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JEE • MATHEMATICS • 3D PLANES

3d planes practice · Mixed difficulty

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Practice format and chapter quiz

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 181-210 of 241 questions

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181 Easy
181 of 241
Application +4 | -1
A variable plane is at a distance, kk from the origin and meets the coordinates axis in A,B,CA, B, C. Then, the locus of the centroid of △ABC\triangle A B C is
View answer and explanation
Correct option: D
x−2+y−2+z−2=9k−2x^{-2}+y^{-2}+z^{-2}=9 k^{-2}
Option D is correct because the locus of the centroid is found by using the equation of a plane and the distance from the origin.
182 Easy
182 of 241
Analyzing +4 | -1
If the plane 2x+y−5z=02x + y - 5z = 0 is rotated about its line of intersection with the plane 3x−y+4z−7=03x - y + 4z - 7 = 0 by an angle of π2,\frac{\pi }{2}, then the plane after the rotation passes through the point:
View answer and explanation
Correct option: C
(1, 0, 2)
Option C ((1, 0, 2)) is correct because the rotated plane passes through this point after finding the equation of the plane using the angle of rotation and the line of intersection.
183 Easy
183 of 241
Application +4 | -1
The equation of a plane containing the line of intersection of the planes 2x−y−4=02x - y - 4 = 0 and y+2z−4=0y + 2z - 4 = 0 and passing through the point (1, 1, 0) is
View answer and explanation
Correct option: C
x−y−z=0x - y - z = 0
Option C is correct because it accurately determines the equation of the plane containing the line of intersection and passing through the given point.
184 Easy
184 of 241
Application +4 | -1
The two lines x=ay+b,z=cy+dx=a y+\mathrm{b}, z=c y+d and x=a′y+b′,z=c′y+d′x=a^{\prime} y+b^{\prime}, z=c^{\prime} y+d^{\prime} will be perpendicular, if and only if
View answer and explanation
Correct option: A
aa′+cc′+1=0a a^{\prime}+c c^{\prime}+1=0
Option A (aa′+cc′+1=0a a^{\prime}+c c^{\prime}+1=0) is correct because it represents the condition for perpendicularity of two lines in 3D space, derived from the dot product of their direction vectors.
185 Easy
185 of 241
Application +4 | -1
If the plane x+ay+z=5x+a y+z=5 has equal intercepts on axes, then the value of aa, is
View answer and explanation
Correct option: B
1
Option B (1) is correct because equal intercepts on the axes imply that the coefficients of x, y, and z (after normalization) must be equal, leading to a = 1.
186 Easy
186 of 241
Analyzing +4 | -1
The image of the line x−13=y−31=z−4−5\frac{x-1}{3}=\frac{y-3}{1}=\frac{z-4}{-5} in the plane 2x−y+z+3=02 x-y+z+3=0 is the line
View answer and explanation
Correct option: A
x+33=y−51=z−2−5\frac{x+3}{3}=\frac{y-5}{1}=\frac{z-2}{-5}
Option A (x+33=y−51=z−2−5\frac{x+3}{3}=\frac{y-5}{1}=\frac{z-2}{-5}) is correct because it represents the image of the given line in the specified plane.
187 Easy
187 of 241
Application +4 | -1
The distance of the point (2,3,4)(2,3,4) from the line 1−x=y2=13(1+z)1-x=\frac{y}{2}=\frac{1}{3}(1+z) is
View answer and explanation
Correct option: D
3735\frac{3}{7} \sqrt{35}
Option D (3735\frac{3}{7} \sqrt{35}) is correct because it represents the perpendicular distance from the point to the line, calculated using the formula involving direction ratios and coordinate differences.
188 Easy
188 of 241
Analyzing +4 | -1
Consider the three planes P1:3x+15y+21z=9,  P2:x−3y−z=5{P_1}:3x + 15y + 21z = 9,\,\,{P_2}:x - 3y - z = 5 and P3:2x+10y+14z=5{P_3}:2x + 10y + 14z = 5
Then, which one of the following is true?
View answer and explanation
Correct option: B
P1{P_1} and P3{P_3} are parallel
Option B is correct. Planes P1{P_1} and P3{P_3} are parallel because their normal vectors are proportional, meaning they have the same direction ratios.
189 Easy
189 of 241
Understanding +4 | -1
Assertion: A line divides the cartesian plane in two part(s).

Reason: If a point P(α,β){P}(\alpha, \beta) on the line ax+by=c{ax}+{by}={c}, then aα+bβ=ca \alpha+b \beta=c.
View answer and explanation
Correct option: B
Assertion is correct, reason is correct; reason is not a
correct explanation for assertion
Option B (Assertion is correct, reason is correct; reason is not a correct explanation for assertion) is correct because both the assertion and the reason are correct statements, but the reason does not explain the assertion.
190 Easy
190 of 241
Analyzing +4 | -1
Statement -1: The point A(3,1,6)\mathrm{A}(3,1,6) is the mirror image of the point B(1,3,4)\mathrm{B}(1,3,4) in the plane x−y+z=5x-y+z=5.
Statement -2: The plane x−y+z=5x-y+z=5 bisects the line segment joining A(3,1,6)\mathrm{A}(3,1,6) and B(1,3,4)\mathrm{B}(1,3,4).
View answer and explanation
Correct option: A
Statement -1 is true, Statement - 2 is true ; Statement 2 is not a correct explanation for Statement -1 .
Option A (Statement -1 is true, Statement - 2 is true ; Statement 2 is not a correct explanation for Statement -1) is correct because both statements are true, but the second statement doesn't fully explain why A is the mirror image of B.
191 Easy
191 of 241
Application +4 | -1
The points i^−j^+3k^\hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}} and 3i^+3j^+3k^3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} are equidistant from the plane r\mathbf{r}. (5i^+2j^−7k^)+9=0(5 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-7 \hat{\mathbf{k}})+9=0, then they are
View answer and explanation
Correct option: C
on the opposite sides of the plane
Option C (on the opposite sides of the plane) is correct because the perpendicular distances from the points to the plane have opposite signs, indicating they lie on opposite sides.
192 Easy
192 of 241
Application +4 | -1
If the line x−23=y+12=z−1−1\frac{{x - 2}}{3} = \frac{{y + 1}}{2} = \frac{{z - 1}}{{ - 1}} intersects the plane 2x+3y−z+13=02x + 3y - z + 13 = 0 at a point P and the plane 3x+y+4z=163x + y + 4z = 16 at a point Q, then PQ is equal to
View answer and explanation
Correct option: D
2142\sqrt 14
Option D (2142\sqrt {14}) is correct because it is the distance between the points P and Q, where the given line intersects the two planes.
193 Easy
193 of 241
Application +4 | -1
The mirror image of the point (1, 2, 3) in a plane is (−73, −43, −13).\left( { - \frac{7}{3},\, - \frac{4}{3},\, - \frac{1}{3}} \right). Which of the following points lies on the plane?
View answer and explanation
Correct option: A
(1, -1, 1)
Option D (100 12100\,\frac{1}{2}) is correct because it is the sum of the first 200 terms of the A.P. calculated using the given information.
194 Easy
194 of 241
Application +4 | -1
The image of the point (−1,3,4)(-1,3,4) in the plane x−2y=0x-2 y=0 is
View answer and explanation
Correct option: D
(95,−135,4)\left(\frac{9}{5},-\frac{13}{5}, 4\right)
Option D (95,−135,4)\left(\frac{9}{5},-\frac{13}{5}, 4\right) is correct because it is the result of applying the image formula for a point in a plane.
195 Easy
195 of 241
Application +4 | -1
A plane passes through the points P(4,0,0)P(4,0,0) and Q(0,0,4)Q(0,0,4) and is parallel to the YY-axis. The distance of the plane from the origin is
View answer and explanation
Correct option: D
222 \sqrt{2}
Option D (222 \sqrt{2}) is correct because the distance of the plane x+z=4 from the origin (0,0,0) is calculated correctly using the point-to-plane distance formula.
196 Medium
196 of 241
Application +4 | -1
If equation of the plane through the straight line x−12=y+2−3=z5\frac{x-1}{2}=\frac{y+2}{-3}=\frac{z}{5} and perpendicular to the plane x−y+z+2=0\mathrm{x}-\mathrm{y}+\mathrm{z}+2=0 is ax−by+cz+4=0ax - by + c z+4=0, then find the value of 103a+102 b+10c342\frac{10^3 a+10^2 \mathrm{~b}+10 \mathrm{c}}{342}.
View answer and explanation
Correct option: A
5
Option A (5) is correct because solving the system of equations derived from the line and plane conditions yields the correct values for a, b, and c, and thus the correct answer.
197 Easy
197 of 241
Analyzing +4 | -1
A plane passing through (1,1,1)(1,1,1) cuts positive direction of coordinate axes at A,BA, B and CC, then the volume of tetrahedron OABCO A B C satisfies
View answer and explanation
Correct option: B
V≥92V \geq \frac{9}{2}
Option B (V≥92V \geq \frac{9}{2}) is correct because using AM-GM inequality, the minimum volume of the tetrahedron is found to be 92\frac{9}{2}.
198 Easy
198 of 241
Application +4 | -1
The distance of the point (1,3,−7)\left( {1,3, - 7} \right) from the plane passing through the point (1,−1,−1)\left( {1, - 1, - 1} \right) having normal perpendicular to both the lines x−11=y+2−2=z−43\frac{{x - 1}}{1} = \frac{{y + 2}}{{ - 2}} = \frac{{z - 4}}{3}\mathrm{ } and x−22=y+1−1=z+7−1\frac{{x - 2}}{2} = \frac{{y + 1}}{{ - 1}} = \frac{{z + 7}}{{ - 1}}, is
View answer and explanation
Correct option: B
1083\frac{{10}}{{\sqrt {83} }} units
Option B (1083\frac{{10}}{{\sqrt {83} }} units) is correct because it represents the perpendicular distance of the point from the plane, calculated using the formula for distance from a point to a plane.
199 Medium
199 of 241
Application +4 | -1
If the distance of the point (1,−2,3)(1,-2,3) from the plane x+2y−3z+10=0x+2 y-3 z+10=0 measured parallel to the line x−13=2−ym=z+31\frac{x-1}{3}=\frac{2-y}{m}=\frac{z+3}{1} is 72\sqrt{\frac{7}{2}} then the value of ∣m∣=|m|=
View answer and explanation
Correct option: B
2
Option B (2) is correct because it is the absolute value of the parameter 'm' that satisfies the given distance condition between the point and the plane, measured parallel to the given line.
200 Easy
200 of 241
Application +4 | -1
Let the acute angle bisector of the two planes x−2y−2z+1=0   and  2x−3y−6z+1=0x - 2y - 2z + 1 = 0\,\,\,\mathrm{ and }\,\,2x - 3y - 6z + 1 = 0 be the plane PP Then, which of the following points lies on PP ?
View answer and explanation
Correct option: B
(−2,  0,  −12)\left( { - 2,\,\,0,\,\, - \frac{1}{2}} \right)
Option B (−2, 0, −12)\left( { - 2,\,0,\, - \frac{1}{2}} \right) is correct because it satisfies the equation of the acute angle bisector plane.
201 Easy
201 of 241
Analyzing +4 | -1
If the lines x−21=y−31=z−4−k\frac{{x - 2}}{1} = \frac{{y - 3}}{1} = \frac{{z - 4}}{{ - k}} and x−1k=y−42=z−51\frac{{x - 1}}{k} = \frac{{y - 4}}{2} = \frac{{z - 5}}{1} are coplanar, then k can have
View answer and explanation
Correct option: C
exactly two values
Option C (exactly two values) is correct because the coplanarity condition gives a quadratic equation in k, which has two solutions.
202 Easy
202 of 241
Application +4 | -1
A symmetrical form of the line of intersection of the planes x=ay+bx=a y+b and z=cy+dz=c y+d is
View answer and explanation
Correct option: B
x−b−aa=y−11=z−d−cc\frac{x-b-a}{a}=\frac{y-1}{1}=\frac{z-d-c}{c}
Option B (x−b−aa=y−11=z−d−cc\frac{{x-b-a}}{{a}}=\frac{{y-1}}{{1}}=\frac{{z-d-c}}{{c}}) is correct because it represents the symmetrical form of the line of intersection of the given planes.
203 Medium
203 of 241
Application +4 | -1
If the lines x−12=y−23=z+31\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+3}{1} and x−114=y−92=z−43\frac{x-11}{4}=\frac{y-9}{2}=\frac{z-4}{3} intersects at point pp, then perpendicular distance of pp from plane 3x+2y+6z=103 x+2 y+6 z=10 is
View answer and explanation
Correct option: B
37\frac{3}{7}
Option B (3/7) is correct because it is the perpendicular distance of the intersection point (3, 5, -2) from the plane 3x + 2y + 6z = 10, calculated using the point-to-plane distance formula.
204 Easy
204 of 241
Analyzing +4 | -1
The foot of the perpendicular drawn from the point (4, 2, 3) to the line joining the points (1, - 2, 3) and (1, 1, 0) lies on the plane
View answer and explanation
Correct option: A
2x+y−z=12x + y - z = 1
Option A 2x+y−z=12x + y - z = 1 is correct because the foot of the perpendicular from the given point to the line lies on this plane.
205 Easy
205 of 241
Analyzing +4 | -1
A plane 2x+3y+5z=12 x+3 y+5 z=1 has a point PP which is at minimum distance from line joining A(1,0,−3),B(1,−5,7)A(1,0,-3), B(1,-5,7), then distance APA P is equal to
View answer and explanation
Correct option: B
252 \sqrt{5}
Option B (252 \sqrt{5}) is correct because it calculates the distance AP where P is the point on the plane closest to the line AB.
206 Easy
206 of 241
Application +4 | -1
The equation of the plane which passes through the line of intersection of the planes r. n1=q1,r⋅n2=q2\mathbf{n}_1=q_1, \mathbf{r} \cdot \mathbf{n}_2=q_2 and is parallel to the line of intersection of the planes
r. n3=q3\mathbf{n}_3=q_3 and r⋅n4=q4\mathbf{r} \cdot \mathbf{n}_4=q_4, is
View answer and explanation
Correct option: A
[n2 n3 n4](r⋅n1−q1)=[n1 n3 n4](r⋅n2−q2)\left[ {{n_2}\,{n_3}\,{n_4}} \right]\left( {r \cdot {n_1} - {q_1}} \right) = \left[ {{n_1}\,{n_3}\,{n_4}} \right](r \cdot {n_2} - {q_2})
Option A is correct because it represents the equation of a plane passing through the intersection of two planes and parallel to the line of intersection of another two planes, which is derived by setting the scalar triple product to zero.
207 Easy
207 of 241
Application +4 | -1
The distance between the line r=2i^−2j^+3k^+λ(i^−j^+4k^)\mathbf{r}=2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}+\lambda(\hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}}) and the plane r⋅(i^+5j^+k^)=5\mathbf{r} \cdot(\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+\hat{\mathbf{k}})=5 is
View answer and explanation
Correct option: A
1033\frac{10}{3 \sqrt{3}}
Option A (1033\frac{{10}}{{3 \sqrt{{3}}}}) is correct because it represents the distance between the line and the plane, calculated using the formula for the distance between a point on the line and the plane, given that the line and plane are parallel.
208 Easy
208 of 241
Application +4 | -1
The distance of the point (−1,−5,−10)(-1,-5,-10) from the point of intersection of the line x−22=y+14=z−212\frac{x-2}{2}=\frac{y+1}{4}=\frac{z-2}{12} and the plane x−y+z=5x-y+z=5 is
View answer and explanation
Correct option: C
13
Option C (13) is correct because by finding the intersection point of the line and the plane, the distance between that point and (-1, -5, -10) is calculated to be 13.
209 Easy
209 of 241
Application +4 | -1
Let two vertices of a triangle ABCA B C be (2,4,6)(2,4,6) and (0,−2,−5)(0,-2,-5), and its centroid be (2,1,−1)(2,1,-1). If the image of the third vertex in the plane x+2y+4z=x+2 y+4 z= 11 is (α,β,γ)(\alpha, \beta, \gamma), then αβ+βγ+γα\alpha \beta+\beta \gamma+\gamma \alpha is equal to
View answer and explanation
Correct option: C
74
Option C (74) is correct because it involves finding the third vertex of a triangle given two vertices and the centroid, then finding the image of the third vertex in a plane, and finally calculating the required expression.
210 Medium
210 of 241
Analyzing +4 | -1
Let the equation of plane passing through the line of intersection of the planes x+2y+az=2x+2 y+a z=2 and x−y+z=3x-y+z=3 be 5x−11y+bz=6a−15 x-11 y+b z=6 a-1. For c∈Zc \in \mathbb{Z}, if the distance of this plane from the point (a,−c,c)(a,-c, c) is 2a\frac{2}{\sqrt{a}}, then a+bc\frac{a+b}{c} is equal to
View answer and explanation
Correct option: C
−4-4
Option C (−4-4) is correct because after finding the values of a and b, and using the distance formula, the value of c is determined, leading to the correct ratio.
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