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JEE • MATHEMATICS • 3D PLANES

3d planes practice · Easy

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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 1-30 of 177 questions
1 Easy
1 of 177
Application +4 | -1
Let x−23=y+1−2=z+3−1\frac{x-2}{3}=\frac{y+1}{-2}=\frac{z+3}{-1} lie on the plane px−qy+p x-q y+ z=5z=5, for some p,q∈Rp, q \in \mathbb{R}. The shortest distance of the plane from the origin is :
View answer and explanation
Correct option: B
5142\sqrt{\frac{5}{142}}
Option B (5142\sqrt{\frac{5}{142}}) is correct because after finding the equation of the plane, the shortest distance from the origin is calculated correctly.
2 Easy
2 of 177
Application +4 | -1
A plane which passes through the point (3,2,0)(3,2,0) and the line x−41=y−75=z−44\frac{x-4}{1}=\frac{y-7}{5}=\frac{z-4}{4} is
View answer and explanation
Correct option: A
x−y+z=1x-y+z=1
Option A (x−y+z=1x-y+z=1) is correct because the point (3,2,0) and the line satisfy this equation.
3 Easy
3 of 177
Application +4 | -1
A variable plane passes through a fixed point (3,2,1)(3,2,1) and meets x,yx, y and zz axes at A,BA, B and CC respectively. A plane is drawn parallel to yzy z-plane through AA, a second plane is drawn parallel zxz x-plane through BB and a third plane is drawn parallel to xyx y-plane through CC. Then the locus of the point of intersection of these three planes, is
View answer and explanation
Correct option: C
3x+2y+1z=1\frac{3}{x}+\frac{2}{y}+\frac{1}{z}=1
Option C (3x+2y+1z=1\frac{{3}}{{x}} + \frac{{2}}{{y}} + \frac{{1}}{{z}} = 1) is correct because it represents the locus of the point of intersection of the three planes, derived from the equation of a plane passing through a fixed point.
4 Easy
4 of 177
Application +4 | -1
The ratio in which the line joining the points (a,b,c)(a, b, c) and (−a,−c,−b)(-a,-c,-b) is divided by the xyx y-plane is
View answer and explanation
Correct option: D
c:bc: b
Option D (c:bc: b) is correct because the ratio is found by setting the z-coordinate of the divided point to zero, as it lies on the xy-plane.
5 Easy
5 of 177
Analyzing +4 | -1
If the plane PP passes through the intersection of two mutually perpendicular planes 2x+ky−5z=12 x+k y-5 z=1 and 3kx−ky+z=5,k<33 k x-k y+z=5, k<3 and intercepts a unit length on positive xx-axis, then the intercept made by the plane PP on the yy-axis is
View answer and explanation
Correct option: D
77
Option D (77) is the correct answer because after finding the value of k and lambda, the y-intercept of the plane P is calculated to be 7.
6 Easy
6 of 177
Application +4 | -1
Let the plane ax+by+cz=da x+b y+c z=d pass through (2,3,−5)(2,3,-5) and is perpendicular to the planes
2x+y−5z=102 x+y-5 z=10 and
3x+5y−7z=12.3 x+5 y-7 z=12.

If a,b,c,da, b, c, d are integers d>0d>0 and gcd⁡(∣a∣,∣b∣,∣c∣\operatorname{gcd}(|a|,|b|,|c|, d)=1d)=1, then the value of a+7b+c+20da+7 b+c+20 d is equal to:
View answer and explanation
Correct option: D
22
Option D (22) is correct because it's the result of finding the plane equation and substituting the coefficients into the given expression.
7 Easy
7 of 177
Application +4 | -1
The two lines x=ay+b,z=cy+dx=a y+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} are perpendicular to each other, if
View answer and explanation
Correct option: D
aa′+cc′=−1a a^{\prime}+c c^{\prime}=-1
Option D (aa′+cc′=−1a a^{\prime}+c c^{\prime}=-1) is correct because it is the condition for the two lines to be perpendicular, derived from the dot product of their direction vectors being zero.
8 Easy
8 of 177
Analyzing +4 | -1
If the line of intersection of the planes ax+by=3a x+b y=3 and ax+by+cz=0,a>0a x+b y+c z=0, a>0 makes an angle 30∘30^{\circ} with the plane y−z+2=0y-z+2=0, then the direction cosines of the line are:
View answer and explanation
Correct option: B
12,−12,0\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}, 0
Option B (12,−12,0\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}, 0) is correct because after finding the normal vectors and using the angle condition, the direction cosines of the line are determined to be 12,−12,0\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}, 0.
9 Easy
9 of 177
Application +4 | -1
A plane EE is perpendicular to the two planes 2x−2y+z=02 x-2 y+z=0 and x−y+2z=4x-y+2 z=4, and passes through the point P(1,−1,1)P(1,-1,1). If the distance of the plane EE from the point Q(a,a,2)Q(a, a, 2) is 323 \sqrt{2}, then (PQ)2(P Q)^2 is equal to
View answer and explanation
Correct option: C
21
Option C (21) is correct because finding the normal vector to the plane, then the equation of the plane, and finally the distance condition allows us to find the coordinates of point Q and calculate (PQ)2.
10 Easy
10 of 177
Analyzing +4 | -1
If for a>0a > 0, the feet of perpendiculars from the points A(a, −2a, 3) A(a,\, - 2a,\,3)\, and B(0, 4, 5)B(0,\,4,\,5) on the plane lx + my +nz = 0lx\, + \,my\, + nz\, = \,0 are points C(0, −a, −1)C(0,\, - a,\, - 1) and DD respectively, then the length of line segment CDCD is equal to:
View answer and explanation
Correct option: D
66\sqrt {66}
Option D (66\sqrt{66}) is correct because it is the calculated length of the line segment CD based on the given geometric conditions and vector calculations.
11 Easy
11 of 177
Analyzing +4 | -1
The shortest distance between the lines x−10=y+1−1=z1\frac{{x - 1}}{0} = \frac{{y + 1}}{{ - 1}} = \frac{z}{1} and x+y+z+1=0,x + y + z + 1 = 0, 2x−y+z+3=02x - y + z + 3 = 0 is
View answer and explanation
Correct option: B
13\frac{1}{{\sqrt 3 }}
Option B (13\frac{{1}}{{\sqrt 3 }}) is correct because it represents the shortest distance calculated using the formula for the distance between a point and a plane.
12 Easy
12 of 177
Application +4 | -1
The distance of the origin from the point of intersection of the line x2=y−23=z−34\frac{x}{2}=\frac{y-2}{3}=\frac{z-3}{4} and the plane 2x+y−z=22 x+y-z=2 is
View answer and explanation
Correct option: D
(78)\sqrt{(78)}
Option D ((78)\sqrt{(78)}) is correct because it is the distance of the origin from the point of intersection of the line and the plane, calculated by finding the point of intersection first.
13 Easy
13 of 177
Application +4 | -1
Let (α,β,γ)(\alpha, \beta, \gamma) be the image of point P(2,3,5)P(2,3,5) in the plane 2x+y−3z=62 x+y-3 z=6. Then α+β+γ\alpha+\beta+\gamma is equal to
View answer and explanation
Correct option: B
10
Option B (10) is correct because it is the sum of the coordinates of the image point, calculated using the formula for the image of a point in a plane.
14 Easy
14 of 177
Application +4 | -1
Let A(0,0,1),B(0,1,0)\mathrm{A}(0,0,1), \mathrm{B}(0,1,0) and C(1,1,1)\mathrm{C}(1,1,1) are the points in a plane, then the equation of the plane perpendicular to plane ABC\mathrm{ABC} and passing through A\mathrm{A} and B\mathrm{B} is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the equation of the plane perpendicular to plane ABC and passing through A and B requires finding the normal vector and applying the plane equation, which doesn't match any of the given options.
15 Easy
15 of 177
Analyzing +4 | -1
If the lines r⃗=(i^−j^+k^)+λ(3j^−k^)\vec r = \left( {\hat i - \hat j + \hat k} \right) + \lambda \left( {3\hat j - \hat k} \right) and r⃗=(αi^−j^)+μ(2j^−3k^)\vec r = \left( {\alpha \hat i - \hat j} \right) + \mu \left( {2\hat j - 3\hat k} \right) are co-planar, then the distance of the plane containing these two lines from the point (α, 0, 0)\left( {\alpha ,\,0,\,0} \right) is:
View answer and explanation
Correct option: B
211\frac{2}{{11}}
Option B (211\frac{2}{{11}}) is correct because it represents the distance of the plane containing the coplanar lines from the point (α,0,0)(\alpha, 0, 0) after finding the value of α\alpha and the equation of the plane.

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16 Easy
16 of 177
Application +4 | -1
The image of the point (3,2,1)(3,2,1) in the plane 2x−y+3z=72 x-y+3 z=7 is:
View answer and explanation
Correct option: C
(3,2,1)(3,2,1)
Option C (3,2,1) is correct because the point (3,2,1) already lies on the plane 2x - y + 3z = 7, so its image in the plane is itself.
17 Easy
17 of 177
Application +4 | -1
A vector n⃗\vec{n} is inclined to xx-axis at 45∘45^{\circ}, to yy-axis at 60∘60^{\circ} and at an acute angle to zz-axis. If n⃗\vec{n} is a normal to a plane passing through the point (2,−1,1)(\sqrt{2},-1,1) then the equation of the plane is:
View answer and explanation
Correct option: B
2x+y+2z=22+12 x+y+2 z=2 \sqrt{2}+1
Option B (2x+y+2z=22+12 x+y+2 z=2 \sqrt{{2}}+1) is correct because it represents the equation of the plane with the given normal vector and passing through the specified point.
18 Easy
18 of 177
Application +4 | -1
The angle between the lines 2x=3y=−z2 x=3 y=-z and 6x=−y=−4z6 x=-y=-4 z is
View answer and explanation
Correct option: D
90∘90^{\circ}
Option D (90∘90^{\circ}) is correct because the dot product of the direction vectors of the two lines is zero, indicating they are perpendicular.
19 Easy
19 of 177
Application +4 | -1
The angle between the lines 2x=3y=−z2 x=3 y=-z and 6x=−y=−4z6 x=-y=-4 z is
View answer and explanation
Correct option: C
90∘90^{\circ}
The angle between the lines is 90∘90^{\circ} because the dot product of their direction ratios is zero, indicating perpendicularity.
20 Easy
20 of 177
Application +4 | -1
The length of projection of the line segment joining the points (1,0,−1)(1,0,-1) and (−1,2,2)(-1,2,2) on the plane x+3y−5z=6x+3 y-5 z=6, is equal to
View answer and explanation
Correct option: D
47435\sqrt{\frac{474}{35}}
Option D (47435\sqrt{\frac{474}{35}}) is correct because it is the length of the projection calculated using direction ratios and the angle between the line and the plane.
21 Easy
21 of 177
Application +4 | -1
The intercept made by the plane r.n=q\mathbf{r} . \mathbf{n}=q on the xx-axis is
View answer and explanation
Correct option: A
q i.n \frac{q}{\text { i.n }}
Option A (q i.n \frac{q}{\text { i.n }}) is correct because it represents the x-coordinate where the plane intersects the x-axis, derived from the plane equation.
22 Easy
22 of 177
Application +4 | -1
If the angle between the line x=y−12=z−3λx=\frac{y-1}{2}=\frac{z-3}{\lambda} and the plane x+2y+3z=4x+2 y+3 z=4 is cos⁡−1(514)\cos ^{-1}\left(\sqrt{\frac{5}{14}}\right), then λ\lambda equals
View answer and explanation
Correct option: D
23\frac{2}{3}
Option D (23\frac{2}{3}) is correct because it is the value of λ\lambda that satisfies the given condition for the angle between the line and the plane, found by using the formula relating the angle to the direction ratios of the line and the normal to the plane.
23 Easy
23 of 177
Application +4 | -1
The point at which the line joining the points (2,−3,1)(2,-3,1) and (3,−4,−5)(3,-4,-5) intersects the plane 2x+y+z=72 x+y+z=7 is
View answer and explanation
Correct option: B
(1,−2,7)(1,-2,7)
Option B ((1,−2,7)(1,-2,7)) is correct because it is found by determining the ratio in which the line intersects the plane and then finding the point of intersection.
24 Easy
24 of 177
Application +4 | -1
Let L1L_1 be the line r1=2i^+j^−k^+λ(i^+2k^)\mathbf{r}_1=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}+\lambda(\hat{\mathbf{i}}+2 \hat{\mathbf{k}}) and let L2L_2 be the another line r2=3i^+j^+μ(i^+j^−k^)\mathbf{r}_2=3 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\mu(\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}). Let π\pi be the plane which contains the line L1L_1 and is parallel to L2L_2. The distance of the plane π\pi from the origin is
View answer and explanation
Correct option: A
27\sqrt{\frac{2}{7}}
Option A (27\sqrt{\frac{2}{7}}) is correct because it represents the distance of the plane from the origin, calculated using the formula for the distance of a point from a plane.
25 Easy
25 of 177
Analyzing +4 | -1
Equation of the plane which passes through the point of intersection oflines
x−13=y−21=z−32\frac{x-1}{3}=\frac{y-2}{1}=\frac{z-3}{2} and
x−31=y−12=z−23\frac{x-3}{1}=\frac{y-1}{2}=\frac{z-2}{3}
and has the largest distance from the origin is:
View answer and explanation
Correct option: C
4x+3y+5z=504 x+3 y+5 z=50
Option C (4x+3y+5z=504 x+3 y+5 z=50) is correct because it represents the equation of the plane passing through the intersection of the given lines and having the largest distance from the origin.
26 Easy
26 of 177
Application +4 | -1
The equation of the plane containing the line 2x–5y+z=32x – 5y + z = 3, x+y+4z=5x + y + 4z = 5 and parallel to the plane x+3y+6z=1x + 3y + 6z = 1 is
View answer and explanation
Correct option: C
x+3y+6z=7x + 3y + 6z = 7
Option C (x+3y+6z=7x + 3y + 6z = 7) is correct because the plane equation is found by combining the given lines and ensuring parallelism with the specified plane.
27 Easy
27 of 177
Application +4 | -1
Let the plane P:8x+a1y+a2z+12=0P: 8 x+a_{1} y+a_{2} z+12=0 be parallel to the line L:x+22=y−33=z+45L: \frac{x+2}{2}=\frac{y-3}{3}=\frac{z+4}{5}. If the intercept of PP on the yy-axis is 1 , then the distance between PP and LL is:
View answer and explanation
Correct option: D
14\sqrt{14}
Option D (14\sqrt{14}) is correct because it involves finding the equation of the plane, using the fact that it's parallel to the line, and then calculating the distance between the plane and the line.
28 Easy
28 of 177
Understanding +4 | -1
The equation xy=0\mathrm{x y} = 0 in three dimensional space represents -
View answer and explanation
Correct option: C
A pair of planes at right angles
Option C (A pair of planes at right angles) is correct because the equation xy = 0 represents the union of the x = 0 plane and the y = 0 plane, which are perpendicular.
29 Easy
29 of 177
Application +4 | -1
The distance of the point (1,−2,4)(1,-2,4) from the plane passing through the point (1,2,2)(1,2,2) and perpendicular to the planes x−y+2z=3x-y+2 z=3 and 2x−2y+z+12=02 x-2 y+z+12=0, is :
View answer and explanation
Correct option: C
222 \sqrt{2}
Option C (222 \sqrt{2}) is correct because it is the distance calculated from the point to the plane, where the plane is perpendicular to the given planes.
30 Easy
30 of 177
Analyzing +4 | -1
The number of planes that are equidistant from four non-coplanar points is
View answer and explanation
Correct option: C
7
Option C (7) is correct because there are 4 planes with one point on one side and the other three on the other side, and 3 planes with two points on each side.
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