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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 121-140 of 140 questions

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121 Easy
121 of 140
Application +4 | -1
Locus of centroid of the 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 tt 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(3x - 1)^2 + (3y)^2 = a^2 + b^2) is correct because it represents the locus of the centroid, derived by eliminating the parameter t.
122 Easy
122 of 140
Application +4 | -1
The intercepts of the plane 2x−3y+4z=122 x-3 y+4 z=12 on the coordinate axes are given by
View answer and explanation
Correct option: B
6,−4,36,-4,3
Option B (6,−4,36,-4,3) is correct because it represents the points where the plane intersects the x, y, and z axes.
123 Easy
123 of 140
Application +4 | -1
The ratio in which the line x+y=4x + y = 4 divides the line joining the points (1,  −1)\left( {1,\; - 1} \right) and (5, 7) is
View answer and explanation
Correct option: A
1:21:2
Option A (1:21:2) is correct because by using the section formula and substituting the point of intersection into the equation of the line, we find the ratio to be 1:21:2.
124 Easy
124 of 140
Application +4 | -1
The number of lines that can be drawn through the point (4,−5)(4,-5) at a distance 12 from the point (−2,3)(-2,3) is
View answer and explanation
Correct option: A
0
Option A (0) is correct because the distance between the given points is less than the required distance, so no such line can be drawn.
125 Easy
125 of 140
Application +4 | -1
If the vertices AA and BB of a triangle ABCA B C are given by (2,5)(2,5) and (4,−11)(4,-11) and CC moves along the line L1:9x+7y+4=0L_1: 9 x+7 y+4=0, the locus of the centroid of the triangle ABCA B C is a straight line parallel to
View answer and explanation
Correct option: D
L1L_1
Option D (L1L_1) is correct because the locus of the centroid will be a line parallel to the line on which C moves.
126 Medium
126 of 140
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.
127 Easy
127 of 140
Application +4 | -1
The equation of the sphere whose centre has the position vector (3i^+6j^−4k^)(3 \hat{i}+6 \hat{j}-4 \hat{k}) and which touches the plane r⋅(2i^−2j^−k^)=10{r} \cdot(2 \hat{i}-2 \hat{j}-\hat{k})=10 is
View answer and explanation
Correct option: A
∣r−(3i^+6j^−4k^)∣=4|{r}-(3 \hat{i}+6 \hat{j}-4 \hat{k})|=4
Option A (∣r−(3i^+6j^−4k^)∣=4|{r}-(3 \hat{i}+6 \hat{j}-4 \hat{k})|=4) is correct because it represents the equation of the sphere with the given center and radius.
128 Medium
128 of 140
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.
129 Easy
129 of 140
Application +4 | -1
The coordinates of the point of intersection of the line x+11=y+33=z−2−2\frac{x+1}{1}=\frac{y+3}{3}=\frac{z-2}{-2} with the plane 3x+4y+5z=253 x+4 y+5 z=25 are
View answer and explanation
Correct option: B
(5,15,−10)(5,15,-10)
Option B ((5,15,−10)(5,15,-10)) is correct because these coordinates satisfy both the line equation and the plane equation.
130 Medium
130 of 140
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.
131 Medium
131 of 140
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.
132 Medium
132 of 140
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.
133 Medium
133 of 140
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.
134 Medium
134 of 140
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.
135 Easy
135 of 140
Application +4 | -1
If a straight line makes an angle of π4\frac{\pi}{4} with each of xx axis and yy-axis, then the angle made by the line with zz-axis is
View answer and explanation
Correct option: C
π2\frac{\pi}{2}
Option C (π2\frac{{\pi}}{{2}}) is correct because the sum of squares of direction cosines must equal 1.
136 Easy
136 of 140
Application +4 | -1
The projection of the line segment joining the points (−1,0,3)(-1,0,3) and (2,5,1)(2,5,1) on the line whose direction ratios are 6,2,36,2,3, is
View answer and explanation
Correct option: A
227\frac{22}{7}
Option A (227\frac{{22}}{{7}}) is correct because it is the projection of the line segment on the line with given direction ratios.
137 Medium
137 of 140
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.
138 Easy
138 of 140
Analyzing +4 | -1
Points (−2,4,7),(3,−6,−8)(-2,4,7),(3,-6,-8) and (1,−2,−2)(1,-2,-2) are
View answer and explanation
Correct option: A
Collinear
Option A (Collinear) is correct because the ratios of the differences in coordinates are equal, indicating that the points lie on the same line.
139 Medium
139 of 140
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.
140 Medium
140 of 140
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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