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JEE • MATHEMATICS • 3D PLANES

3d planes practice · Mixed difficulty

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Showing 1-30 of 241 questions
1 Easy
1 of 241
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 Medium
2 of 241
Analyzing +4 | -1
If the planes x−cy−bz=0,cx−y+az=0x-c y-b z=0, c x-y+a z=0 and bx+ay−z=0b x+a y-z=0 pass through a straight line, then the value of a2+b2+c2+2abca^2+b^2+c^2+2 a b c is:
View answer and explanation
Correct option: D
1
The correct answer is D (1) because if the planes pass through a straight line, the determinant of the coefficients of x, y, and z must be zero, leading to the given expression equaling 1.
3 Medium
3 of 241
Analyzing +4 | -1
Let the plane 2x−3y+9z=02 x-3 y+9 z=0 is PP, the equation of line passing through (0,1,−1)(0,1,-1) and lying-in plane 4x−2y+7z+9=04 x-2 y+7 z+9=0 & parallel to PP is:
View answer and explanation
Correct option: B
x−3=y−122=z+18\quad \frac{x}{-3}=\frac{y-1}{22}=\frac{z+1}{8}
Option B (x−3=y−122=z+18\frac{x}{-3}=\frac{y-1}{22}=\frac{z+1}{8}) is correct because it represents the line passing through the given point and parallel to the plane P, lying in the plane 4x−2y+7z+9=04x - 2y + 7z + 9 = 0.
4 Easy
4 of 241
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.
5 Hard
5 of 241
Understanding +4 | -1
The plane ax+a x+ by =0=0 is rotated through an angle α\alpha about its line of intersection with the plane z=0z=0. If the equation of plane in new position is ax+by±za2+b2tan⁡α+c=0\mathrm{ax}+\mathrm{by} \pm \mathrm{z} \sqrt{\mathrm{a}^2+\mathrm{b}^2} \tan \alpha+\mathrm{c}=0. Find c.
View answer and explanation
Correct option: A
0
Option A (0) is correct because the equation of the rotated plane passes through the line of intersection of the original plane and z=0, resulting in c=0.
6 Easy
6 of 241
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.
7 Easy
7 of 241
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.
8 Easy
8 of 241
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.
9 Easy
9 of 241
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.
10 Medium
10 of 241
Application +4 | -1
Distance of point (−2,3,4)(-2,3,4) from the line x+23=2y+34=3x+45\frac{x+2}{3}=\frac{2 y+3}{4}=\frac{3 x+4}{5} measured parallel to the plane 8x+6y−9z+1=08 x+6 y-9 z+1=0 is
View answer and explanation
Correct option: B
4012\frac{\sqrt{401}}{2}
Option B (4012\frac{\sqrt{401}}{2}) is correct because it represents the distance calculated by finding the intersection point of the line and a line parallel to the plane.
11 Easy
11 of 241
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.
12 Easy
12 of 241
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.
13 Easy
13 of 241
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.
14 Medium
14 of 241
Application +4 | -1
A variable plane is at a constant distance 'rr' from the origin and meets the axes in A,B,CA, B, C then, the locus of the centroid of the tetrahedron OABC{OABC} is-
View answer and explanation
Correct option: A
1x2+1y2+1z2=16r2\frac{1}{{x}^2}+\frac{1}{{y}^2}+\frac{1}{{z}^2}=\frac{16}{{r}^2}
Option A (1x2+1y2+1z2=16r2\frac{1}{{x}^2}+\frac{1}{{y}^2}+\frac{1}{{z}^2}=\frac{16}{{r}^2}) is correct because it represents the locus of the centroid of the tetrahedron, derived from the equation of the plane and the centroid coordinates.
15 Medium
15 of 241
Application +4 | -1
If the lines x−11=y−22=z+31\frac{x-1}{1}=\frac{y-2}{2}=\frac{z+3}{1} and x−a2=y+23=z−31\frac{x-a}{2}=\frac{y+2}{3}=\frac{z-3}{1} intersect at the point PP, then the distance of the point PP from the plane z=az=a is
View answer and explanation
Correct option: A
28
Option A (28) is correct because after finding the intersection point P and the value of a, the distance from P to the plane z=a is calculated correctly.

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16 Easy
16 of 241
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.
17 Medium
17 of 241
Application +4 | -1
If the greatest and the least resultants of two forces are P{P} and Q{Q} respectively, then the resultant of these forces, when acting at right angles, will be
View answer and explanation
Correct option: D
(P2+Q2)2\sqrt{\frac{\left({P}^2+{Q}^2\right)}{2}}
The correct answer is D ((P2+Q2)2\sqrt{\frac{\left({P}^2+{Q}^2\right)}{2}}) because it correctly calculates the resultant of two forces acting at right angles, given their greatest and least resultants.
18 Easy
18 of 241
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.
19 Easy
19 of 241
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.
20 Easy
20 of 241
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.
21 Medium
21 of 241
Application +4 | -1
A ray of light passing through the point A(1,2,3)A(1,2,3), strikes the plane x+y+z=12x+y+z=12 at BB and on reflection passes through point C(3,5,9){C}(3,5,9). Coordinates of point B{B} is-
View answer and explanation
Correct option: B
(−7,0,19)(-7,0,19)
Option B ((-7,0,19)) is correct because it is the point of intersection of the reflected ray BC with the plane x+y+z=12, calculated using the image of point A.
22 Easy
22 of 241
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.
23 Medium
23 of 241
Application +4 | -1
Let the image of the point P(2,−1,3)P(2,-1,3) in the plane x+2y−z=0x+2 y-z=0 be Q{Q}. Then the distance of the plane 3x+2y+z+29=03 x+2 y+z+29=0 from the point QQ is
View answer and explanation
Correct option: D
3143 \sqrt{14}
Option D (3143 \sqrt{14}) is correct because it represents the distance of the plane from the image point Q, calculated using the formula for the distance between a point and a plane.
24 Medium
24 of 241
Application +4 | -1
If the equation of the plane that contains the point (−2,3,5)(-2,3,5) and is perpendicular to each of the planes 2x+4y+5z=82 x+4 y+5 z=8 and 3x−2y+3z=53 x-2 y+3 z=5 is αx+βy+γz+97=0\alpha x+\beta y+\gamma z+97=0 then α+β+γ=\alpha+\beta+\gamma=
View answer and explanation
Correct option: A
15
Option A (15) is correct because it is the sum of the coefficients of x, y, and z in the plane's equation, determined by using the point-normal form and the given perpendicularity conditions.
25 Easy
25 of 241
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.
26 Medium
26 of 241
Application +4 | -1
The equation of a plane passing through the line of intersection of the planes x+2y+3z=2x+2 y+3 z=2 and x−y+z=3x-y+z=3 and at a distance 23\frac{2}{\sqrt{3}} from the point (3,1,−1)(3,1,-1) is
View answer and explanation
Correct option: A
5x−11y+z=175 x-11 y+z=17
Option A (5x - 11y + z = 17) is correct because it represents a plane passing through the intersection of the given planes and satisfies the distance condition from the given point.
27 Easy
27 of 241
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.
28 Medium
28 of 241
Application +4 | -1
Find the foot of perpendicular from the point (4,6,2)(4,6,2) on the line r⃗=2(i^+j^+k^)+t(3i^+2j^+k^)\vec{r}=2(\hat{i}+\hat{j}+\hat{k})+t(3 \hat{i}+2 \hat{j}+\hat{k}).
View answer and explanation
Correct option: C
(5,4,3)(5,4,3)
Option C is correct because the foot of the perpendicular is found by finding a point on the line such that the direction vector from the given point to that point is perpendicular to the line's direction vector.
29 Medium
29 of 241
Application +4 | -1
If P,Q,R,SP, Q, R, S are four coplanar points on the sides AB,BCA B, B C, CD,DAC D, D A of a skew quadrilateral, then APPB⋅BQQC⋅CRRD⋅DSSA\frac{A P}{P B} \cdot \frac{B Q}{Q C} \cdot \frac{C R}{R D} \cdot \frac{D S}{S A} equals
View answer and explanation
Correct option: A
1
Option A (1) is correct because the product of the ratios of segments on the sides of a skew quadrilateral with coplanar points is 1.
30 Easy
30 of 241
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.
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