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JEE • MATHEMATICS • 3D PLANES

3d planes 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 211-240 of 241 questions

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211 Easy
211 of 241
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 is zero, indicating perpendicular lines.
212 Easy
212 of 241
Application +4 | -1
The plane passing through the points (1, 2, 1), (2, 1, 2) and parallel to the line, 2x=3y,z=12x = 3y,z = 1 also passes through the point
View answer and explanation
Correct option: A
(−2,0,1)\left( { - 2,0,1} \right)
Option A ((-2,0,1)) is correct because the plane equation derived from the given points and parallel line condition satisfies this point.
213 Easy
213 of 241
Application +4 | -1
The equation of the plane containing the line 2x−5y+z=3,x+y+4z=52 x-5 y+z=3, x+y+4 z=5 and parallel to the plane x+3y+6z=1x+3 y+6 z=1, is
View answer and explanation
Correct option: C
x+3y+6z=7x+3 y+6 z=7
Option C (x+3y+6z=7x+3 y+6 z=7) is correct because it represents the equation of the plane containing the line and parallel to the given plane.
214 Easy
214 of 241
Application +4 | -1
A ray of light is sent through the point P(1,2,3)P(1,2,3) and is reflected on the XYX Y-plane. If the reflected ray passes through the point Q(3,2,5)Q(3,2,5), then the equation of the reflected ray is
View answer and explanation
Correct option: C
x−31=y−20=z−54\frac{x-3}{1}=\frac{y-2}{0}=\frac{z-5}{4}
Option C (x−31=y−20=z−54\frac{x-3}{1}=\frac{y-2}{0}=\frac{z-5}{4}) is correct because it represents the equation of the reflected ray derived from the image of point P and point Q.
215 Medium
215 of 241
Application +4 | -1
The angle between the line x−22=y+1−1=z−32\frac{x-2}{2}=\frac{y+1}{-1}=\frac{z-3}{2} and the plane 3x+6y−2z+5=03 x+6 y-2 z+5=0 is
View answer and explanation
Correct option: D
sin⁡−1(421)\sin ^{-1}\left(\frac{4}{21}\right)
The correct answer is D (sin⁡−1(421)\sin ^{-1}(\frac{4}{21})) because it is calculated using the formula for the angle between a line and a plane.
216 Easy
216 of 241
Analyzing +4 | -1
Equation of plane which passes through the point of intersection of lines 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 at greatest distance from the point (0,0,0)(0,0,0) is
View answer and explanation
Correct option: B
4x+3y+5z=504 x+3 y+5 z=50
Option B (4x+3y+5z=504 x+3 y+5 z=50) is correct because the plane passes through the intersection of the two lines and is perpendicular to the line connecting the origin and the intersection point, maximizing the distance from the origin.
217 Easy
217 of 241
Application +4 | -1
A plane PP contains the line of intersection of the plane r⃗⋅(i^+j^+k^)=6\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=6 and r⃗⋅(2i^+3j^+4k^)=−5\vec{r} \cdot(2 \hat{i}+3 \hat{j}+4 \hat{k})=-5. If PP passes through the point (0,2,−2)(0,2,-2), then the square of distance of the point (12,12,18)(12,12,18) from the plane PP is
View answer and explanation
Correct option: A
620
Option A (620) is correct because it is the square of the distance calculated using the formula for the distance of a point from a plane.
218 Easy
218 of 241
Application +4 | -1
Let the equation of the plane, that passes through the point (1, 4, -3) and contains the line of intersection of the planes 3x−2y+4z−7 = 03x - 2y + 4z - 7\, = \,0 and x+5y−2z+9 = 0x + 5y - 2z + 9\, = \,0 be αx+βy+γz+3 = 0,\alpha x + \beta y + \gamma z + 3\, = \,0, then α+β+γ \alpha + \beta + \gamma \, is equal to
View answer and explanation
Correct option: A
-23
Option A (-23) is correct because it represents the sum of the coefficients α\alpha, β\beta, and γ\gamma in the equation of the plane that passes through the given point and contains the line of intersection of the two planes.
219 Easy
219 of 241
Application +4 | -1
If the angle between the lines, x2=y2=z1\frac{x}{2}=\frac{y}{2}=\frac{z}{1} and 5−x−2=7y−14P=z−34\frac{5-x}{-2}=\frac{7 y-14}{P}=\frac{z-3}{4} is cos⁡−1(23)\cos ^{-1}\left(\frac{2}{3}\right), then PP is equal to
View answer and explanation
Correct option: D
72\frac{7}{2}
Option D (7/2) is correct because substituting the direction cosines into the angle formula and solving for P yields P = 7/2.
220 Easy
220 of 241
Application +4 | -1
Let the foot of the perpendicular from the point (1, 2, 4) on the line x+24=y−12=z+13\frac{{x + 2}}{4} = \frac{{y - 1}}{2} = \frac{{z + 1}}{3} be P,P, then the distance of PP from the plane 3x+4y+12z+23=03x + 4y + 12z + 23 = 0 is
View answer and explanation
Correct option: A
5
Option A (5) is correct because after finding the foot of the perpendicular P, its distance from the plane is calculated to be 5.
221 Easy
221 of 241
Application +4 | -1
Let PP be the plane passing through the point (1, 2, 3) and the line of intersection of the planes
r⋅(i^+j^+4k^)=16r \cdot \left( {\hat i + \hat j + 4\hat k} \right) = 16 and (−i^+j^+k^)=6\left( { - \hat i + \hat j + \hat k} \right) = 6. Then which of the following points does not lie on PP?
View answer and explanation
Correct option: C
(4, 2, 2)
Option C ((4, 2, 2)) is correct because it does not satisfy the equation of the plane formed by the intersection of the two given planes and passing through the point (1, 2, 3).
222 Easy
222 of 241
Application +4 | -1
If a plane cuts off intercepts −6,3,4-6,3,4 from the co-ordinate axes, then the length of the perpendicular from the origin to the plane is:
View answer and explanation
Correct option: C
1229\frac{12}{\sqrt{29}}
Option C (1229\frac{12}{\sqrt{29}}) is correct because it is the length of the perpendicular from the origin to the plane, calculated using the formula involving intercepts.
223 Easy
223 of 241
Application +4 | -1
The distance of the point (1,−2, 3)\left( {1, - 2,\,3} \right) from the plane x−y+z=5x - y + z = 5 measured parallel to the line x2=y3=z−6\frac{x}{2} = \frac{y}{3} = \frac{z}{{ - 6}} is
View answer and explanation
Correct option: B
1
The correct answer is B (1) because it represents the distance of the point from the plane, measured parallel to the given line, calculated by finding the intersection point and then the distance.
224 Easy
224 of 241
Application +4 | -1
The equation of a plane containing the line x+1−3=y−32=z+21\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1} and the point (0,7,−7)(0,7,-7) is
View answer and explanation
Correct option: A
x+y+z=0x+y+z=0
Option A (x+y+z=0x+y+z=0) is correct because it represents the equation of the plane that contains the given line and the point (0,7,-7).
225 Medium
225 of 241
Application +4 | -1
If line x−12=y+αα=z−β2\frac{x-1}{2}=\frac{y+\alpha}{\alpha}=\frac{z-\beta}{2} lies in plane 2x+y+z=52 x+y+z=5, then α+β\alpha+\beta is
View answer and explanation
Correct option: D
−9-9
Option D (-9) is correct because by substituting the parametric form of the line into the plane equation, we can solve for α and β, and then find their sum.
226 Easy
226 of 241
Analyzing +4 | -1
The plane x−y−z=4x-y-z=4 is rotated through an angle 90∘90^{\circ} about its line of intersection with the plane x+y+2z=4x+y+2 z=4. Then the equation of the plane in its new position is
View answer and explanation
Correct option: D
5x+y+4z=205 x+y+4 z=20
The equation of the rotated plane is found by taking a linear combination of the original plane and the plane of intersection, then using the perpendicularity condition, resulting in 5x+y+4z=205 x+y+4 z=20.
227 Easy
227 of 241
Application +4 | -1
A plane bisects the line segment joining the points (1,2,3)(1,2,3) and (−3,4,5)(-3,4,5) at right angles. Then this plane also passes through the point.
View answer and explanation
Correct option: A
(−3,2,1)(-3,2,1)
Option A ((-3,2,1)) is correct because it satisfies the equation of the plane that bisects the line segment at right angles.
228 Medium
228 of 241
Application +4 | -1
The plane x−2y+3z=17x-2 y+3 z=17 divides the line joining the points (−2,4,7)(-2,4,7) and (3,−5,8)(3,-5,8) in the ratio
View answer and explanation
Correct option: B
3:103: 10
Option B (3:103: 10) is correct because it represents the ratio in which the plane divides the line segment, calculated by substituting the section formula into the plane equation.
229 Easy
229 of 241
Application +4 | -1
The distance of the point (1,0,2)(1,0,2) from the point of intersection of the line x−23=y+14=z−212\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12} and the plane x−y+z=16x-y+z=16, is
View answer and explanation
Correct option: D
13
Option D (13) is correct because it is the distance between the given point and the intersection point of the line and plane.
230 Easy
230 of 241
Application +4 | -1
The Cartesian equation of the plane perpendicular to the line x−12=y−3−1=z−42\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{2} and passing through the origin is
View answer and explanation
Correct option: C
2x−y+2z=02 x-y+2 z=0
Option C (2x−y+2z=02x - y + 2z = 0) is correct because it represents the equation of a plane that is perpendicular to the given line and passes through the origin.
231 Easy
231 of 241
Application +4 | -1
If the angle θ\theta between the line x+11=y−12=z−22\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2} and the plane 2x−y+λz+4=02 x-y+\sqrt{\lambda} z+4=0 is such that sin⁡θ=13\sin \theta=\frac{1}{3}. The value of λ\lambda is
View answer and explanation
Correct option: D
53\frac{5}{3}
The correct answer is 53\frac{5}{3} because it is the value of λ\lambda that satisfies the given condition sin⁡θ=13\sin \theta = \frac{1}{3} using the formula for the angle between a line and a plane.
232 Easy
232 of 241
Application +4 | -1
The angle between the pair of planes represented by equation 2x2−2y2+4z2+6xz+2yz+3xy=02 x^2-2 y^2+4 z^2+6 x z+2 y z+3 x y=0 is
View answer and explanation
Correct option: C
cos⁡−1(49)\cos ^{-1}\left(\frac{4}{9}\right)
Option C (cos⁡−1(49)\cos^{{-1}}(\frac{{4}}{{9}})) is correct because it is the angle between the pair of planes represented by the given equation.
233 Medium
233 of 241
Application +4 | -1
The plane, passing through the points (0,−1,2)(0,-1,2) and (−1,2,1)(-1,2,1) and parallel to the line passing through (5,1,−7)(5,1,-7) and (1,−1,−1)(1,-1,-1), also passes through the point
View answer and explanation
Correct option: A
(−2,5,0)(-2,5,0)
Option A ((-2,5,0)) is correct because after finding the equation of the plane, substituting the coordinates of the point (-2,5,0) satisfies the equation.
234 Easy
234 of 241
Application +4 | -1
If the line x−12=y+13=z−14\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4} and x−31=y−k2=z1\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1} intersect, then kk is equal to
View answer and explanation
Correct option: C
92\frac{9}{2}
Option C (92\frac{9}{2}) is correct because it is the value of k for which the two lines intersect, found by equating the coordinates of a general point on each line and solving the resulting system of equations.
235 Easy
235 of 241
Application +4 | -1
If the line, x−32=y+2−1=z+43\frac{x-3}{2}=\frac{y+2}{-1}=\frac{z+4}{3} lies in the plane, lx+my−z=9l x+m y-z=9, then l2+m2l^2+m^2 is equal to
View answer and explanation
Correct option: D
2
Option D (2) is correct because it is the value of l2+m2l^2 + m^2 calculated from the conditions that the line lies in the plane.
236 Easy
236 of 241
Analyzing +4 | -1
An angle between the plane x+y+z=5x+y+z=5 and the line of intersection of planes 3x+4y+z−1=03 x+4 y+z-1=0 and 5x+8y+2z+14=05 x+8 y+2 z+14=0 is
View answer and explanation
Correct option: D
sin⁡−1317\sin ^{-1} \sqrt{\frac{3}{17}}
Option D (sin⁡−1317\sin^{-1} \sqrt{\frac{3}{17}}) is correct because the angle between the plane and the line of intersection is found using the dot product of the normal vector of the plane and the direction vector of the line.
237 Easy
237 of 241
Application +4 | -1
The join of the points (−3,−4)(-3,-4) and (1,−2)(1,-2) is divided by yy-axis in the ratio.
View answer and explanation
Correct option: C
3:13: 1
Option C (3:13:1) is correct because the y-axis divides the line segment in the ratio λ:1\lambda:1 where the x-coordinate of the point of division is 0.
238 Easy
238 of 241
Application +4 | -1
Let the plane passing through the point (-1, 0, - 2) and perpendicular to each of the planes 2x+y−z=22x + y - z = 2 and x−y−z=3x - y - z = 3 be ax+by+cz+8=0ax + by + cz + 8 = 0. Then the value of a+b+ca + b + c is equal to
View answer and explanation
Correct option: D
4
Option D (4) is correct because by finding the normal vector to the plane and using the point (-1, 0, -2), we can determine the equation of the plane and then calculate a + b + c.
239 Easy
239 of 241
Analyzing +4 | -1
In a three dimensional coordinate system P,QP, Q and RR are images of a point A(a,b,c)A(a, b, c) in the XYX Y they YZY Z and the ZXZ X planes respectively. If GG is the centroid of triangle PQRP Q R, then area of triangle AOGA O G is ( OO is the origin)
View answer and explanation
Correct option: A
0
Option A (0) is correct because the points A, O, and G are collinear, resulting in a triangle with zero area.
240 Medium
240 of 241
Analyzing +4 | -1
Let the plane containing the line of intersection of the planes P1:x+(λ+4)y+z=1P 1: x+(\lambda+4) y+z=1 and P2:2x+yP 2: 2 x+y +z=2+z=2 pass through the points (0,1,0)(0,1,0) and (1,0,1).(1,0, 1). Then the distance of the point (2λ,λ,−λ)(2 \lambda, \lambda,-\lambda) from the plane P2P 2 is
View answer and explanation
Correct option: B
363 \sqrt{6}
Option B (363 \sqrt{{6}}) is correct because after finding λ, the distance from the point to the plane P2 is calculated using the distance formula.
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