Previous year questions: Three Dimensional Geometry

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Six items are real JEE Main 2026 questions from the official National Testing Agency papers and four are from the NCERT Exemplar for Class 11 Mathematics. Each heading shows the source and the marks. The JEE questions on this chapter twist the three-dimensional distance formula with a line or a plane condition, which is a step beyond the NCERT exercises but uses only coordinates, distances and centroids. Stems were checked against the official paper pages, and every numerical answer was re-derived.

Q1 (JEE Main 2026, 2 Apr Shift 1, 4 marks)

The point of intersection of the lines (x plus 1) by 3 equal to (y plus a) by 5 equal to (z plus b plus 1) by 7 and (x minus 2) by 1 equal to (y minus b) by 4 equal to (z minus 2a) by 7 lies on the xy-plane. Find a plus b.

Answer: On the xy-plane z is zero. Writing the first line with parameter t gives z equal to minus b minus 1 plus 7t equal to 0, so t is (b plus 1) by 7. For the second line with parameter s, z equal to 2a plus 7s equal to 0 gives s equal to minus 2a by 7. Equating the x and y coordinates from the two expressions gives 3b plus 2a equal to 18 and a plus 5 equal to 2b. Solving gives b equal to 4 and a equal to 3, so a plus b is 7. The official key is 7.

Q2 (JEE Main 2026, 4 Apr Shift 1, 4 marks)

A line with direction ratios 1, minus 1, 2 intersects the lines x by 2 equal to y by 3 equal to (z plus 1) by 3 and (x plus 1) by minus 1 equal to (y minus 2) by 1 equal to z by 4 at P and Q. If the length PQ is alpha, find 225 alpha squared.

Answer: Write P on the first line as (2s, 3s, 3s minus 1) and Q on the second line as (minus t minus 1, t plus 2, 4t). The vector PQ must be parallel to (1, minus 1, 2), so its three components equal the same multiple k. Solving the three equations gives s equal to 1 by 5 and t equal to minus 8 by 15, with k equal to minus 13 by 15. Then alpha is the absolute value of k times root 6, which is 13 root 6 by 15. Squaring and multiplying by 225 gives 1014. The official key is 1014.

Q3 (JEE Main 2026, 5 Apr Shift 1, 4 marks)

Find the square of the distance of the point P(5, 6, 7) from the line (x minus 2) by 2 equal to (y minus 5) by 3 equal to (z minus 2) by 4.

Answer: The line passes through A(2, 5, 2) with direction (2, 3, 4). The vector AP is (3, 1, 5). Its dot product with the direction is 6 plus 3 plus 20 equal to 29, and the squared direction length is 4 plus 9 plus 16 equal to 29, so the projection of AP on the line has parameter 1. The foot of the perpendicular is therefore A plus the direction, which is (4, 8, 6). The squared distance from P to the foot is 1 plus 4 plus 1 equal to 6. The official key is 6.

Q4 (JEE Main 2026, 5 Apr Shift 2, 4 marks)

A triangle PQR has P and Q on the line (x plus 3) by 8 equal to (y minus 4) by 2 equal to (z plus 1) by 2, both at a distance of 6 units from R(1, 2, 3). If (alpha, beta, gamma) is the centroid of the triangle, find alpha plus beta plus gamma.

Answer: A general point on the line is (minus 3 plus 8t, 4 plus 2t, minus 1 plus 2t). Its squared distance from R is 72t squared minus 72t plus 36, and setting this equal to 36 gives t equal to 0 or 1. The two points are (minus 3, 4, minus 1) and (5, 6, 1). The centroid of P, Q and R is (1, 4, 1), so alpha plus beta plus gamma is 6. The official key is 6.

Q5 (JEE Main 2026, 5 Apr Shift 2, 4 marks)

The distance of the point (a, 2, 5) from the image of the point (1, 2, 7) in the line x by 1 equal to (y minus 1) by 1 equal to (z minus 2) by 2 is 4. Find the sum of all possible values of a.

Answer: The foot of the perpendicular from (1, 2, 7) to the line is found by projecting: the point is (t, 1 plus t, 2 plus 2t) and the dot product condition gives t equal to 2, so the foot is (2, 3, 6). The image is twice the foot minus the point, which is (3, 4, 5). The distance from (a, 2, 5) to (3, 4, 5) is the square root of (a minus 3) squared plus 4, and setting this equal to 4 gives (a minus 3) squared equal to 12, so a is 3 plus or minus 2 root 3. The sum of the two values is 6. The official key is 6.

Q6 (JEE Main 2026, 8 Apr Shift 2, 4 marks)

The foot of the perpendicular from the point (lambda, 2, 3) on the line (x minus 4) by 1 equal to (y minus 9) by 2 equal to (z minus 5) by 1 is the point (1, mu, 2). Find the distance between the lines (x minus 1) by 2 equal to (y minus 2) by 3 equal to (z plus 4) by 6 and (x minus lambda) by 2 equal to (y minus mu) by 3 equal to (z plus 5) by 6.

Answer: The point (1, mu, 2) lies on the first line, so its parameter is minus 3 and mu is 3. The vector from (lambda, 2, 3) to (1, 3, 2) is perpendicular to the line direction (1, 2, 1), which gives 2 minus lambda equal to 0, so lambda is 2. With lambda equal to 2 and mu equal to 3 the two given lines are parallel, both with direction (2, 3, 6), and their distance is the length of the projection of the vector between a point on each line, (1, 1, minus 1), on the normal to the direction. The cross product of (1, 1, minus 1) and (2, 3, 6) is (9, minus 8, 1), whose length is root 146, and the direction length is 7, so the distance is root 146 by 7. The official key is root 146 by 7.

Q7 (NCERT Exemplar, 3 marks)

How far apart are the points (2, 0, 0) and (minus 3, 0, 0)?

Answer: Both points lie on the x-axis, so the distance is the absolute difference of the x-coordinates, which is 5 units. Source: NCERT Exemplar Class 11 Mathematics, Chapter 12, Exercise 12.2, Short Answer Type, Q5.

Q8 (NCERT Exemplar, 3 marks)

Find the distance from the origin to the point (6, 6, 7).

Answer: The distance is the square root of 36 plus 36 plus 49, which is the square root of 121, equal to 11 units. Source: NCERT Exemplar Class 11 Mathematics, Chapter 12, Exercise 12.2, Short Answer Type, Q6.

Q9 (NCERT Exemplar, 3 marks)

Show that the triangle with vertices A(0, 4, 1), B(2, 3, minus 1) and C(4, 5, 0) is right angled.

Answer: The squared side lengths are AB squared equal to 4 plus 1 plus 4 equal to 9, BC squared equal to 4 plus 4 plus 1 equal to 9 and AC squared equal to 16 plus 1 plus 1 equal to 18. Since 9 plus 9 equals 18, the triangle is right angled at B. Source: NCERT Exemplar Class 11 Mathematics, Chapter 12, Exercise 12.2, Short Answer Type, Q10.

Q10 (NCERT Exemplar, 3 marks)

Find the third vertex of a triangle whose centroid is the origin and whose two vertices are (2, 4, 6) and (0, minus 2, minus 5).

Answer: Since the centroid is the origin, the sum of the three vertices is zero. The third vertex is the negative of the sum of the two given vertices, which is (minus 2, minus 2, minus 1). Source: NCERT Exemplar Class 11 Mathematics, Chapter 12, Exercise 12.2, Short Answer Type, Q11.