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JEE • MATHEMATICS • 3D GEOMETRY SYSTEM

3d geometry system practice · Mixed difficulty

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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 91-120 of 140 questions

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91 Easy
91 of 140
Application +4 | -1
If the projection of a line segment on x,yx, y and zz axes are respectively 3,4 and 5 , then the length of the line segment is
View answer and explanation
Correct option: B
525 \sqrt{2}
Option B (525 \sqrt{2}) is correct because the length of the line segment is the square root of the sum of the squares of its projections on the axes.
92 Easy
92 of 140
Application +4 | -1
If the origin is the centroid of a ΔABC\Delta A B C having vertices A(a,1,3),B(−2,b,−5)A(a, 1,3), B(-2, b,-5) and C(4,7,c)C(4,7, c), then
View answer and explanation
Correct option: A
a=−2a=-2
Option A (a=−2a=-2) is correct because the centroid of a triangle is the average of the coordinates of its vertices, and the origin (0,0,0) is given as the centroid.
93 Easy
93 of 140
Application +4 | -1
Given that A(3,2,−4),B(5,4−6){A}(3,2,-4), {B}(5,4-6) and C(9,8,−10){C}(9,8,-10) are collinear. Ratio in which BB divides ACAC is 1:m1:{m}. The value of m{m} is
View answer and explanation
Correct option: A
2
Option A (2) is correct because B divides AC in the ratio 1:2, calculated using the section formula for collinear points.
94 Easy
94 of 140
Analyzing +4 | -1
Vertices of a variable triangle are (3,4),(5cos⁡θ,5sin⁡θ)(3,4),(5 \cos \theta, 5 \sin \theta) and (5sin⁡θ,−5cos⁡θ)(5 \sin \theta,-5 \cos \theta), where θ∈R\theta \in R. Locus of its orthocentre is
View answer and explanation
Correct option: D
x2+y2−6x−8y−25=0x^2+y^2-6 x-8 y-25=0
Option D (x2+y2−6x−8y−25=0x^2+y^2-6 x-8 y-25=0) is correct because it represents the locus of the orthocenter derived from the given vertices and their relationships to the circumcenter and centroid.
95 Easy
95 of 140
Application +4 | -1
If the projections of a line on the axes are 6,2,36,2,3, then its direction cosines are
View answer and explanation
Correct option: D
67,27,37\frac{6}{7}, \frac{2}{7}, \frac{3}{7}
Option D (67,27,37\frac{{6}}{{7}}, \frac{{2}}{{7}}, \frac{{3}}{{7}}) is correct because it represents the direction cosines calculated from the given projections on the axes.
96 Easy
96 of 140
Application +4 | -1
The locus of a points which moves such that the sum of the squares of its distance from three vertices of the triangle is constant is a/an
View answer and explanation
Correct option: A
Circle
Option A (Circle) is correct because the locus of the point satisfying the given condition results in the equation of a circle.
97 Easy
97 of 140
Understanding +4 | -1
The reflection of the point (α,β,γ)(\alpha, \beta, \gamma) in the XY-plane is
View answer and explanation
Correct option: D
(α,β,−γ)(\alpha, \beta,-\gamma)
Option D (α,β,−γ\alpha, \beta,-\gamma) is correct because reflection in the XY-plane changes the sign of the z-coordinate while leaving the x and y coordinates unchanged.
98 Easy
98 of 140
Application +4 | -1
Assertion: The coordinates of the point which divides the join of A(2,−1,4){A}(2,-1,4) and B(4,3,2){B}(4,3,2) in the ratio 2:32: 3 externally is C(−2,−9,8)C(-2,-9,8)

Reason: If P(x1,y1,z1){P}\left({x}_1, {y}_1, {z}_1\right) and Q(x2,y2,z2){Q}\left({x}_2, {y}_2, {z}_2\right) be two points, and let R{R} be a point on PQ{PQ} produced dividing it externally in the ratio m1:m2{m}_1: {m}_2. Then the coordinates of R{R} are

(m1x2−m2x1m1−m2,m1y2−m2y1m1−m2,m1z2−m2z1m1−m2)\left(\frac{{m}_1 {x}_2-{m}_2 {x}_1}{{m}_1-{m}_2}, \frac{{m}_1 {y}_2-{m}_2 {y}_1}{{m}_1-{m}_2}, \frac{{m}_1 {z}_2-{m}_2 {z}_1}{{m}_{{1}}-{m}_2}\right)
View answer and explanation
Correct option: A
Assertion is correct, reason is correct; reason is a correct explanation for assertion
Option A is correct because the assertion is true, the reason is true, and the reason correctly explains the assertion about external division of a line segment in 3D space.
99 Easy
99 of 140
Application +4 | -1
The ratio in which the xx-axis divides the line segment joining (3,6)\left( {3,6} \right) and (4,−3)\left( {4, - 3} \right) is
View answer and explanation
Correct option: A
2  :12\;:1
Option A (2  :12\;:1) is correct because the x-axis divides the line segment in the ratio of the negative of the y-coordinates, which simplifies to 2  :12\;:1.
100 Medium
100 of 140
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.
101 Medium
101 of 140
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}.
102 Easy
102 of 140
Application +4 | -1
The locus of a point PP which divides the line joining (1,0)(1,0) and (2cos⁡θ,2sin⁡θ)(2 \cos \theta, 2 \sin \theta) internally in the ratio 2:32: 3 for all θ\theta is
View answer and explanation
Correct option: B
a circle
Option B (a circle) is correct because the locus of the point P can be shown to satisfy the equation of a circle after eliminating the parameter θ\theta.
103 Medium
103 of 140
Application +4 | -1
The distance between the points (1,4,5)(1, 4, 5) and (2,2,3)(2, 2, 3) is___________.
View answer and explanation
Correct option: A
3
Option A (3) is correct because the distance between the two points is calculated using the 3D distance formula.
104 Easy
104 of 140
Application +4 | -1
The line joining A(bcos⁡α,bsin⁡α)A\left( {b\cos \mathrm{\alpha },b\sin \mathrm{\alpha }} \right) and B(acos⁡β,asin⁡β)B\left( {a\cos \mathrm{\beta },a\sin \mathrm{\beta }} \right) is produced to the point M(x,y)M\left( {x,y} \right) so that AM:MB=b:aAM:MB = b:a, then

xcos⁡(α+β2)+ysin⁡(α+β2)x\cos \left( {\frac{{\mathrm{\alpha } + \mathrm{\beta }}}{2}} \right) + y\sin \left( {\frac{{\mathrm{\alpha } + \mathrm{\beta }}}{2}} \right) is
View answer and explanation
Correct option: B
0
Option B (0) is correct because by using the section formula and trigonometric identities, the expression simplifies to 0.
105 Easy
105 of 140
Application +4 | -1
If the centroid of the triangle formed by the points (0, 0), (cos⁡θ,sin⁡θ)\left( {\cos \mathrm{\theta },\sin \mathrm{\theta }} \right) and (sin⁡θ,  −cos⁡θ)\left( {\sin \mathrm{\theta },\; - \cos \mathrm{\theta }} \right) lies on the line y=2xy = 2x, then θ\mathrm{\theta } is equal to
View answer and explanation
Correct option: C
tan⁡−1(−3){\tan ^{ - 1}}\left( { - 3} \right)
Option C (tan⁡−1(−3){\tan ^{ - 1}}\left( { - 3} \right)) is correct because the centroid coordinates satisfy the equation of the line y = 2x, leading to the solution tan⁡θ=−3\tan{\theta} = -3.
106 Easy
106 of 140
Application +4 | -1
The distance between the parallel lines given by (x+7y)2+42(x+7y)−42=0(x+7 y)^2+4 \sqrt{2}(x+7 y)-42=0 is
View answer and explanation
Correct option: C
2
Option C (2) is correct because the distance between the parallel lines is calculated by rewriting the equation and using the formula for the distance between parallel lines.
107 Medium
107 of 140
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.
108 Easy
108 of 140
Application +4 | -1
If the distance between (2, 3) and (−5,  2)\left( { - 5,\;2} \right) is equal to the distance between (x,  2)\left( {x,\;2} \right) and (1, 3), then the values of xx are
View answer and explanation
Correct option: A
−6,  8- 6,\;8
Option A (-6, 8) is correct because it satisfies the given condition that the distance between (2, 3) and (-5, 2) is equal to the distance between (x, 2) and (1, 3).
109 Easy
109 of 140
Application +4 | -1
The direction ratios of the line x−y+z−5=0=x−3y−6x-y+z-5=0=x-3 y-6 are:
View answer and explanation
Correct option: A
3,1,−23,1,-2
Option A (3,1,−23,1,-2) is correct because it satisfies the condition for the direction ratios of the line formed by the intersection of the two planes.
110 Medium
110 of 140
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.
111 Medium
111 of 140
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.
112 Medium
112 of 140
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.
113 Easy
113 of 140
Application +4 | -1
Let A(a, 0),  B(b, 2b+1)A(a,\,0),\,\,B(b,\,2b + 1) and C(0, b),  b≠0,  ∣b∣≠1C(0,\,b),\,\,b \ne 0,\,\,\left| b \right| \ne 1, be points such that the area of ΔABC\Delta ABC is 1 sq. unit, then the sum of all possible values of aa is
View answer and explanation
Correct option: D
−2b2b+1\frac{{ - 2{b^2}}}{{b + 1}}
Option D (−2b2b+1\frac{{ - 2{b^2}}}{{b + 1}}) is correct because it represents the sum of all possible values of 'a' calculated using the area of the triangle ABC.
114 Medium
114 of 140
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.
115 Easy
115 of 140
Application +4 | -1
The ordered pair (λ,μ)(\lambda, \mu) such that the points (λ,μ,−6),(3,2,−4)(\lambda, \mu,-6),(3,2,-4) and (9,8,−10)(9,8,-10) become collinear is
View answer and explanation
Correct option: B
(5,4)(5,4)
Option B ((5,4)) is correct because for collinear points, the ratios of the differences in their coordinates must be equal.
116 Easy
116 of 140
Application +4 | -1
If the points (−1,2,−3),(4,a,1)(-1,2,-3),(4, a, 1) and (b,8,5)(b, 8,5) are collinear, then aa and bb are respectively equal to
View answer and explanation
Correct option: C
5 and 9
Option C (5 and 9) is correct because collinear points have proportional differences in their coordinates, allowing us to solve for a and b.
117 Easy
117 of 140
Application +4 | -1
Locus of the centroid of triangle whose vertices are (acos⁡  t,  asin⁡t),(bsin⁡t,−bcos⁡t)\left( {a\cos \;t,\;a\sin t} \right),\left( {b\sin t, - b\cos t} \right) and (1,0), where t is a parameter, is
View answer and explanation
Correct option: B
(3x−1)2+(3y)2=a2+b2{\left( {3x - 1} \right)^2} + {\left( {3y} \right)^2} = {a^2} + {b^2}
Option B ((3x−1)2+(3y)2=a2+b2{\left( {3x - 1} \right)^2} + {\left( {3y} \right)^2} = {a^2} + {b^2}) is correct because it represents the locus of the centroid derived from the given vertices.
118 Easy
118 of 140
Application +4 | -1
If the centroid of the triangle formed by the points (a,b),(b,c)\left( {a,b} \right),\left( {b,c} \right) and (c,a)\left( {c,a} \right) is at the origin, then a3+b3+c3={a^3} + {b^3} + {c^3} =
View answer and explanation
Correct option: C
3abc3abc
Option C (3abc3abc) is correct because if the centroid of the triangle is at the origin, then a+b+c=0a + b + c = 0, which implies a3+b3+c3=3abc{a^3} + {b^3} + {c^3} = 3abc.
119 Medium
119 of 140
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.
120 Easy
120 of 140
Application +4 | -1
A(3,2,0),B(5,3,2)A(3,2,0), B(5,3,2) and C(−9,6,−3)C(-9,6,-3) are the vertices of a triangle ABCA B C. If the bisector of ∠ABC\angle A B C meets BC\mathrm{BC} at D\mathrm{D}, then coordinates of DD are
View answer and explanation
Correct option: A
(198,5716,1716)\left(\frac{19}{8}, \frac{57}{16}, \frac{17}{16}\right)
Option A is correct because the coordinates of D are found by using the section formula with the ratio AB:AC.
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