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Page 1 JEE Main Previous Year Questions (2025): 3D Geometry Q1: Let ?? ?? : ?? -?? ?? = ?? -?? -?? = ?? +?? ?? and ?? ?? : ?? -?? ?? = ?? ?? = ?? +?? ?? ,?? ? ?? , be two lines, which intersect at the point ?? . If ?? is the foot of perpendicular from the point ?? (?? ,?? ,-?? ) on ?? ?? , then the value of ???? ?? ( ???? ) ?? is ____ - JEE Main 2025 (Online) 22nd January Morning Shift Ans: 216 Solution: Explanation To find the value of 26?? ( PB) 2 , we proceed as follows: Intersection Point ?? of L 1 and L 2 We are given the equations of the lines: L 1 : ?? - 1 3 = ?? - 1 -1 = ?? + 1 0 L 2 : ?? - 2 2 = ?? 0 = ?? + 4 ?? For L 1 ,?? + 1 = 0, which implies ?? = -1. For L 2 , since ?? = 0, the direction ratios can be matched using the parameter ?? : ?? = 3?? + 1,?? = -?? + 1,?? = -1, ?? = 2?? + 2,?? = 0,?? = ???? - 4. Setting the coordinates equal for intersection: 3?? + 1 = 2?? + 2, -?? + 1 = 0, -1 = ???? - 4. From -?? + 1 = 0, we find ?? = 1. Substituting ?? = 1 into 3?? + 1 = 2?? + 2 gives: 3(1)+ 1 = 2?? + 2 ? ?? = 1. Also, substituting ?? = 1 into -1 = ???? - 4 gives: -1 = ?? (1)- 4 ? ?? = 3. So, the intersection point ?? is: ?? (4,0,-1). Foot of Perpendicular from ?? to L 2 Page 2 JEE Main Previous Year Questions (2025): 3D Geometry Q1: Let ?? ?? : ?? -?? ?? = ?? -?? -?? = ?? +?? ?? and ?? ?? : ?? -?? ?? = ?? ?? = ?? +?? ?? ,?? ? ?? , be two lines, which intersect at the point ?? . If ?? is the foot of perpendicular from the point ?? (?? ,?? ,-?? ) on ?? ?? , then the value of ???? ?? ( ???? ) ?? is ____ - JEE Main 2025 (Online) 22nd January Morning Shift Ans: 216 Solution: Explanation To find the value of 26?? ( PB) 2 , we proceed as follows: Intersection Point ?? of L 1 and L 2 We are given the equations of the lines: L 1 : ?? - 1 3 = ?? - 1 -1 = ?? + 1 0 L 2 : ?? - 2 2 = ?? 0 = ?? + 4 ?? For L 1 ,?? + 1 = 0, which implies ?? = -1. For L 2 , since ?? = 0, the direction ratios can be matched using the parameter ?? : ?? = 3?? + 1,?? = -?? + 1,?? = -1, ?? = 2?? + 2,?? = 0,?? = ???? - 4. Setting the coordinates equal for intersection: 3?? + 1 = 2?? + 2, -?? + 1 = 0, -1 = ???? - 4. From -?? + 1 = 0, we find ?? = 1. Substituting ?? = 1 into 3?? + 1 = 2?? + 2 gives: 3(1)+ 1 = 2?? + 2 ? ?? = 1. Also, substituting ?? = 1 into -1 = ???? - 4 gives: -1 = ?? (1)- 4 ? ?? = 3. So, the intersection point ?? is: ?? (4,0,-1). Foot of Perpendicular from ?? to L 2 The point ?? on L 2 is given by: ?? (2?? + 2,0,3?? - 4). For vector ???? ????? = (2?? + 1,-1,3?? - 3), since ???? is perpendicular to L 2 , the dot product should be zero: (2?? + 1)· 2+ (-1)· 0 + (3?? - 3)· 3 = 0. Simplifying gives: 4?? + 2+ 9?? - 9 = 0 ? 13?? = 7 ? ?? = 7 13 . So, the coordinates of point ?? are: ?? ( 40 13 ,0, -31 13 ). Distance PB and Calculation The vector ???? ????? is: ???? ????? = (4 - 40 13 ,0 - 0,-1- ( -31 13 )). Calculating the components: ???? ????? = ( 12 13 ,0, 18 13 ). The square of the distance (???? ) 2 is: ( 12 13 ) 2 + (0) 2 + ( 18 13 ) 2 = 144 169 + 324 169 = 468 169 . Thus, 26?? (???? ) 2 is: 26× 3 × 468 169 = 216 . So, the final value is 216 . Q2: Let ?? be the image of the point ?? (?? ,-?? ,?? ) in the line ?? : ?? -?? ?? = ?? +?? ?? = ?? ?? and ?? (?? ,?? ,?? ) be a point on ?? . Then the square of the area of ? ?????? is ____ . JEE Main 2025 (Online) 24th January Evening Shift Ans: 957 Solution: Page 3 JEE Main Previous Year Questions (2025): 3D Geometry Q1: Let ?? ?? : ?? -?? ?? = ?? -?? -?? = ?? +?? ?? and ?? ?? : ?? -?? ?? = ?? ?? = ?? +?? ?? ,?? ? ?? , be two lines, which intersect at the point ?? . If ?? is the foot of perpendicular from the point ?? (?? ,?? ,-?? ) on ?? ?? , then the value of ???? ?? ( ???? ) ?? is ____ - JEE Main 2025 (Online) 22nd January Morning Shift Ans: 216 Solution: Explanation To find the value of 26?? ( PB) 2 , we proceed as follows: Intersection Point ?? of L 1 and L 2 We are given the equations of the lines: L 1 : ?? - 1 3 = ?? - 1 -1 = ?? + 1 0 L 2 : ?? - 2 2 = ?? 0 = ?? + 4 ?? For L 1 ,?? + 1 = 0, which implies ?? = -1. For L 2 , since ?? = 0, the direction ratios can be matched using the parameter ?? : ?? = 3?? + 1,?? = -?? + 1,?? = -1, ?? = 2?? + 2,?? = 0,?? = ???? - 4. Setting the coordinates equal for intersection: 3?? + 1 = 2?? + 2, -?? + 1 = 0, -1 = ???? - 4. From -?? + 1 = 0, we find ?? = 1. Substituting ?? = 1 into 3?? + 1 = 2?? + 2 gives: 3(1)+ 1 = 2?? + 2 ? ?? = 1. Also, substituting ?? = 1 into -1 = ???? - 4 gives: -1 = ?? (1)- 4 ? ?? = 3. So, the intersection point ?? is: ?? (4,0,-1). Foot of Perpendicular from ?? to L 2 The point ?? on L 2 is given by: ?? (2?? + 2,0,3?? - 4). For vector ???? ????? = (2?? + 1,-1,3?? - 3), since ???? is perpendicular to L 2 , the dot product should be zero: (2?? + 1)· 2+ (-1)· 0 + (3?? - 3)· 3 = 0. Simplifying gives: 4?? + 2+ 9?? - 9 = 0 ? 13?? = 7 ? ?? = 7 13 . So, the coordinates of point ?? are: ?? ( 40 13 ,0, -31 13 ). Distance PB and Calculation The vector ???? ????? is: ???? ????? = (4 - 40 13 ,0 - 0,-1- ( -31 13 )). Calculating the components: ???? ????? = ( 12 13 ,0, 18 13 ). The square of the distance (???? ) 2 is: ( 12 13 ) 2 + (0) 2 + ( 18 13 ) 2 = 144 169 + 324 169 = 468 169 . Thus, 26?? (???? ) 2 is: 26× 3 × 468 169 = 216 . So, the final value is 216 . Q2: Let ?? be the image of the point ?? (?? ,-?? ,?? ) in the line ?? : ?? -?? ?? = ?? +?? ?? = ?? ?? and ?? (?? ,?? ,?? ) be a point on ?? . Then the square of the area of ? ?????? is ____ . JEE Main 2025 (Online) 24th January Evening Shift Ans: 957 Solution: Let R(2?? + 1,3?? - 1,4?? ) 2?? + 1 = 5 ?? = 2 R(5,5,8) let T(2?? + 1,3?? - 1,4?? ) QT ????? = (2?? - 6)i ˆ + (3?? + 1)j ˆ + (4?? - 5)k ˆ b ? = 2i ˆ + 3j ˆ + 4k ˆ QT ????? · b ? = 0 4?? - 12+ 9?? + 3 + 16?? - 20 = 0 ?? = 1 T(3,2,4) QT = v33 RT = v29 ( area of ? PQR) 2 = ( 1 2 v29· 2v33) 2 = 957 Q3: Let the area of the triangle formed by the lines ?? + ?? = ?? - ?? = ?? , ?? -?? ?? = ?? -?? = ?? -?? ?? and ?? -?? = ?? -?? ?? = ?? -?? ?? be ?? . Then ?? ?? is equal to ____ . JEE Main 2025 (Online) 8th April Evening Shift Ans: 56 Solution: ?? 1 :?? + 2 = ?? - 1 = ?? = l ?? 2 : ?? - 3 5 = ?? -1 = ?? - 1 1 = ?? ?? 3 : ?? -3 = ?? - 3 5 = ?? - 2 1 = ?? Page 4 JEE Main Previous Year Questions (2025): 3D Geometry Q1: Let ?? ?? : ?? -?? ?? = ?? -?? -?? = ?? +?? ?? and ?? ?? : ?? -?? ?? = ?? ?? = ?? +?? ?? ,?? ? ?? , be two lines, which intersect at the point ?? . If ?? is the foot of perpendicular from the point ?? (?? ,?? ,-?? ) on ?? ?? , then the value of ???? ?? ( ???? ) ?? is ____ - JEE Main 2025 (Online) 22nd January Morning Shift Ans: 216 Solution: Explanation To find the value of 26?? ( PB) 2 , we proceed as follows: Intersection Point ?? of L 1 and L 2 We are given the equations of the lines: L 1 : ?? - 1 3 = ?? - 1 -1 = ?? + 1 0 L 2 : ?? - 2 2 = ?? 0 = ?? + 4 ?? For L 1 ,?? + 1 = 0, which implies ?? = -1. For L 2 , since ?? = 0, the direction ratios can be matched using the parameter ?? : ?? = 3?? + 1,?? = -?? + 1,?? = -1, ?? = 2?? + 2,?? = 0,?? = ???? - 4. Setting the coordinates equal for intersection: 3?? + 1 = 2?? + 2, -?? + 1 = 0, -1 = ???? - 4. From -?? + 1 = 0, we find ?? = 1. Substituting ?? = 1 into 3?? + 1 = 2?? + 2 gives: 3(1)+ 1 = 2?? + 2 ? ?? = 1. Also, substituting ?? = 1 into -1 = ???? - 4 gives: -1 = ?? (1)- 4 ? ?? = 3. So, the intersection point ?? is: ?? (4,0,-1). Foot of Perpendicular from ?? to L 2 The point ?? on L 2 is given by: ?? (2?? + 2,0,3?? - 4). For vector ???? ????? = (2?? + 1,-1,3?? - 3), since ???? is perpendicular to L 2 , the dot product should be zero: (2?? + 1)· 2+ (-1)· 0 + (3?? - 3)· 3 = 0. Simplifying gives: 4?? + 2+ 9?? - 9 = 0 ? 13?? = 7 ? ?? = 7 13 . So, the coordinates of point ?? are: ?? ( 40 13 ,0, -31 13 ). Distance PB and Calculation The vector ???? ????? is: ???? ????? = (4 - 40 13 ,0 - 0,-1- ( -31 13 )). Calculating the components: ???? ????? = ( 12 13 ,0, 18 13 ). The square of the distance (???? ) 2 is: ( 12 13 ) 2 + (0) 2 + ( 18 13 ) 2 = 144 169 + 324 169 = 468 169 . Thus, 26?? (???? ) 2 is: 26× 3 × 468 169 = 216 . So, the final value is 216 . Q2: Let ?? be the image of the point ?? (?? ,-?? ,?? ) in the line ?? : ?? -?? ?? = ?? +?? ?? = ?? ?? and ?? (?? ,?? ,?? ) be a point on ?? . Then the square of the area of ? ?????? is ____ . JEE Main 2025 (Online) 24th January Evening Shift Ans: 957 Solution: Let R(2?? + 1,3?? - 1,4?? ) 2?? + 1 = 5 ?? = 2 R(5,5,8) let T(2?? + 1,3?? - 1,4?? ) QT ????? = (2?? - 6)i ˆ + (3?? + 1)j ˆ + (4?? - 5)k ˆ b ? = 2i ˆ + 3j ˆ + 4k ˆ QT ????? · b ? = 0 4?? - 12+ 9?? + 3 + 16?? - 20 = 0 ?? = 1 T(3,2,4) QT = v33 RT = v29 ( area of ? PQR) 2 = ( 1 2 v29· 2v33) 2 = 957 Q3: Let the area of the triangle formed by the lines ?? + ?? = ?? - ?? = ?? , ?? -?? ?? = ?? -?? = ?? -?? ?? and ?? -?? = ?? -?? ?? = ?? -?? ?? be ?? . Then ?? ?? is equal to ____ . JEE Main 2025 (Online) 8th April Evening Shift Ans: 56 Solution: ?? 1 :?? + 2 = ?? - 1 = ?? = l ?? 2 : ?? - 3 5 = ?? -1 = ?? - 1 1 = ?? ?? 3 : ?? -3 = ?? - 3 5 = ?? - 2 1 = ?? Point of intersection of ?? 1 and ?? 2 l - 2 = 5 m + 3 l + 1 = -m l = m + 1 }l = 0, m = -1 A(-2,1,0) Point of intersection of ?? 2 and ?? 3 5?? + 3 = -3?? -?? = 3?? + 3 ?? + 1 = ?? + 2 }?? = 0,?? = -1,?? (3,0,1) Point of intersection ?? 3 and ?? 4 -3n = l - 2 3n + 3 = l + 1 n +2 = l }l = 2,n = 0,C(0,3,2) Ar(? ABC) = | 1 2 | i ˆ j ˆ k ˆ -5 1 -1 -3 3 1 | A = 1 2 |i ˆ(4)- j ˆ(-8)+ k ˆ (-12)| A = 1 2 v16+ 64+ 144= v56 A 2 = 56 Q4: Let ?? ?? : ?? -?? ?? = ?? -?? ?? = ?? -?? ?? and ?? ?? : ?? -?? ?? = ?? -?? ?? = ?? -?? ?? be two lines. Then which of the following points lies on the line of the shortest distance between ?? ?? and ?? ?? ? JEE Main 2025 (Online) 22nd January Morning Shift Options: A. ( 14 3 ,-3, 22 3 ) B. (2,3, 1 3 ) C. ( 8 3 ,-1, 1 3 ) D. (- 5 3 ,-7,1) Solution: ?? 1 : ?? - 1 2 = ?? - 2 3 = ?? - 3 4 ?? 2 : ?? - 2 3 = ?? - 4 4 = ?? - 5 5 Page 5 JEE Main Previous Year Questions (2025): 3D Geometry Q1: Let ?? ?? : ?? -?? ?? = ?? -?? -?? = ?? +?? ?? and ?? ?? : ?? -?? ?? = ?? ?? = ?? +?? ?? ,?? ? ?? , be two lines, which intersect at the point ?? . If ?? is the foot of perpendicular from the point ?? (?? ,?? ,-?? ) on ?? ?? , then the value of ???? ?? ( ???? ) ?? is ____ - JEE Main 2025 (Online) 22nd January Morning Shift Ans: 216 Solution: Explanation To find the value of 26?? ( PB) 2 , we proceed as follows: Intersection Point ?? of L 1 and L 2 We are given the equations of the lines: L 1 : ?? - 1 3 = ?? - 1 -1 = ?? + 1 0 L 2 : ?? - 2 2 = ?? 0 = ?? + 4 ?? For L 1 ,?? + 1 = 0, which implies ?? = -1. For L 2 , since ?? = 0, the direction ratios can be matched using the parameter ?? : ?? = 3?? + 1,?? = -?? + 1,?? = -1, ?? = 2?? + 2,?? = 0,?? = ???? - 4. Setting the coordinates equal for intersection: 3?? + 1 = 2?? + 2, -?? + 1 = 0, -1 = ???? - 4. From -?? + 1 = 0, we find ?? = 1. Substituting ?? = 1 into 3?? + 1 = 2?? + 2 gives: 3(1)+ 1 = 2?? + 2 ? ?? = 1. Also, substituting ?? = 1 into -1 = ???? - 4 gives: -1 = ?? (1)- 4 ? ?? = 3. So, the intersection point ?? is: ?? (4,0,-1). Foot of Perpendicular from ?? to L 2 The point ?? on L 2 is given by: ?? (2?? + 2,0,3?? - 4). For vector ???? ????? = (2?? + 1,-1,3?? - 3), since ???? is perpendicular to L 2 , the dot product should be zero: (2?? + 1)· 2+ (-1)· 0 + (3?? - 3)· 3 = 0. Simplifying gives: 4?? + 2+ 9?? - 9 = 0 ? 13?? = 7 ? ?? = 7 13 . So, the coordinates of point ?? are: ?? ( 40 13 ,0, -31 13 ). Distance PB and Calculation The vector ???? ????? is: ???? ????? = (4 - 40 13 ,0 - 0,-1- ( -31 13 )). Calculating the components: ???? ????? = ( 12 13 ,0, 18 13 ). The square of the distance (???? ) 2 is: ( 12 13 ) 2 + (0) 2 + ( 18 13 ) 2 = 144 169 + 324 169 = 468 169 . Thus, 26?? (???? ) 2 is: 26× 3 × 468 169 = 216 . So, the final value is 216 . Q2: Let ?? be the image of the point ?? (?? ,-?? ,?? ) in the line ?? : ?? -?? ?? = ?? +?? ?? = ?? ?? and ?? (?? ,?? ,?? ) be a point on ?? . Then the square of the area of ? ?????? is ____ . JEE Main 2025 (Online) 24th January Evening Shift Ans: 957 Solution: Let R(2?? + 1,3?? - 1,4?? ) 2?? + 1 = 5 ?? = 2 R(5,5,8) let T(2?? + 1,3?? - 1,4?? ) QT ????? = (2?? - 6)i ˆ + (3?? + 1)j ˆ + (4?? - 5)k ˆ b ? = 2i ˆ + 3j ˆ + 4k ˆ QT ????? · b ? = 0 4?? - 12+ 9?? + 3 + 16?? - 20 = 0 ?? = 1 T(3,2,4) QT = v33 RT = v29 ( area of ? PQR) 2 = ( 1 2 v29· 2v33) 2 = 957 Q3: Let the area of the triangle formed by the lines ?? + ?? = ?? - ?? = ?? , ?? -?? ?? = ?? -?? = ?? -?? ?? and ?? -?? = ?? -?? ?? = ?? -?? ?? be ?? . Then ?? ?? is equal to ____ . JEE Main 2025 (Online) 8th April Evening Shift Ans: 56 Solution: ?? 1 :?? + 2 = ?? - 1 = ?? = l ?? 2 : ?? - 3 5 = ?? -1 = ?? - 1 1 = ?? ?? 3 : ?? -3 = ?? - 3 5 = ?? - 2 1 = ?? Point of intersection of ?? 1 and ?? 2 l - 2 = 5 m + 3 l + 1 = -m l = m + 1 }l = 0, m = -1 A(-2,1,0) Point of intersection of ?? 2 and ?? 3 5?? + 3 = -3?? -?? = 3?? + 3 ?? + 1 = ?? + 2 }?? = 0,?? = -1,?? (3,0,1) Point of intersection ?? 3 and ?? 4 -3n = l - 2 3n + 3 = l + 1 n +2 = l }l = 2,n = 0,C(0,3,2) Ar(? ABC) = | 1 2 | i ˆ j ˆ k ˆ -5 1 -1 -3 3 1 | A = 1 2 |i ˆ(4)- j ˆ(-8)+ k ˆ (-12)| A = 1 2 v16+ 64+ 144= v56 A 2 = 56 Q4: Let ?? ?? : ?? -?? ?? = ?? -?? ?? = ?? -?? ?? and ?? ?? : ?? -?? ?? = ?? -?? ?? = ?? -?? ?? be two lines. Then which of the following points lies on the line of the shortest distance between ?? ?? and ?? ?? ? JEE Main 2025 (Online) 22nd January Morning Shift Options: A. ( 14 3 ,-3, 22 3 ) B. (2,3, 1 3 ) C. ( 8 3 ,-1, 1 3 ) D. (- 5 3 ,-7,1) Solution: ?? 1 : ?? - 1 2 = ?? - 2 3 = ?? - 3 4 ?? 2 : ?? - 2 3 = ?? - 4 4 = ?? - 5 5 ?? (2?? + 1,3?? + 2,4?? + 3) ?? (3?? + 2,4?? + 4,5?? + 5) Dr's of ???? < 2?? - 3?? - 1,3?? - 4?? - 2, 4?? - 5?? - 2 > ???? = | ??ˆ ??ˆ ?? ˆ 2 3 4 3 4 5 | = -??ˆ + 2??ˆ - ?? ˆ ? 2?? - 3?? - 1 -1 = 3?? - 4?? - 2 2 = 4?? - 5?? - 2 -1 ? ?? = 1 3 ?? = -1 6 ? ?? ( 5 3 ,3, 13 3 ) ?? ( 3 2 , 10 3 , 25 6 ) Dr's ???? ?1,-2,1? ? Line ?? - 5 3 1 = ?? - 3 -2 = ?? - 13 3 1 Q5: The perpendicular distance, of the line ?? -?? ?? = ?? +?? -?? = ?? +?? ?? from the point ?? (?? ,-???? ,?? ), is : JEE Main 2025 (Online) 22nd January Evening Shift Options: A. 6 B. 4v3 C. 3v5 D. 5v2 Ans: CRead More
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FAQs on JEE Main Previous Year Questions (2025): 3D Geometry - Mathematics (Maths) for JEE Main & Advanced
| 1. What are the key concepts involved in 3D geometry that students should focus on for JEE preparation? |  |
Ans. Students should focus on the following key concepts in 3D geometry for JEE preparation: coordinates of points in space, distance between two points, section formula, midpoint formula, direction cosines, direction ratios, equations of lines and planes, angles between lines and planes, and the intersection of lines and planes. Understanding these concepts is crucial for solving problems related to the geometry of three-dimensional space.
| 2. How do you find the distance between two points in 3D space? | |
Ans. The distance between two points A(x₁, y₁, z₁) and B(x₂, y₂, z₂) in 3D space can be calculated using the distance formula:
Distance (AB) = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²]. This formula derives from the Pythagorean theorem applied in three dimensions.
| 3. What is the section formula in 3D geometry and how is it used? | |
Ans. The section formula in 3D geometry is used to find the coordinates of a point that divides the line segment joining two points in a given ratio. If point P divides the line segment joining points A(x₁, y₁, z₁) and B(x₂, y₂, z₂) in the ratio m:n, the coordinates of P can be expressed as:
P(x, y, z) = ((mx₂ + nx₁) / (m + n), (my₂ + ny₁) / (m + n), (mz₂ + nz₁) / (m + n)). This is useful in various applications, including finding centroids and midpoints.
| 4. How can one determine the angle between two lines in 3D geometry? | |
Ans. The angle θ between two lines can be determined using their direction ratios. If the direction ratios of the two lines are given as l₁, m₁, n₁ and l₂, m₂, n₂, then the cosine of the angle between the lines can be calculated using the formula:
cos(θ) = (l₁l₂ + m₁m₂ + n₁n₂) / (√(l₁² + m₁² + n₁²) * √(l₂² + m₂² + n₂²)). This formula is derived from the dot product of vectors.
| 5. What are the equations of a plane in 3D geometry, and how can one find the equation given a point and a normal vector? | |
Ans. The equation of a plane in 3D geometry can be expressed in the form Ax + By + Cz + D = 0, where (A, B, C) are the components of the normal vector to the plane. To find the equation of a plane given a point P(x₀, y₀, z₀) and a normal vector N(A, B, C), you can use the point-normal form, which is
A(x - x₀) + B(y - y₀) + C(z - z₀) = 0. Expanding this will yield the standard form of the plane's equation.
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