3D Geometry · JEE Main previous year questions

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2025 13 questions

Q162025 · 8 Apr 2025 · Shift 2
There are 12 points in a plane, no three of which are in the same straight line, except 5 points which are collinear. Then the total number of triangles that can be formed with the vertices at any three of these 12 points is
1. A.230
2. B.210
3. C.200
4. D.220
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Answer: B
Q22025 · 4 Apr 2025 · Shift 1
Let the shortest distance between the lines $\frac{x - 3}{3} = \frac{y - α}{- 1} = \frac{z - 3}{1}$ and $\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - β}{4}$ be $330$ . Then the positive value of $5 α + β$ is
1. A. $42$
2. B. $40$
3. C. $48$
4. D. $46$
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Answer: D
Q132025 · 3 Apr 2025 · Shift 1
Line $L_{1}$ passes through the point $(1, 2, 3)$ and is parallel to $z$ -axis. Line $L_{2}$ passes through the point $(λ, 5, 6)$ and is parallel to $y$ -axis. Let for $λ = λ_{1}, λ_{2}$ , $λ_{2} < λ_{1}$ , the shortest distance between the two lines be 3 . Then the square of the distance of the point $(λ_{1}, λ_{2}, 7)$ from the line $L_{1}$ is
1. A.25
2. B.32
3. C.40
4. D.37
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Answer: A
Q62025 · 3 Apr 2025 · Shift 2
The shortest distance between the curves $y^{2} = 8 x$ and $x^{2} + y^{2} + 12 y + 35 = 0$ is:
1. A. $23 - 1$
2. B. $22 - 1$
3. C. $32 - 1$
4. D. $2$
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Answer: B
Q82025 · 2 Apr 2025 · Shift 1
Let $A B C D$ be a tetrahedron such that the edges $A B, A C$ and $A D$ are mutually perpendicular. Let the areas of the triangles $A B C, A C D$ and $A D B$ be $5, 6$ and $7$ square units respectively. Then the area (in square units) of the $△ B C D$ is equal to :
1. A. $110$
2. B.12
3. C. $340$
4. D. $73$
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Answer: A
Q82025 · 2 Apr 2025 · Shift 2
If the image of the point $P (1, 0, 3)$ in the line joining the points $A (4, 7, 1)$ and $B (3, 5, 3)$ is $Q (α, β, γ)$ , then $α + β + γ$ is equal to :
1. A. $\frac{46}{3}$
2. B.18
3. C.13
4. D. $\frac{47}{3}$
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Answer: A
Q132025 · 2 Apr 2025 · Shift 2
The line $L_{1}$ is parallel to the vector $a = - 3 \hat{i} + 2 \hat{j} + 4 \hat{k}$ and passes through the point $(7, 6, 2)$ and the line $L_{2}$ is parallel to the vector $b = 2 \hat{i} + \hat{j} + 3 \hat{k}$ and passes through the point $(5, 3, 4)$ . The shortest distance between the lines $L_{1}$ and $L_{2}$ is :
1. A. $\frac{23}{38}$
2. B. $\frac{21}{38}$
3. C. $\frac{23}{57}$
4. D. $\frac{21}{57}$
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Answer: A
Q32025 · 29 Jan 2025 · Shift 2
Let a straight line $L$ pass through the point $P (2, - 1, 3)$ and be perpendicular to the lines $\frac{x - 1}{2} = \frac{y + 1}{1} = \frac{z - 3}{- 2}$ and $\frac{x - 3}{1} = \frac{y - 2}{3} = \frac{z + 2}{4}$ . If the line $L$ intersects the $y z$ -plane at the point $Q$ , then the distance between the points $P$ and $Q$ is:
1. A. $10$
2. B. $2$
3. C. $23$
4. D. $3$
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Answer: D
Q182025 · 29 Jan 2025 · Shift 2
Let P be the foot of the perpendicular from the point $(1, 2, 2)$ on the line $L : \frac{x - 1}{1} = \frac{y + 1}{- 1} = \frac{z - 2}{2}$ . Let the line $r = (- \hat{i} + \hat{j} - 2 \hat{k}) + λ (\hat{i} - \hat{j} + \hat{k}), λ \in R$ , intersect the line L at Q . Then $2 (PQ)^{2}$ is equal to :
1. A. $25$
2. B. $27$
3. C. $19$
4. D. $29$
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Answer: B
Q12025 · 28 Jan 2025 · Shift 1
Let $A (x, y, z)$ be a point in $x y$ -plane, which is equidistant from three points $(0, 3, 2)$ , $(2, 0, 3)$ and $(0, 0, 1)$ . Let $B = (1, 4, - 1)$ and $C = (2, 0, - 2)$ . Then among the statements (S1) : $△ ABC$ is an isosceles right angled triangle, and (S2) : the area of $△ ABC$ is $\frac{9 2}{2}$ ,
1. A.both are false
2. B.only (S2) is true
3. C.only (S1) is true
4. D.both are true
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Answer: C
Q132025 · 28 Jan 2025 · Shift 1
If the image of the point $(4, 4, 3)$ in the line $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 1}{3}$ is $(α, β, γ)$ , then $α + β + γ$ is equal to
1. A.12
2. B.9
3. C.7
4. D.8
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Answer: B
Q212025 · 24 Jan 2025 · Shift 2· Numerical answer
Let P be the image of the point $Q (7, - 2, 5)$ in the line $L : \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$ and $R (5, p, q)$ be a point on $L$ . Then the square of the area of $△ P QR$ is ________.
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Answer: 957
Q42025 · 22 Jan 2025 · Shift 1
Let $L_{1} : \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $L_{2} : \frac{x - 2}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}$ be two lines. Then which of the following points lies on the line of the shortest distance between $L_{1}$ and $L_{2}$ ?
1. A. $(\frac{14}{3}, - 3, \frac{22}{3})$
2. B. $(2, 3, \frac{1}{3})$
3. C. $(\frac{8}{3}, - 1, \frac{1}{3})$
4. D. $(- \frac{5}{3}, - 7, 1)$
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Answer: A