JEE Mathematics Three Dimensional Geometry MCQs Set 11

Mathematics Objective Questions and Answers: Three Dimensional Geometry

Access targeted multiple-choice questions for Three Dimensional Geometry designed to align with the latest JEE academic syllabus for JEE Mathematics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

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View or download the dedicated Three Dimensional Geometry MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Mathematics evaluations.

Multiple Choice Questions

Question 1: If P = (0, 1, 0) and Q = (0, 0, 1), then the projection of PQ on the plane \( x+y+z=3 \) is:
(a) 2
(b) \( \sqrt2 \)
(c) 3
(d) \( \sqrt3 \)
Show Answer & Explanation

Answer: (b) \( \sqrt2 \)

Explanation:
1. PQ = (0, -1, 1) with length √(2).
2. The unit normal is (1, 1, 1)/√(3), and PQ dot normal = 0, so PQ is parallel to the plane.
3. Its projection is PQ itself, of length √(2).

Teacher's Note:
1. Projection onto a plane removes the component along the normal.
2. Here that component is zero.

Question 2: A parallelepiped is formed by the planes drawn through the points \( (2,3,5) \) and \( (5,9,7) \) parallel to the coordinate planes. The length of the diagonal of the parallelepiped is:
(a) 7
(b) \( \sqrt{38} \)
(c) \( \sqrt{155} \)
(d) \( \sqrt7 \)
Show Answer & Explanation

Answer: (a) 7

Explanation:
1. Edge lengths: 5 - 2 = 3, 9 - 3 = 6, 7 - 5 = 2.
2. Diagonal = √(9 + 36 + 4) = √(49).
3. This equals 7.

Teacher's Note:
1. The planes through the two points parallel to the coordinate planes give a box whose opposite vertices are the two points.

Question 3: Match the following. List-I: (a) the image of \( (2,1,4) \) in the plane \( 2x-y+z+5=0 \); (b) the foot of the perpendicular from \( (2,1,4) \) to the same plane; (c) the point lying on the plane. List-II: (i) \( (-2,3,2) \); (ii) \( (-4,1,4) \); (iii) \( (-6,5,0) \). The correct matching is:
(a) a-iii, b-i, c-ii
(b) a-i, b-iii, c-ii
(c) a-ii, b-iii, c-i
(d) a-iii, b-ii, c-i
Show Answer & Explanation

Answer: (a) a-iii, b-i, c-ii

Explanation:
1. Value of the plane expression at (2, 1, 4): 4 - 1 + 4 + 5 = 12, and the normal (2, -1, 1) has length squared 6.
2. Image = (2, 1, 4) - 2(12/6)(2, -1, 1) = (-6, 5, 0), which is item (iii).
3. Foot = mid-point of the point and its image = (-2, 3, 2), which is item (i).
4. Check (-4, 1, 4): 2(-4) - 1 + 4 + 5 = 0, so it lies on the plane and is item (ii).
5. Hence a-iii, b-i, c-ii.

Teacher's Note:
1. The foot is the mid-point of the point and its image.
2. A point lies on the plane when it makes the plane expression zero.

Question 4: \( \alpha,\beta,\gamma \) are the angles made by the normal of the plane \( 2x-3y+6z-6=0 \) with the axes X, Y, Z. Match: List-I (a) \( \cos\alpha \), (b) \( \cos\beta \), (c) \( \cos\gamma \); List-II (i) \( 6/7 \), (ii) \( -3/7 \), (iii) \( 2/7 \). The correct matching is:
(a) a-i, b-ii, c-iii
(b) a-ii, b-i, c-iii
(c) a-iii, b-ii, c-i
(d) a-ii, b-iii, c-i
Show Answer & Explanation

Answer: (c) a-iii, b-ii, c-i

Explanation:
1. The normal is (2, -3, 6) with length 7.
2. cos(α) = 2/7, cos(β) = -3/7, cos(γ) = 6/7.
3. So a-iii, b-ii, c-i.

Teacher's Note:
1. Direction cosines are the normal components divided by its length.

Question 5: The point P is \( (1,2,-1) \). Match the plane in List-I with its perpendicular distance from P in List-II. List-I: (a) \( 2x+y-z+2=0 \); (b) \( x-y+6z=0 \); (c) \( 3x+2y+6z+6=0 \). List-II: (i) 1; (ii) \( 7/\sqrt6 \); (iii) \( 7/\sqrt{38} \). The correct matching is:
(a) a-iii, b-i, c-ii
(b) a-ii, b-iii, c-i
(c) a-i, b-ii, c-iii
(d) a-iii, b-ii, c-i
Show Answer & Explanation

Answer: (b) a-ii, b-iii, c-i

Explanation:
1. (a) |2 + 2 + 1 + 2|/√(6) = 7/√(6).
2. (b) |1 - 2 - 6|/√(38) = 7/√(38).
3. (c) |3 + 4 - 6 + 6|/7 = 1.
4. So a-ii, b-iii, c-i.

Teacher's Note:
1. Use the distance formula for each plane separately.

Question 6: Statement I: The image of \( (0,0,0) \) in the plane \( 2x-3y+6z-49=0 \) is \( (4,-6,12) \). Statement II: The foot of the perpendicular from \( (1,2,3) \) to the plane \( x+y-2z-9=0 \) is \( (3,4,-1) \). Which of the following is correct?
(a) I is true, II is true
(b) I is true, II is false
(c) Both I and II are false
(d) I is false, II is true
Show Answer & Explanation

Answer: (a) I is true, II is true

Explanation:
1. I: the plane expression at the origin is -49, normal length squared 49, so image = 0 - 2(-49/49)(2, -3, 6) = (4, -6, 12), which is true.
2. II: expression at (1, 2, 3) is 1 + 2 - 6 - 9 = -12, length squared 6, so foot = (1, 2, 3) + 2(1, 1, -2) = (3, 4, -1), which is also true.

Teacher's Note:
1. Compute the image and the foot with the same shift formula.
2. The foot uses one times the shift, the image twice.

Question 7: Assertion (A): The equation of the plane through the intersection of the planes \( x+y+z=6 \) and \( 2x+3y+4z+5=0 \) and the point \( (4,4,4) \) is \( 29x+23y+17z=276 \). Reason (R): The equation of the plane through the line of intersection of \( P_1=0 \) and \( P_2=0 \) is \( P_1+\lambda P_2=0\ (\lambda\ne0) \). Choose the correct option.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
Show Answer & Explanation

Answer: (a) Both A and R are true and R is the correct explanation of A

Explanation:
1. Use (x + y + z - 6) + k(2x + 3y + 4z + 5) = 0.
2. At (4, 4, 4): 6 + 41k = 0, so k = -6/41.
3. Then 41(x + y + z - 6) - 6(2x + 3y + 4z + 5) = 29x + 23y + 17z - 276 = 0, so A is true, and R is exactly the method used.

Teacher's Note:
1. The family P1 + k P2 = 0 explains the construction directly.

Question 8: Assertion (A): The points \( (2,1,5) \) and \( (3,4,3) \) lie on opposite sides of the plane \( 2x+2y-2z-1=0 \). Reason (R): The algebraic perpendicular distances from the points to the plane have opposite signs. Choose the correct option.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
Show Answer & Explanation

Answer: (a) Both A and R are true and R is the correct explanation of A

Explanation:
1. At (2, 1, 5): 4 + 2 - 10 - 1 = -5. At (3, 4, 3): 6 + 8 - 6 - 1 = 7.
2. The values have opposite signs, so the points are on opposite sides, and A is true.
3. The sign of the expression is exactly the sign of the algebraic distance, so R explains A.

Teacher's Note:
1. The sign of the plane expression tells the side of the plane.

Question 9: Two planes are given: \( P_1: 2x-y+z=2 \) and \( P_2: x+2y-z=3 \). The equation of the plane passing through the intersection of \( P_1 \) and \( P_2 \) and the point \( (3,2,1) \) is:
(a) \( 3x-y+2z-9=0 \)
(b) \( x-3y+2z+1=0 \)
(c) \( 2x-3y+z-1=0 \)
(d) \( 4x-3y+2z-8=0 \)
Show Answer & Explanation

Answer: (b) \( x-3y+2z+1=0 \)

Explanation:
1. Use (2x - y + z - 2) + k(x + 2y - z - 3) = 0.
2. At (3, 2, 1): 3 + 3k = 0, so k = -1.
3. The plane is x - 3y + 2z + 1 = 0.

Teacher's Note:
1. Fix the parameter k by substituting the given point.

Question 10: For the planes \( P_1: 2x-y+z=2 \) and \( P_2: x+2y-z=3 \), the equation of the plane through the point \( (-1,3,2) \) and perpendicular to each of them is:
(a) \( x+3y-5z+2=0 \)
(b) \( x+3y+5z-18=0 \)
(c) \( x-3y-5z+20=0 \)
(d) \( x-3y+5z=0 \)
Show Answer & Explanation

Answer: (c) \( x-3y-5z+20=0 \)

Explanation:
1. Normal = (2, -1, 1) × (1, 2, -1) = (-1, 3, 5).
2. Plane: -(x + 1) + 3(y - 3) + 5(z - 2) = 0, i.e. -x + 3y + 5z - 20 = 0.
3. Equivalent form: x - 3y - 5z + 20 = 0.

Teacher's Note:
1. The normal of the required plane is the cross product of the two normals.

Question 11: For the planes \( P_1: 2x-y+z=2 \) and \( P_2: x+2y-z=3 \), the equation of the bisector of the acute angle between them is:
(a) \( x-3y+2z+1=0 \)
(b) \( 3x+y-5=0 \)
(c) \( x+3y-2z+1=0 \)
(d) \( 3x+z+7=0 \)
Show Answer & Explanation

Answer: (a) \( x-3y+2z+1=0 \)

Explanation:
1. Write both with positive constants: -2x + y - z + 2 = 0 and -x - 2y + z + 3 = 0.
2. The dot product of the normals is 2 - 2 - 1 = -1 < 0, so the bisector containing the origin is the acute one.
3. (-2x + y - z + 2) = (-x - 2y + z + 3) gives x - 3y + 2z + 1 = 0.

Teacher's Note:
1. Make both constant terms positive first.
2. The sign of the dot product of the normals then decides which bisector is acute.

Question 12: For the planes \( P_1: 2x-y+z=2 \) and \( P_2: x+2y-z=3 \), the equation of the bisector of the angle between them which does not contain the origin is:
(a) \( x-3y+2z+1=0 \)
(b) \( x+3y=5 \)
(c) \( x+3y+2z+2=0 \)
(d) \( 3x+y=5 \)
Show Answer & Explanation

Answer: (d) \( 3x+y=5 \)

Explanation:
1. With positive constants: -2x + y - z + 2 = 0 and -x - 2y + z + 3 = 0.
2. The bisector not containing the origin uses the negative sign: (-2x + y - z + 2) = -(-x - 2y + z + 3).
3. This gives -3x - y + 5 = 0, i.e. 3x + y = 5.

Teacher's Note:
1. The bisector containing the origin has the same sign on both sides; the other uses the opposite sign.

Question 13: For the planes \( P_1: 2x-y+z=2 \) and \( P_2: x+2y-z=3 \), the image of the plane \( P_1 \) in the plane mirror \( P_2 \) is:
(a) \( 7x+y+9z+9=0 \)
(b) \( x+7y-4z+5=0 \)
(c) \( 3x+4y-5z+9=0 \)
(d) \( 7x-y+2z-9=0 \)
Show Answer & Explanation

Answer: (d) \( 7x-y+2z-9=0 \)

Explanation:
1. Use 2(n1 . n2)(P2) = |n2|^2 (P1) with n1 . n2 = 2 - 2 - 1 = -1 and |n2|^2 = 6.
2. So -2(x + 2y - z - 3) = 6(2x - y + z - 2).
3. Simplifying gives 14x - 2y + 4z - 18 = 0, i.e. 7x - y + 2z - 9 = 0.

Teacher's Note:
1. The image of a plane in a mirror plane uses the same reflection formula as a point.
2. Divide out the common factor at the end.

Question 14: The distance between the parallel planes \( 2x-2y+z+3=0 \) and \( 4x-4y+2z+5=0 \) is:
(a) \( \frac13 \)
(b) \( \frac56 \)
(c) \( \frac16 \)
(d) \( \frac12 \)
Show Answer & Explanation

Answer: (c) \( \frac16 \)

Explanation:
1. Write the second plane as 2x - 2y + z + 5/2 = 0.
2. Distance = |3 - 5/2| / √(4 + 4 + 1) = (1/2)/3.
3. This is 1/6.

Teacher's Note:
1. Make the coefficients equal before subtracting the constants.

Question 15: The equation of the plane lying midway between the parallel planes \( 2x-3y+6z-7=0 \) and \( 2x-3y+6z+7=0 \) is:
(a) \( 2x-3y+6z-14=0 \)
(b) \( 2x-3y+6z+14=0 \)
(c) \( 2x-3y+6z=7 \)
(d) \( 2x-3y+6z=0 \)
Show Answer & Explanation

Answer: (d) \( 2x-3y+6z=0 \)

Explanation:
1. The mid-way plane has the same coefficients and the average constant.
2. (-7 + 7)/2 = 0.
3. So 2x - 3y + 6z = 0.

Teacher's Note:
1. The middle plane has constant (d1 + d2)/2.

Question 16: If the image of the point \( (-1,3,4) \) in the plane \( x-2y=0 \) is \( (x_1,y_1,z_1) \), then \( z_1= \):
(a) 3
(b) 4
(c) 0
(d) -4
Show Answer & Explanation

Answer: (b) 4

Explanation:
1. The normal (1, -2, 0) has no z-component.
2. Moving along the normal does not change z.
3. So z1 = 4.

Teacher's Note:
1. The image differs from the point only along the normal direction.

Question 17: If the plane \( 2x-3y+5z-2=0 \) divides the line segment joining \( (1,2,3) \) and \( (2,1,k) \) in the ratio \( 9:11 \), then \( k= \):
(a) -1
(b) 2
(c) -2
(d) 1
Show Answer & Explanation

Answer: (c) -2

Explanation:
1. The ratio is -(value at first point) : (value at second point) = 9 : 11.
2. First point: 2 - 6 + 15 - 2 = 9. Second point: 4 - 3 + 5k - 2 = 5k - 1.
3. -9/(5k - 1) = 9/11 gives 5k - 1 = -11, so k = -2.

Teacher's Note:
1. Apply the ratio formula with the values of the plane expression at the end points.

Question 18: If the equation of the plane \( 2x-3y+6z=7 \) in the normal form is \( lx+my+nz=p \), then \( l+p= \):
(a) \( \frac{11}7 \)
(b) \( \frac57 \)
(c) \( \frac97 \)
(d) \( \frac27 \)
Show Answer & Explanation

Answer: (c) \( \frac97 \)

Explanation:
1. The normal length is √(4 + 9 + 36) = 7.
2. Normal form: (2/7)x - (3/7)y + (6/7)z = 1, so l = 2/7 and p = 1.
3. l + p = 9/7.

Teacher's Note:
1. In the normal form the right side is the distance from the origin and must be positive.

Question 19: The area of the triangle formed by the plane \( 2x+3y+6z+9=0 \) with the Y-axis and Z-axis (in square units) is:
(a) \( \frac92 \)
(b) \( \frac32 \)
(c) \( \frac{27}{4} \)
(d) \( \frac94 \)
Show Answer & Explanation

Answer: (d) \( \frac94 \)

Explanation:
1. The y-intercept is -9/3 = -3 and the z-intercept is -9/6 = -3/2.
2. Area = (1/2)|(-3)(-3/2)|.
3. This equals 9/4.

Teacher's Note:
1. The triangle with two axes is a right triangle with legs equal to the absolute intercepts.

Question 20: The plane \( \frac x2+\frac y3+\frac z4=1 \) cuts the axes at A, B, C. The area of the triangle ABC (in square units) is:
(a) \( \sqrt{61} \)
(b) \( \sqrt{29} \)
(c) \( 2\sqrt{61} \)
(d) \( \frac{\sqrt{61}}{2} \)
Show Answer & Explanation

Answer: (a) \( \sqrt{61} \)

Explanation:
1. Here a = 2, b = 3, c = 4, so ab = 6, bc = 12, ca = 8.
2. Area = (1/2) √((ab)^2 + (bc)^2 + (ca)^2) = (1/2) √(36 + 144 + 64).
3. This is (1/2) √(244) = √(61).

Teacher's Note:
1. Use the formula for the area of the intercept triangle.

Question 21: If the angle between the planes \( x+2y+4z+7=0 \) and \( x-3y+2z+6=0 \) is \( \cos^{-1}\left(\sqrt k\right) \), then \( k= \):
(a) \( \frac{1}{98} \)
(b) \( \frac{3}{98} \)
(c) \( \frac{9}{98} \)
(d) \( \frac{3}{14} \)
Show Answer & Explanation

Answer: (b) \( \frac{3}{98} \)

Explanation:
1. cos(θ) = |1 - 6 + 8| / (√(21) √(14)) = 3/√(294).
2. So √(k) = 3/√(294), and k = 9/294.
3. This simplifies to 3/98.

Teacher's Note:
1. Square the cosine value to find k.

Question 22: If the planes \( 2x-y-3z-7=0 \) and \( 4x-2y+5kz+9=0 \) are parallel, then \( 5k^2+6= \):
(a) \( \frac{36}{5} \)
(b) \( 12 \)
(c) \( \frac{66}{5} \)
(d) \( 11 \)
Show Answer & Explanation

Answer: (c) \( \frac{66}{5} \)

Explanation:
1. Parallel planes: 2/4 = -1/-2 = -3/(5k).
2. So 5k = -6 and k = -6/5.
3. 5k^2 + 6 = 5(36/25) + 6 = 36/5 + 6 = 66/5.

Teacher's Note:
1. Equate the ratios of coefficients and solve for k first.

Question 23: If the planes \( x+2y+kz=0 \) and \( 2x+y-2z=0 \) are at right angles, then \( k= \):
(a) 2
(b) -2
(c) 4
(d) 1
Show Answer & Explanation

Answer: (a) 2

Explanation:
1. Perpendicular normals: 1(2) + 2(1) + k(-2) = 0.
2. 4 - 2k = 0, so k = 2.

Teacher's Note:
1. Dot product of normals must be zero.

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