 3t) dot (5, 2, 3) = 0, which gives -15 + 25t - 2 + 4t - 21 + 9t = 0, that is 38t - 38 = 0 and t = 1. The foot is (2, 3, -1). The length of the perpendicular is the distance from (0, 2, 3) to (2, 3, -1), which is the square root of 4 + 1 + 16 = root 21.

Q6 (CBSE Board 2023, 5 marks)

A line l passes through (-1, 3, -2). Two given lines have directions (1, 2, 3) and (-3, 2, 5), and l is perpendicular to both of them. Write the vector equation of l and then its distance from the origin.

Answer: The directions are (1, 2, 3) and (-3, 2, 5). Their cross product is (4, -14, 8), which simplifies to (2, -7, 4). The line through (-1, 3, -2) with this direction has vector equation r = -i + 3j - 2k + lambda(2i - 7j + 4k). For the distance from the origin, use the point vector a = (-1, 3, -2) and direction d = (2, -7, 4). Then a cross d is (-2, 0, 1) with magnitude root 5, and the length of d is root 69. The distance is root 5 divided by root 69, that is root of 5/69.

Q7 (CBSE SQP 2022-23, 2 marks)

Find the direction ratios and the direction cosines of a line parallel to the line whose equation is 6x - 12 = 3y + 9 = 2z - 2.

Answer: Write each expression as a multiple of a coordinate difference. From 6x - 12 = 6(x - 2), 3y + 9 = 3(y + 3) and 2z - 2 = 2(z - 1), the common value gives (x - 2)/(1/6) = (y + 3)/(1/3) = (z - 1)/(1/2). So the direction ratios are proportional to (1/6, 1/3, 1/2), that is to (1, 2, 3). The length of (1, 2, 3) is root 14, so the direction cosines are 1/root 14, 2/root 14 and 3/root 14.

Q8 (CBSE SQP 2024-25, 3 marks)

Find the vector and cartesian equations of the line through (-1, 2, 7) that is perpendicular to the lines r = 2i + j - 3k + lambda(i + 2j + 5k) and r = 3i + 3j - 7k + mu(3i - 2j + 5k).

Answer: The two directions are (1, 2, 5) and (3, -2, 5). Their cross product is (20, 10, -8), which simplifies to (10, 5, -4). This is the direction of the required line. Its vector equation is r = -i + 2j + 7k + lambda(10i + 5j - 4k), and the cartesian equation is (x + 1)/10 = (y - 2)/5 = (z - 7)/(-4).

Q9 (CBSE SQP 2024-25, 5 marks)

The lines l1: r = (-i - j - k) + lambda(7i - 6j + k) and l2: r = (3i + 5j + 7k) + mu(i - 2j + k) are given. Work out the shortest distance between them.

Answer: The directions are (7, -6, 1) and (1, -2, 1), and their cross product is (-4, -6, -8) with magnitude 2 root 29. The joining vector between the base points is (4, 6, 8). Its dot product with the cross product is negative 116, whose modulus is 116. The shortest distance is 116 divided by 2 root 29, which is 2 root 29 after rationalising.

Q10 (CBSE SQP 2025-26, 5 marks)

Two lines with equations (x - 1)/3 = -y, z + 1 = 0 and (-x)/2 = (y + 1)/2 = z + 2 meet where the y-coordinate is 1. Work out the coordinates of that meeting point, then write the vector equation of a line through it that is perpendicular to both given lines.

Answer: Put y = 1 in the first line. From -y = (x - 1)/3 and z + 1 = 0 we get (x - 1)/3 = -1 and z = -1, so x = -2 and the point is (-2, 1, -1). The same point lies on the second line, which confirms the intersection. The first line has direction (3, -1, 0) and the second has direction (-2, 2, 1). Their cross product is (-1, -3, 4), so a line perpendicular to both through the intersection point has equation r = -2i + j - k + lambda(-i - 3j + 4k).

Six of these are board questions across four board exam years. Note the pattern: every solution starts by naming the point and the direction, then chooses between the angle formula and the distance formula. The one-mark items are read-offs, and the five-mark items only look long because of the arithmetic. Keep the working compact: write the direction vectors, then the formula, then substitute, and leave the final answer in a simplified surd form. Where a question has an OR part, only one option has to be attempted, but both options usually test the same base skills, so a mistake in the base step costs marks in either choice. Check every answer against a second test where one exists: the dot product for perpendicularity, the ratio of direction ratios for parallelism, and the sign of the triple product for skew lines. A five-mark question usually has one setup step and two short computations, so most of the marks are earned before the final arithmetic. Write the conclusion sentence as well, because a bare number does not say whether the lines are perpendicular, parallel or skew.