JEE Mathematics Three Dimensional Geometry MCQs Set 10

Download JEE MCQs for JEE Mathematics: Three Dimensional Geometry

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Chapter-wise Objective Questions: Three Dimensional Geometry

Navigate directly to the 50 objective questions for Three Dimensional Geometry using the digital viewer below. Each practice set includes verified answer keys, allowing students to instantly cross-check their work and identify areas requiring further revision.

Multiple Choice Questions

Question 1: The planes \( 2x-y+4z=5 \) and \( 5x-2.5y+10z=6 \) are:
(a) perpendicular
(b) parallel
(c) intersecting on the y-axis
(d) such that the second passes through \( (0,0,\frac54) \)
Show Answer & Explanation

Answer: (b) parallel

Explanation:
1. Coefficient ratios: 5/2 = 2.5/1 = 10/4 = 2.5.
2. The ratios are equal but the constant ratio 6/5 is different.
3. So the planes are parallel and distinct.

Teacher's Note:
1. Equal ratios of x, y, z coefficients give parallel planes; the constants decide whether they coincide.

Question 2: A plane passes through the points \( (1,2,3) \) and \( (-1,2,0) \) and is perpendicular to the ZX-plane. If its equation is \( ax+by+cz+d=0\ (a>0) \), then:
(a) \( a=0\ \text{and}\ c=0 \)
(b) \( a+d=0 \)
(c) \( c+d-5=0 \)
(d) \( a+c+d-4=0 \)
Show Answer & Explanation

Answer: (d) \( a+c+d-4=0 \)

Explanation:
1. Perpendicular to the ZX-plane means the normal has no y-component, so b = 0.
2. Substituting the points: a + 3c + d = 0 and -a + d = 0, so d = a and c = -2a/3.
3. Take a = 3: c = -2, d = 3 and a + c + d = 4.

Teacher's Note:
1. A plane is perpendicular to a coordinate plane if the normal lies in that plane.
2. Work with ratios, since a, c, d can be scaled.

Question 3: A plane \( \pi \) passes through \( (1,1,1) \). If \( b,c,a \) are the direction ratios of a normal to the plane, where \( a(a) \( 29x+31y+3z=63 \)
(b) \( 23x+29y-29z=23 \)
(c) \( 23x+29y+3z=55 \)
(d) \( 31x+27y+3z=71 \)
Show Answer & Explanation

Answer: (c) \( 23x+29y+3z=55 \)

Explanation:
1. 2001 = 3 × 23 × 29, so a = 3, b = 23, c = 29.
2. The normal ratios (b, c, a) = (23, 29, 3), so the plane is 23(x - 1) + 29(y - 1) + 3(z - 1) = 0.
3. Hence 23x + 29y + 3z = 55.

Teacher's Note:
1. Factorise 2001 first, then assign the prime factors in the stated order.
2. Use point-normal form through (1, 1, 1).

Question 4: Let \( A(1,1,1),B(2,3,5) \) and \( C(-1,0,2) \) be three points. The equation of a plane parallel to the plane ABC which is at a distance 2 units from it is:
(a) \( 2x-3y+z+2\sqrt{14}=0 \)
(b) \( 2x-3y+z-\sqrt{14}=0 \)
(c) \( 2x-3y+z+2=0 \)
(d) \( 2x-3y+z-2=0 \)
Show Answer & Explanation

Answer: (a) \( 2x-3y+z+2\sqrt{14}=0 \)

Explanation:
1. AB = (1, 2, 4) and AC = (-2, -1, 1); the normal is AB x AC = (6, -9, 3), i.e. (2, -3, 1).
2. Plane ABC: 2x - 3y + z = 0. A parallel plane is 2x - 3y + z = k.
3. Distance |k|/√(14) = 2 gives k = +-2 √(14); the option with 2 √(14) is the first.

Teacher's Note:
1. Find the plane through three points first, then shift it along its normal.
2. The distance between parallel planes is |k|/|n|.

Question 5: A variable plane is at a constant distance \( p \) from the origin and meets the axes at A, B, C. The locus of the centroid of the tetrahedron OABC is:
(a) \( \frac1{x^2}+\frac1{y^2}+\frac1{z^2}=\frac1{p^2} \)
(b) \( \frac1{x^2}+\frac1{y^2}+\frac1{z^2}=\frac4{p^2} \)
(c) \( \frac1{x^2}+\frac1{y^2}+\frac1{z^2}=\frac{16}{p^2} \)
(d) \( \frac1{x^2}+\frac1{y^2}+\frac1{z^2}=\frac{1}{16p^2} \)
Show Answer & Explanation

Answer: (c) \( \frac1{x^2}+\frac1{y^2}+\frac1{z^2}=\frac{16}{p^2} \)

Explanation:
1. Let the plane be x/a + y/b + z/c = 1, with 1/a^2 + 1/b^2 + 1/c^2 = 1/p^2.
2. The centroid of OABC is (a/4, b/4, c/4) = (x, y, z), so a = 4x, b = 4y, c = 4z.
3. Therefore 1/(16x^2) + 1/(16y^2) + 1/(16z^2) = 1/p^2, i.e. the sum equals 16/p^2.

Teacher's Note:
1. For a tetrahedron OABC the centroid is one quarter of the sum of the vertices.
2. Use the distance formula in intercept form.

Question 6: The equation of the plane through the line of intersection of the planes \( 2x+y+3z-2=0 \) and \( x-y+z+4=0 \), such that the plane is at a distance 2 units from the origin, is:
(a) \( x+y+2z+13=0\ \text{or}\ x+y+z-3=0 \)
(b) \( 2x+y-2z+3=0\ \text{or}\ x-2y-2z-3=0 \)
(c) \( 15x-12y+16z+50=0\ \text{or}\ x+2y+2z-6=0 \)
(d) \( x-y+2z-13=0\ \text{or}\ x+y-z-3=0 \)
Show Answer & Explanation

Answer: (c) \( 15x-12y+16z+50=0\ \text{or}\ x+2y+2z-6=0 \)

Explanation:
1. The family is (2 + k)x + (1 - k)y + (3 + k)z + (4k - 2) = 0.
2. Distance 2: (4k - 2)^2 = 4(3k^2 + 8k + 14), which gives k^2 - 12k - 13 = 0, so k = 13 or -1.
3. k = 13: 15x - 12y + 16z + 50 = 0. k = -1: x + 2y + 2z - 6 = 0.

Teacher's Note:
1. Use P1 + k P2 = 0 for planes through the line of intersection.
2. Impose the distance condition to find k.

Question 7: The sum of the intercepts of the plane which bisects the line segment joining \( (0,1,2) \) and \( (2,3,0) \) perpendicularly is:
(a) 2
(b) 4
(c) 6
(d) 12
Show Answer & Explanation

Answer: (a) 2

Explanation:
1. The mid-point is (1, 2, 1) and the normal is (2, 2, -2), i.e. (1, 1, -1).
2. The plane is x + y - z = 1 + 2 - 1 = 2.
3. Intercepts: 2, 2, -2, whose sum is 2.

Teacher's Note:
1. A perpendicular bisector plane has the segment as its normal and passes through its mid-point.
2. Signs of intercepts matter: z-intercept is negative.

Question 8: A plane meets the coordinate axes at A, B, C so that the centroid of the triangle ABC is \( (1,2,4) \). Then the equation of the plane is:
(a) \( x+2y+4z=12 \)
(b) \( 4x+2y+z=12 \)
(c) \( x+2y+4z=3 \)
(d) \( 4x+2y+z=3 \)
Show Answer & Explanation

Answer: (b) \( 4x+2y+z=12 \)

Explanation:
1. Centroid (a/3, b/3, c/3) = (1, 2, 4) gives a = 3, b = 6, c = 12.
2. The plane is x/3 + y/6 + z/12 = 1.
3. Multiply by 12: 4x + 2y + z = 12.

Teacher's Note:
1. The centroid of the intercept triangle is one third of each intercept.
2. Convert to integer coefficients at the end.

Question 9: The reflection of the plane \( 2x-3y+4z-3=0 \) in the plane \( x-y+z-3=0 \) is:
(a) \( 4x-3y+2z-15=0 \)
(b) \( x-3y+2z-15=0 \)
(c) \( 4x+3y-2z+15=0 \)
(d) \( 4x+3y+2z+15=0 \)
Show Answer & Explanation

Answer: (a) \( 4x-3y+2z-15=0 \)

Explanation:
1. The reflection of a1x + b1y + c1z + d1 = 0 in ax + by + cz + d = 0 is 2(aa1 + bb1 + cc1)(ax + by + cz + d) = (a^2 + b^2 + c^2)(a1x + b1y + c1z + d1).
2. Here the dot product is 2 + 3 + 4 = 9 and a^2 + b^2 + c^2 = 3.
3. So 18(x - y + z - 3) = 3(2x - 3y + 4z - 3), which simplifies to 4x - 3y + 2z - 15 = 0.

Teacher's Note:
1. The mirror formula comes from reflecting the normal and the constant term together.
2. Simplify by dividing out the common factor 3.

Question 10: The vertices of a tetrahedron are \( A(3,4,2),B(1,2,1),C(4,1,3) \) and \( D(-1,-1,3) \). The height of A above the base BCD is:
(a) \( \frac{27}{\sqrt{237}} \)
(b) \( \frac{23}{\sqrt{237}} \)
(c) \( \frac{20}{\sqrt{237}} \)
(d) \( \frac{27}{\sqrt{247}} \)
Show Answer & Explanation

Answer: (b) \( \frac{23}{\sqrt{237}} \)

Explanation:
1. BC = (3, -1, 2) and BD = (-2, -3, 2). Their cross product is (4, -10, -11).
2. Plane BCD: 4(x - 1) - 10(y - 2) - 11(z - 1) = 0, i.e. 4x - 10y - 11z + 27 = 0.
3. Height = |12 - 40 - 22 + 27| / √(16 + 100 + 121) = 23/√(237).

Teacher's Note:
1. The height of a vertex above the opposite face is its distance from the plane of that face.

Question 11: The equation of the plane which passes through the line of intersection of the planes \( 2x-y=0 \) and \( 3z-y=0 \) and is perpendicular to the plane \( 4x+5y-3z=8 \) is:
(a) \( 28x-17y+9z=0 \)
(b) \( 28x+17y+9z=0 \)
(c) \( 2x+17y-9z=0 \)
(d) \( 2x-y-z=0 \)
Show Answer & Explanation

Answer: (a) \( 28x-17y+9z=0 \)

Explanation:
1. The family is (2x - y) + k(3z - y) = 0, with normal (2, -1 - k, 3k).
2. Perpendicular to (4, 5, -3): 8 - 5 - 5k - 9k = 0, so k = 3/14.
3. Multiply by 14: 28x - 14y + 9z - 3y = 28x - 17y + 9z = 0.

Teacher's Note:
1. Planes through the line of intersection form a one-parameter family.
2. Use the perpendicularity condition on the normals.

Question 12: The plane \( 2x+3y+kz-7=0 \) is parallel to the line whose direction ratios are \( (2,-3,1) \). Then \( k= \):
(a) 5
(b) 8
(c) 1
(d) 0
Show Answer & Explanation

Answer: (a) 5

Explanation:
1. A plane is parallel to a line when the normal is perpendicular to the line direction.
2. 2(2) + 3(-3) + k(1) = 0.
3. So k = 5.

Teacher's Note:
1. Normal dot direction is zero when a line is parallel to a plane.

Question 13: A variable plane is at a constant distance \( 3p \) from the origin and meets the axes in A, B and C. The locus of the centroid of the triangle ABC is:
(a) \( x^{-2}+y^{-2}+z^{-2}=p^{-2} \)
(b) \( x^{-2}+y^{-2}+z^{-2}=4p^{-2} \)
(c) \( x^{-2}+y^{-2}+z^{-2}=16p^{-2} \)
(d) \( x^{-2}+y^{-2}+z^{-2}=9p^{-2} \)
Show Answer & Explanation

Answer: (a) \( x^{-2}+y^{-2}+z^{-2}=p^{-2} \)

Explanation:
1. With intercepts a, b, c: 1/a^2 + 1/b^2 + 1/c^2 = 1/(9p^2).
2. The centroid of triangle ABC is (a/3, b/3, c/3), so a = 3x, b = 3y, c = 3z.
3. Then 1/(9x^2) + 1/(9y^2) + 1/(9z^2) = 1/(9p^2), giving the sum = 1/p^2.

Teacher's Note:
1. Centroid of the triangle is one third of each intercept.
2. The factor 9 cancels on both sides.

Question 14: The equation of the plane through the line of intersection of the planes \( x-2y+3z-1=0 \) and \( 2x+y+z-2=0 \) and the point \( (1,2,3) \) is:
(a) \( 7x-9y+8z=0 \)
(b) \( 7x+y+8z=0 \)
(c) \( x+3y-2z-1=0 \)
(d) \( x-3y-2z+1=0 \)
Show Answer & Explanation

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

Explanation:
1. The family is (x - 2y + 3z - 1) + k(2x + y + z - 2) = 0.
2. At (1, 2, 3): 5 + 5k = 0, so k = -1.
3. The plane is -x - 3y + 2z + 1 = 0, i.e. x + 3y - 2z - 1 = 0.

Teacher's Note:
1. Use the family P1 + k P2 = 0 and fix k with the given point.

Question 15: If 5, 7, 6 are the sums of the X and Y intercepts, Y and Z intercepts, Z and X intercepts respectively of a plane, then the perpendicular distance from the origin to that plane is:
(a) \( \frac{144}{61} \)
(b) \( \frac{12}{\sqrt{61}} \)
(c) \( \frac{\sqrt{61}}{12} \)
(d) \( \frac{61}{144} \)
Show Answer & Explanation

Answer: (b) \( \frac{12}{\sqrt{61}} \)

Explanation:
1. a + b = 5, b + c = 7, c + a = 6, so a + b + c = 9, giving a = 2, b = 3, c = 4.
2. The plane is x/2 + y/3 + z/4 = 1, i.e. 6x + 4y + 3z = 12.
3. Distance = 12/√(36 + 16 + 9) = 12/√(61).

Teacher's Note:
1. Add the three equations to get the sum of the intercepts.
2. Then find each intercept by subtraction.

Question 16: The equation of the plane through the line of intersection of the planes \( 2x+3y+4z-7=0 \), \( x+y+z-1=0 \) and perpendicular to the plane \( x-5y+3z-2=0 \) is:
(a) \( 7x-y-6z-17=0 \)
(b) \( x-y-6z-27=0 \)
(c) \( x+2y+3z-6=0 \)
(d) \( x+y+6z-27=0 \)
Show Answer & Explanation

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

Explanation:
1. The family is (2x + 3y + 4z - 7) + k(x + y + z - 1) = 0 with normal (2 + k, 3 + k, 4 + k).
2. Perpendicular to (1, -5, 3): (2 + k) - 5(3 + k) + 3(4 + k) = 0, so -1 - k = 0 and k = -1.
3. The plane is x + 2y + 3z - 6 = 0.

Teacher's Note:
1. Apply the perpendicularity condition to the normal of the family member.

Question 17: If the angles made by the normal of the plane \( 2x+3y-4z-16=0 \) with the coordinate axes X, Y, Z are \( \cos^{-1}k_1,\cos^{-1}k_2,\cos^{-1}k_3 \), then \( k_1,k_2,k_3 \) respectively are:
(a) \( \frac2{\sqrt{29}},\frac3{\sqrt{29}},\frac4{\sqrt{29}} \)
(b) \( \frac2{\sqrt{29}},-\frac3{\sqrt{29}},\frac4{\sqrt{29}} \)
(c) \( \frac2{\sqrt{29}},\frac3{\sqrt{29}},-\frac4{\sqrt{29}} \)
(d) \( \frac12,\frac13,\frac14 \)
Show Answer & Explanation

Answer: (c) \( \frac2{\sqrt{29}},\frac3{\sqrt{29}},-\frac4{\sqrt{29}} \)

Explanation:
1. The normal has ratios (2, 3, -4) and length √(29).
2. The direction cosines are 2/√(29), 3/√(29), -4/√(29).
3. They are the cosines of the angles with the axes.

Teacher's Note:
1. The cosines of the angles made with the axes are exactly the direction cosines.

Question 18: The two planes represented by \( 12x^2-2y^2-6z^2+7yz+6zx-2xy=0 \) are:
(a) \( 2x+y+2z=0,\ 6x-2y+3z=0 \)
(b) \( 2x-y+2z=0,\ 6x+2y-3z=0 \)
(c) \( 2x+y+2z+4=0,\ 6x+2y-3z=0 \)
(d) \( 2x-y+2z=0,\ 6x+2y-3z+1=0 \)
Show Answer & Explanation

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

Explanation:
1. Factorise the expression: (2x - y + 2z)(6x + 2y - 3z).
2. Expanding: 12x^2 + 4xy - 6xz - 6xy - 2y^2 + 3yz + 12xz + 4yz - 6z^2 = 12x^2 - 2xy + 6xz - 2y^2 + 7yz - 6z^2.
3. This matches the given expression, so the planes are 2x - y + 2z = 0 and 6x + 2y - 3z = 0.

Teacher's Note:
1. A homogeneous second-degree equation in x, y, z represents two planes through the origin.
2. Verify a factorisation by expanding it.

Question 19: The plane \( 3(x-2)+(y-2)+6(z+3)=0 \) contains the line \( \frac{x-2}{a}=\frac{y-2}{b}=\frac{z+3}{1} \) whose inclination with the X-axis is \( 60^\circ \). Then \( a \) satisfies:
(a) \( 6a^2+36a+37=0 \)
(b) \( 36a^2+37=0 \)
(c) \( 36a^2+37a+36=0 \)
(d) \( a+3=0 \)
Show Answer & Explanation

Answer: (a) \( 6a^2+36a+37=0 \)

Explanation:
1. Line in the plane: 3a + b + 6 = 0, so b = -3a - 6.
2. cos 60 degrees = a/√(a^2 + b^2 + 1) = 1/2 gives b^2 = 3a^2 - 1.
3. Substituting: (3a + 6)^2 = 3a^2 - 1, so 6a^2 + 36a + 37 = 0.

Teacher's Note:
1. Use both the containment condition and the angle condition, and eliminate b.

Question 20: The angle between the planes represented by \( 2x^2-6y^2-12z^2+18yz+2zx+xy=0 \) is:
(a) \( \cos^{-1}\left(\frac{16}{21}\right) \)
(b) \( \cos^{-1}\left(\frac{17}{21}\right) \)
(c) \( \cos^{-1}\left(\frac{19}{21}\right) \)
(d) \( \frac\pi2 \)
Show Answer & Explanation

Answer: (a) \( \cos^{-1}\left(\frac{16}{21}\right) \)

Explanation:
1. Factorise: (2x - 3y + 6z)(x + 2y - 2z).
2. Normals (2, -3, 6) and (1, 2, -2) with lengths 7 and 3.
3. cos(θ) = |2 - 6 - 12|/(7 × 3) = 16/21.

Teacher's Note:
1. Factorise the quadratic form into two planes and then use the angle formula.

Question 21: The equation of the plane passing through the line of intersection of the planes \( ax+by+cz+d=0 \) and \( a_1x+b_1y+c_1z+d_1=0 \) and perpendicular to the XY-plane is \( px+qy+rz+s=0 \). Then \( s= \):
(a) \( dc_1-d_1c \)
(b) \( dc_1+d_1c \)
(c) \( dd_1+cc_1 \)
(d) \( aa_1+bb_1+cc_1 \)
Show Answer & Explanation

Answer: (a) \( dc_1-d_1c \)

Explanation:
1. Take (ax + by + cz + d) + k(a1x + b1y + c1z + d1) = 0.
2. Perpendicular to the XY-plane means the z-coefficient is zero: c + k c1 = 0, so k = -c/c1.
3. Multiply by c1: the constant term is d c1 - d1 c.

Teacher's Note:
1. Perpendicular to the XY-plane means the normal has no z-component.

Question 22: P is a point such that the sum of the squares of its distances from the planes \( x+y+z=0,\ x+y-2z=0,\ x-y=0 \) is 5. Then the locus of P is:
(a) \( x^2+y^2+z^2=10 \)
(b) \( x^2+y^2+z^2=25 \)
(c) \( x^2+y^2+z^2=5 \)
(d) \( x^2+y^2+z^2=50 \)
Show Answer & Explanation

Answer: (c) \( x^2+y^2+z^2=5 \)

Explanation:
1. Sum of squares of distances = (x + y + z)^2/3 + (x + y - 2z)^2/6 + (x - y)^2/2 = 5.
2. Coefficient of x^2: 1/3 + 1/6 + 1/2 = 1, same for y^2 and z^2.
3. All cross terms cancel, so x^2 + y^2 + z^2 = 5.

Teacher's Note:
1. The three planes have mutually perpendicular normals, so they act like new coordinate axes.
2. The sum of squared distances is then the squared distance from the origin.

Question 23: The plane \( ax+by+cz+(-3)=0 \) meets the coordinate axes at A, B, C. The centroid of the triangle ABC is:
(a) \( (3a,3b,3c) \)
(b) \( \left(\frac3a,\frac3b,\frac3c\right) \)
(c) \( \left(\frac a3,\frac b3,\frac c3\right) \)
(d) \( \left(\frac1a,\frac1b,\frac1c\right) \)
Show Answer & Explanation

Answer: (d) \( \left(\frac1a,\frac1b,\frac1c\right) \)

Explanation:
1. The plane is ax + by + cz = 3 with intercepts 3/a, 3/b, 3/c.
2. The centroid is one third of the sum of the vertices: (1/a, 1/b, 1/c).

Teacher's Note:
1. Find the intercepts first, then divide each by 3.

Question 24: The equation of the plane passing through the point \( (-1,3,2) \) and perpendicular to each of the planes \( x+2y+3z=5 \) and \( 3x+3y+z=0 \) is:
(a) \( 7x+8y-3z=0 \)
(b) \( 7x-8y+3z=-37 \)
(c) \( 7x-8y+3z+25=0 \)
(d) \( 7x+8y+3z=23 \)
Show Answer & Explanation

Answer: (c) \( 7x-8y+3z+25=0 \)

Explanation:
1. Normal = (1, 2, 3) × (3, 3, 1) = (2 - 9, 9 - 1, 3 - 6) = (-7, 8, -3).
2. Plane: 7(x + 1) - 8(y - 3) + 3(z - 2) = 0.
3. This is 7x - 8y + 3z + 25 = 0.

Teacher's Note:
1. A plane perpendicular to two planes has a normal along the cross product of their normals.

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