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JEE • MATHEMATICS • 3D GEOMETRY SYSTEM

3d geometry system practice · Medium

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 31-53 of 53 questions
31 Medium
31 of 53
Analyzing +4 | -1
The points (0,7,10),(−1,6,6)(0,7,10),(-1,6,6) and (−4,9,6)(-4,9,6) form
View answer and explanation
Correct option: A
A right angled isosceles triangle
Option A (A right angled isosceles triangle) is correct because the sides PQ and QR are equal, and PQ2+QR2=PR2PQ^2 + QR^2 = PR^2, satisfying the Pythagorean theorem.
32 Medium
32 of 53
Remembering +4 | -1
The perpendicular distance of the point P(6,7,8)P(6,7,8) from XYX Y-plane is ____.
View answer and explanation
Correct option: A
8
Option A (8) is correct because the perpendicular distance of a point from the XY-plane is the absolute value of its z-coordinate.
33 Medium
33 of 53
Application +4 | -1
The equation of the plane through the points (2,−1,0)(2,-1,0), (3,−4,5)(3,-4,5) and parallel to the line 2x=3y=4z2 x=3 y=4 z is
View answer and explanation
Correct option: B
29(x−2)−27(y+1)−22z=029(x-2)-27(y+1)-22 z=0
Option B (29(x−2)−27(y+1)−22z=029(x-2)-27(y+1)-22 z=0) is the correct equation of the plane, derived by finding the normal vector and using the point-normal form of a plane equation.
34 Medium
34 of 53
Application +4 | -1
The perpendicular distance of the point (6,5,8)(6, 5, 8) from Y-axis is_________.
View answer and explanation
Correct option: A
10
Option A (10) is correct because the perpendicular distance from the Y-axis is calculated using the x and z coordinates: x2+z2\sqrt{x^2 + z^2}.
35 Medium
35 of 53
Analyzing +4 | -1
The position of a moving point in the xy plane at time t is given by (ucos⁡α.t,  usin⁡α.t−12g  t2),\left( {u\cos \alpha .t,\;u\sin \alpha .t - \frac{1}{2}g\;{t^2}} \right), where u,α,gu,\alpha ,g are constants. The locus of the moving point is
View answer and explanation
Correct option: B
A parabola
Option B (A parabola) is correct because the locus of the moving point is a parabolic trajectory, which is characteristic of projectile motion.
36 Medium
36 of 53
Application +4 | -1
The perimeter of triangle with vertices (1,0,0),(0,1,0)(1, 0, 0), (0, 1, 0) and (0,0,1)(0, 0, 1) is k2,k \sqrt{2}, where the value of kk is _______.
View answer and explanation
Correct option: A
323 \sqrt{2}
Option B (3) is correct because the perimeter is calculated by summing the lengths of the three sides, each of which is 2\sqrt{2}.
37 Medium
37 of 53
Application +4 | -1
The distance between the points (1,4,5)(1, 4, 5) and (2,2,3)(2, 2, 3) is___________.
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Correct option: A
3
Option A (3) is correct because the distance between the two points is calculated using the 3D distance formula.
38 Medium
38 of 53
Application +4 | -1
The equation of the plane containing the points A(1,0,1){A}(1,0,1) and B(3,1,2){B}(3,1,2) and parallel to the line joining the origin to the point C(1,−1,2){C}(1,-1,2) is
View answer and explanation
Correct option: D
x−y−z=0x-y-z=0
Option D (x−y−z=0x-y-z=0) is correct because it is the equation of the plane containing the given points and parallel to the given line.
39 Medium
39 of 53
Application +4 | -1
LL is the foot of the perpendicular drawn from a point P(3,4,5)P(3,4,5) on the XYX Y-plane. The coordinates of point LL are
View answer and explanation
Correct option: D
None of these
The correct answer is D (None of these) because the foot of the perpendicular from P(3,4,5) on the XY-plane is (3,4,0), which is not listed in the options.
40 Medium
40 of 53
Application +4 | -1
Three vertices of a parallelogram ABCD\mathrm{ABCD} are A(3,−1,2),B(1,2,−4)\mathrm{A}(3,-1,2), \mathrm{B}(1,2,-4) and C(−1,1,2)\mathrm{C}(-1,1,2). The coordinates of fourth vertex D\mathrm{D} are
View answer and explanation
Correct option: B
(1,−2,8)(1,-2,8)
Option B (1,-2,8) is correct because in a parallelogram, the midpoints of the diagonals coincide. By equating the midpoints of AC and BD, we can solve for the coordinates of vertex D.
41 Medium
41 of 53
Application +4 | -1
Coordinates of the vertices BB and CC of the base of a triangle ABCA B C are (−a,0)(-a, 0) and (a,0)(a, 0) respectively. If C−B=π/3C-B=\pi / 3, the vertex AA lies on the curve
View answer and explanation
Correct option: C
3(x2−y2)+2xy−3a2=0\sqrt{3}\left(x^2-y^2\right)+2 x y-\sqrt{3} a^2=0
Option C (3(x2−y2)+2xy−3a2=0\sqrt{3}(x^2-y^2)+2 x y-\sqrt{3} a^2=0) is correct because it represents the locus of point A such that the difference of angles C and B is π/3.
42 Medium
42 of 53
Understanding +4 | -1
A plane is parallel to XZ-plane, so it is perpendicular to
View answer and explanation
Correct option: B
Y-axis
The correct answer is Y-axis because a plane parallel to the XZ-plane is perpendicular to the Y-axis.
43 Medium
43 of 53
Application +4 | -1
The coordinates of three vertices of a quadrilateral in order are (6,1),(7,2) and (-1,0). If the area of the quadrilateral is 4 square units, then the locus of the fourth vertex is
View answer and explanation
Correct option: C
(x−7y)2+14(x−7y)−15=0{\left( {x - 7y} \right)^2} + 14\left( {x - 7y} \right) - 15 = 0
Option D ((x−7y)2+14(x−7y)−15=0{\left( {x - 7y} \right)^2} + 14\left( {x - 7y} \right) - 15 = 0) is correct because it represents the locus of the fourth vertex derived from the area condition of the quadrilateral.
44 Medium
44 of 53
Application +4 | -1
If AA and BB are the points (3,4,5)(3,4,5) and (−1,3,−7)(-1,3,-7) respectively, then the equation of set of point PP such that PA2+PB2=k2P A^2+P B^2=k^2, where kk is constant, is
View answer and explanation
Correct option: A
2(x2+y2+z2)−4x−14y+4z+109−k2=02\left(x^2+y^2+z^2\right)-4 x-14 y+4 z+109-k^2=0
Option A (2(x2+y2+z2)−4x−14y+4z+109−k2=02(x^2+y^2+z^2)-4x-14y+4z+109-k^2=0) is correct because it represents the equation derived from the given condition PA2+PB2=k2PA^2 + PB^2 = k^2.
45 Medium
45 of 53
Application +4 | -1
If G(1,  4)G\left( {1,\;4} \right) is the centroid of triangle ABCABC having its two vertices AA and BB at (4,−3)\left( {4, - 3} \right) and (−9,  7)\left( { - 9,\;7} \right) respectively, then area of the triangle ABCABC in square units, is
View answer and explanation
Correct option: C
1832\frac{{183}}{2}
Option C (1832\frac{{183}}{{2}}) is correct because the area of triangle ABC is three times the area of triangle GAB, where G is the centroid.

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46 Medium
46 of 53
Application +4 | -1
The equation of the set of point PP, the sum of whose distances from A(4,0,0)A(4,0,0) and B(−4,0,0)B(-4,0,0) is equal to 10, is
View answer and explanation
Correct option: B
9x2+25y2+25z2−225=09 x^2+25 y^2+25 z^2-225=0
Option B (9x2+25y2+25z2−225=09 x^2+25 y^2+25 z^2-225=0) is correct because it represents the equation of the set of points where the sum of distances from A(4,0,0) and B(-4,0,0) is 10, derived using the distance formula and algebraic manipulation.
47 Medium
47 of 53
Application +4 | -1
A plane passing through (1,1,1)(1,1,1) intersects the positive direction at A,BA, B and CC, then the volume of the tetrahedron OABCO A B C satisfies (V=Volume)
View answer and explanation
Correct option: B
V≥92V \geq \frac{9}{2}
Option B (V ≥ 9/2) is correct because using the equation of a plane and applying AM-GM inequality, the volume of the tetrahedron is greater than or equal to 9/2.
48 Medium
48 of 53
Application +4 | -1
If the axes rectangular, the distance from the point (3,4,5)(3,4,5) to the point where the line x−31=y−42=z−52\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2} meets the plane x+y+z=17x+y+z=17 is
View answer and explanation
Correct option: C
3
Option C (3) is correct because it represents the distance from the point to the intersection of the line and the plane.
49 Medium
49 of 53
Application +4 | -1
A sphere of constant radius 2k2 k passes through the origin and meets the axes in A,B,CA, B, C. The locus of the centroid of the tetrahedron OABC{OABC} is
View answer and explanation
Correct option: C
x2+y2+z2=k2x^2+y^2+z^2=k^2
Option A (x2+y2+z2=k2x^2+y^2+z^2=k^2) is correct because it represents the locus of the centroid of the tetrahedron OABC, derived from the sphere's equation and the centroid's coordinates.
50 Medium
50 of 53
Application +4 | -1
If the distance between the points P(acos⁡48∘,0)P\left( {a\cos 48^\circ ,0} \right) and Q(0,acos⁡12∘)Q\left( {0,a\cos 12^\circ } \right) is d,d, then d2−a2={d^2} - {a^2} =
View answer and explanation
Correct option: D
a28(5+1)\frac{{{a^2}}}{8}\left( {\sqrt 5 + 1} \right)
The correct answer is D a28(5+1)\frac{{{a^2}}}{8}\left( {\sqrt 5 + 1} \right) because it is obtained by using the distance formula and trigonometric identities to simplify the expression.
51 Medium
51 of 53
Application +4 | -1
If P(3,7)P\left( {3,7} \right) is a point on the line joining A(1,1)A\left( {1,1} \right) and B(6,16),B\left( {6,16} \right), then the harmonic conjugate QQ of point PP has the coordinates
View answer and explanation
Correct option: D
(−9,−29)\left( { - 9, - 29} \right)
Option D ((-9, -29)) is correct because it represents the harmonic conjugate of point P, calculated using the section formula externally.
52 Medium
52 of 53
Application +4 | -1
A plane passes through the point (1,1,1)(1,1,1) cuts the coordinate axes at A,B&C\mathrm{A}, \mathrm{B} \& \mathrm{C} respectively such that OA=OB=OC\mathrm{OA}=\mathrm{OB}=\mathrm{OC}, where O\mathrm{O} is the origin and volume of tetrahedron OABC\mathrm{OABC} is λ\lambda then value of 2λ2 \lambda is equal to
View answer and explanation
Correct option: A
9
Option A (9) is correct because the volume of the tetrahedron is calculated using the intercepts and then doubled.
53 Medium
53 of 53
Application +4 | -1
The vertex AA of the triangle ABCA B C is on the line r⃗=i^+j^+λk^\vec{r}=\hat{i}+\hat{j}+\lambda \hat{k} and the vertices BB and CC have respective position vectors i^\hat{i} and j^\hat{j}. Let Δ\Delta be the area of the triangle and Δ∈[32,332]\Delta \in\left[\frac{3}{2}, \frac{\sqrt{33}}{2}\right], then the range of values λ\lambda corresponding to 'AA' is
View answer and explanation
Correct option: D
[−4,−2]∪[2,4][-4,-2] \cup[2,4]
Option D, [−4,−2]∪[2,4][-4,-2] \cup[2,4], is correct because it represents the range of λ\lambda values that satisfy the given area condition for the triangle.
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