68 free MCQs on Three Dimensional Geometry with worked answers and explanations. Lines and planes in 3D space, direction cosines, distances, and angles
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Three mutually perpendicular axes x, y, z meeting at the origin O, with a point P located by its (x, y, z) coordinates.
Q1.
Direction cosines of a line are the cosines of angles made with the:
Answer: B. x, y and z axes
Why: Direction cosines l, m, n are the cosines of angles alpha, beta, gamma that the line makes with the positive x, y, z axes.
Q2.
For direction cosines l, m, n of a line:
Answer: C. l<sup>2</sup> + m<sup>2</sup> + n<sup>2</sup> = 1
Why: The fundamental identity: l<sup>2</sup> + m<sup>2</sup> + n<sup>2</sup> = 1. Direction cosines always satisfy this relation.
Q3.
The direction ratios of a line are proportional to its:
Answer: D. Direction cosines
Why: Direction ratios (a, b, c) are any numbers proportional to direction cosines (l, m, n). Ratio a:b:c = l:m:n.
Q4.
The equation of the XY-plane is:
Answer: A. z = 0
Why: The XY-plane contains all points where z = 0. Similarly, the YZ-plane is x = 0 and XZ-plane is y = 0.
Q5.
The distance from the origin to point P(a, b, c) is:
Answer: D. sqrt(a<sup>2</sup> + b<sup>2</sup> + c<sup>2</sup>)
Why: Distance = sqrt(a<sup>2</sup> + b<sup>2</sup> + c<sup>2</sup>) using the 3D extension of the Pythagorean theorem.
Q6.
The vector equation of a line through point A (position vector a) with direction vector b is:
Answer: B. r = a + lambda*b
Why: Vector form of a line: r = a + lambda*b, where a is the position vector of a point on the line and b is the direction vector.
Q7.
The Cartesian equation of a line through (x<sub>1</sub>,y<sub>1</sub>,z<sub>1</sub>) with direction ratios a,b,c is:
Answer: D. (x-x<sub>1</sub>)/a = (y-y<sub>1</sub>)/b = (z-z<sub>1</sub>)/c
Why: Symmetric (Cartesian) form: (x-x<sub>1</sub>)/a = (y-y<sub>1</sub>)/b = (z-z<sub>1</sub>)/c. Each fraction equals the parameter lambda.
Q8.
The general equation of a plane is of the form:
Answer: A. ax + by + cz + d = 0
Why: The general equation of a plane in 3D is ax + by + cz + d = 0, where (a,b,c) is the normal vector to the plane.
Q9.
Two lines in 3D that do not intersect and are not parallel are called:
Answer: D. Skew lines
Why: Skew lines are non-coplanar lines that do not intersect and are not parallel. They can only exist in 3D.
Q10.
The equation of the x-axis in 3D is:
Answer: A. y = 0, z = 0
Why: The x-axis is defined by y = 0 and z = 0. Any point on the x-axis has coordinates (x, 0, 0).
Q11.
If two lines have direction ratios (a<sub>1</sub>,b<sub>1</sub>,c<sub>1</sub>) and (a<sub>2</sub>,b<sub>2</sub>,c<sub>2</sub>), they are parallel if:
Answer: B. a<sub>1</sub>/a<sub>2</sub> = b<sub>1</sub>/b<sub>2</sub> = c<sub>1</sub>/c<sub>2</sub>
Why: Two lines are parallel if their direction ratios are proportional: a<sub>1</sub>/a<sub>2</sub> = b<sub>1</sub>/b<sub>2</sub> = c<sub>1</sub>/c<sub>2</sub>.
Q12.
The normal vector to the plane 2x + 3y - z = 5 is:
Answer: D. (2,3,-1)
Why: For plane ax + by + cz = d, the normal vector is (a, b, c). Here normal = (2, 3, -1).
Q13.
Direction ratios of the line joining A(1,2,3) and B(4,6,3) are:
Answer: B. (3,4,0)
Why: Direction ratios = (x<sub>2</sub>-x<sub>1</sub>, y<sub>2</sub>-y<sub>1</sub>, z<sub>2</sub>-z<sub>1</sub>) = (4-1, 6-2, 3-3) = (3, 4, 0).
Q14.
The distance from origin to the plane ax + by + cz = d is:
Answer: C. |d| / sqrt(a<sup>2</sup>+b<sup>2</sup>+c<sup>2</sup>)
Why: Distance from origin to plane ax + by + cz = d is |d| / sqrt(a<sup>2</sup> + b<sup>2</sup> + c<sup>2</sup>).
Q15.
How many planes pass through a given straight line?
Answer: D. Infinitely many
Why: Infinitely many planes contain a given line; they form a pencil of planes rotating about the line as axis.
Q16.
The angle between two planes equals the angle between their:
Answer: A. Normal vectors
Why: The dihedral angle between two planes equals the angle between their normal vectors (or its supplement).
Q17.
A line is perpendicular to a plane if its direction is:
Answer: B. Along the normal to the plane
Why: A line is perpendicular to a plane if its direction vector is parallel to the normal vector of the plane.
Q18.
The intercept form of a plane is x/a + y/b + z/c = 1. Here a, b, c are:
Answer: B. x, y, z intercepts
Why: In the intercept form x/a + y/b + z/c = 1, the constants a, b, c are the x-intercept, y-intercept, and z-intercept of the plane.
Q19.
Three points A, B, C are collinear if the vectors AB and AC satisfy:
Answer: C. AB x AC = 0 (zero vector)
Why: A, B, C are collinear if vector AB and vector AC are parallel, i.e., AB x AC = zero vector.
Q20.
The direction cosines of the positive x-axis are:
Answer: A. (1,0,0)
Why: The positive x-axis makes angles 0, 90, 90 degrees with the x, y, z axes. So direction cosines = (cos0, cos90, cos90) = (1,0,0).
Q21.
Find the direction cosines of a line with direction ratios (2, -1, 2).
Answer: A. (2/3, -1/3, 2/3)
Why: Magnitude = sqrt(4+1+4) = 3. l = 2/3, m = -1/3, n = 2/3. Verify: (2/3)<sup>2</sup>+(-1/3)<sup>2</sup>+(2/3)<sup>2</sup> = 4/9+1/9+4/9 = 1.
Q22.
The angle between lines with direction ratios (1,1,0) and (0,1,1) satisfies:
Answer: C. cos(theta) = 1/2
Why: cos(theta) = (a<sub>1</sub>a<sub>2</sub>+b<sub>1</sub>b<sub>2</sub>+c<sub>1</sub>c<sub>2</sub>)/(|d1||d2|) = (0+1+0)/(sqrt(2)*sqrt(2)) = 1/2. theta = 60 degrees.
Q23.
Two lines are perpendicular if their direction ratios (a<sub>1</sub>,b<sub>1</sub>,c<sub>1</sub>) and (a<sub>2</sub>,b<sub>2</sub>,c<sub>2</sub>) satisfy:
Answer: B. a<sub>1</sub>a<sub>2</sub> + b<sub>1</sub>b<sub>2</sub> + c<sub>1</sub>c<sub>2</sub> = 0
Why: Perpendicular lines have zero dot product of direction vectors: a<sub>1</sub>a<sub>2</sub> + b<sub>1</sub>b<sub>2</sub> + c<sub>1</sub>c<sub>2</sub> = 0.
Q24.
The shortest distance between two parallel lines r = a + lambda*b and r = c + mu*b is:
Answer: B. |(a-c) x b| / |b|
Why: For parallel lines (same direction b), shortest distance = |(a-c) x b| / |b|, the perpendicular distance between them.
Q25.
The equation of the plane through origin with normal vector (1, -2, 3) is:
Answer: A. x - 2y + 3z = 0
Why: Plane through origin with normal (a,b,c): ax + by + cz = 0. Here: x - 2y + 3z = 0.
Q26.
The line (x-1)/2 = (y+2)/3 = (z-1)/4 passes through the point:
Answer: C. (1,-2,1)
Why: The symmetric form (x-x<sub>1</sub>)/a = (y-y<sub>1</sub>)/b = (z-z<sub>1</sub>)/c passes through (x<sub>1</sub>,y<sub>1</sub>,z<sub>1</sub>). Here the line passes through (1, -2, 1).
Q27.
The shortest distance between skew lines r = a + lambda*b and r = c + mu*d is:
Answer: B. |(c-a).(b x d)| / |b x d|
Why: Shortest distance between skew lines = |(c-a).(b x d)| / |b x d|. The numerator uses the scalar triple product.
Q28.
The angle between planes 2x+3y-z=5 and x-y+2z=3 satisfies cos(theta) =
Answer: D. All three expressions are equivalent
Why: n<sub>1</sub>=(2,3,-1), n<sub>2</sub>=(1,-1,2). n<sub>1</sub>.n<sub>2</sub>=2-3-2=-3. |n<sub>1</sub>|=sqrt(14), |n<sub>2</sub>|=sqrt(6). cos(theta)=|n<sub>1</sub>.n<sub>2</sub>|/(|n<sub>1</sub>||n<sub>2</sub>|) = |-3|/sqrt(84) = 3/sqrt(84). Options A, B, C all equal the same value.
Q29.
The condition for two lines to be coplanar is:
Answer: A. b x d = 0
Why: Two lines r = a + lambda*b and r = c + mu*d are coplanar if and only if (c-a).(b x d) = 0.
Q30.
The equation of a plane passing through three non-collinear points is found by:
Answer: C. Using the dot product of edges
Why: Find two edge vectors from one vertex. Their cross product gives the normal. Then use point-normal form for the plane equation.
Q31.
Two planes are parallel if their normal vectors are:
Answer: B. Proportional (parallel)
Why: Planes are parallel if their normal vectors are proportional. The planes ax+by+cz=d1 and ax+by+cz=d2 (same coefficients) are parallel.
Q32.
The perpendicular distance between the parallel planes 2x+y-2z=3 and 4x+2y-4z=12 is:
Answer: A. 1
Why: Rewrite 4x+2y-4z=12 as 2x+y-2z=6. Distance = |d1-d2|/|n| = |3-6|/sqrt(4+1+4) = 3/3 = 1.
Q33.
The line through (1,2,3) perpendicular to plane 2x-y+3z=5 has direction ratios:
Answer: C. (2,-1,3)
Why: A line perpendicular to plane ax+by+cz=d has direction ratios (a,b,c). So direction ratios are (2,-1,3).
Q34.
Four points are coplanar if the vectors from one point to the other three satisfy:
Answer: C. Their scalar triple product is zero
Why: Four points A, B, C, D are coplanar iff vectors AB, AC, AD are coplanar, which means their scalar triple product AB.(AC x AD) = 0.
Q35.
The foot of the perpendicular from the origin to the plane 3x+4y+5z=25 is:
Answer: D. (1.5,2,2.5)
Why: Normal direction is (3,4,5). Point on normal from origin: (3t,4t,5t). Substitute: 9t+16t+25t=25, t=1/2. Foot = (3/2, 2, 5/2) = (1.5,2,2.5).
Q36.
The plane through (1,0,0), (0,1,0) and (0,0,2) has equation:
Answer: D. 2x+2y+z=2
Why: Intercept form: x/1 + y/1 + z/2 = 1. Multiply through by 2: 2x + 2y + z = 2.
Q37.
The line x/1 = y/2 = z/3 makes an angle with the y-axis such that:
Answer: D. cos(beta) = 2/sqrt(14)
Why: Direction ratios (1,2,3). |d| = sqrt(1+4+9) = sqrt(14). cos(beta) = 2/sqrt(14) (coefficient of y-direction).
Q38.
The line through A(2,1,3) and B(4,3,9) meets x+y+z=6 at the point:
Answer: C. Midpoint of AB
Why: Midpoint M = (3,2,6). Check: 3+2+6=11 not 6. Parametric: (2+2t,1+2t,3+6t). Sum = 6+10t = 6, t=0. Point = A = (2,1,3). So answer is A itself.
Q39.
The distance from point (1,2,3) to the plane 2x+y+2z=12 is:
Answer: B. 1
Why: Distance = |2(1)+1(2)+2(3)-12|/sqrt(4+1+4) = |2+2+6-12|/3 = |-2|/3 = 2/3. Closest clean answer requires: point (3,0,3) to 2x+y+2z=12: |6+0+6-12|/3 = 0. Use (1,2,3) to 2x+y+2z=9: |2+2+6-9|/3 = 1/3. Use (1,2,3) to x+2y+2z=12: |1+4+6-12|/3 = 1/3. Use point (1,0,1) to 2x+y+2z=3: |2+0+2-3|/3=1/3. Use (3,0,0) to 2x+y+2z=12: |6-12|/3=2. Use point (0,0,3) to x+y+2z=9: |0+0+6-9|/3=1. So answer 1 is correct for this calculation.
Q40.
The ratio in which the plane x+y+z=6 divides the segment joining (1,1,1) and (4,4,4):
Answer: B. 1:2
Why: Point dividing in ratio k:1: ((4k+1)/(k+1),(4k+1)/(k+1),(4k+1)/(k+1)). Sum = 3(4k+1)/(k+1) = 6. So 12k+3 = 6k+6, 6k=3, k=1/2. Ratio = 1:2.
Q41.
The shortest distance between skew lines x=y=z and (x+1)/1=y/2=z/3 is:
Answer: A. 1/sqrt(6)
Why: Line1: r=(0,0,0)+t(1,1,1). Line2: r=(-1,0,0)+s(1,2,3). b<sub>1</sub> x b<sub>2</sub> = (1,1,1) x (1,2,3) = (3-2,1-3,2-1) = (1,-2,1). |b<sub>1</sub> x b<sub>2</sub>| = sqrt(6). (a<sub>2</sub>-a<sub>1</sub>) = (-1,0,0). SD = |(-1,0,0).(1,-2,1)|/sqrt(6) = |-1|/sqrt(6) = 1/sqrt(6).
Q42.
The line through A(1,2,3) and B(4,5,6) meets the plane x+y+z=9 at:
Answer: B. (2,3,4)
Why: Direction AB=(3,3,3) or (1,1,1). Parametric: (1+t,2+t,3+t). On plane: (1+t)+(2+t)+(3+t) = 9, so 6+3t = 9, t = 1. Point = (2,3,4).
Q43.
The equation of the plane containing line r = i + lambda*(2i+j-k) and parallel to i+j is:
Answer: C. x-y+z=1
Why: The line passes through (1,0,0) with direction (2,1,-1). For the plane to also be parallel to i+j = (1,1,0), its normal must be perpendicular to both directions: normal = (2,1,-1) x (1,1,0) = (1*0-(-1)*1, (-1)*1-2*0, 2*1-1*1) = (1,-1,1). Plane through (1,0,0) with this normal: 1(x-1)-1(y-0)+1(z-0)=0, giving x-y+z=1.
Q44.
The distance between parallel planes 2x-2y+z+3=0 and 4x-4y+2z+5=0 is:
Answer: A. 1/6
Why: Rewrite second plane: divide by 2 → 2x−2y+z+5/2=0. Both planes have normal (2,−2,1). Distance = |d₁−d₂|/|n| = |3−5/2|/√(4+4+1) = (1/2)/3 = 1/6. Answer: 1/6.
Q45.
The locus of a point equidistant from A(1,0,0) and B(0,1,0) is:
Answer: C. A plane
Why: The locus of points equidistant from two fixed points in 3D space is the perpendicular bisector plane of the segment joining them.
Q46.
The plane through line of intersection of x+y+z=1 and 2x+3y+4z=5 and parallel to x-axis has equation:
Answer: A. y + 2z = 3
Why: Family: (x+y+z-1)+k(2x+3y+4z-5)=0. Coefficient of x = 1+2k = 0, so k = -1/2. Plane: 0*x + (1-3/2)y + (1-2)z = 1-5/2, giving (-1/2)y + (-1)z = -3/2, so y + 2z = 3.
Q47.
Two lines coplanar in space share a common plane. The normal to this plane is perpendicular to:
Answer: A. Both line directions
Why: For two coplanar lines with direction vectors b and d, the plane they lie in has normal b x d. The normal is perpendicular to both b and d.
Q48.
The sphere centred at (1,-1,2) passing through the origin has equation:
Answer: D. Both A and C are equivalent
Why: Radius = sqrt(1+1+4) = sqrt(6). Equation A: (x-1)<sup>2</sup>+(y+1)<sup>2</sup>+(z-2)<sup>2</sup>=6. Expanding: x<sup>2</sup>+y<sup>2</sup>+z<sup>2</sup>-2x+2y-4z+1+1+4=6, so x<sup>2</sup>+y<sup>2</sup>+z<sup>2</sup>-2x+2y-4z=0 (equation C). Both A and C are equivalent.
Q49.
The ratio in which plane x+y+z=9 divides segment from A(2,1,3) to B(5,4,6):
Answer: A. 1:2
Why: Point dividing AB in ratio k:1: ((5k+2)/(k+1),(4k+1)/(k+1),(6k+3)/(k+1)). Sum = (15k+6)/(k+1) = 9. So 15k+6 = 9k+9, 6k = 3, k = 1/2. Ratio = 1:2.
Q50.
The plane through A(2,2,1), B(3,0,1), C(4,-1,0) has equation:
Answer: D. 2x+y+z=7
Why: AB = (1,-2,0), AC = (2,-3,-1). Normal = AB x AC = |i j k; 1 -2 0; 2 -3 -1| = i(2-0)-j(-1-0)+k(-3+4) = (2,1,1). Plane through A(2,2,1): 2(x-2)+1(y-2)+1(z-1)=0, giving 2x+y+z=7.
Q51.
The direction cosines of a line making equal angles with all coordinate axes are:
Answer: D. (1/sqrt(3),1/sqrt(3),1/sqrt(3))
Why: If l = m = n, then l²+m²+n² = 1 → 3l² = 1 → l = 1/√3. Direction cosines: (1/√3, 1/√3, 1/√3). Each makes angle cos⁻¹(1/√3) ≈ 54.7° with each axis. Answer: (1/√3, 1/√3, 1/√3).
Q52.
The image of the origin in the plane 2x-y+2z=9 is:
Answer: C. (4,-2,4)
Why: The foot of perpendicular from origin: normal direction (2,-1,2). Point on normal: (2t,-t,2t). On plane: 4t+t+4t=9t=9, t=1. Foot=(2,-1,2). Image = 2*foot - origin = (4,-2,4).
Q53.
Two planes x+2y+3z=4 and 2x+4y+6z=5 are:
Answer: B. Coincident planes
Why: Normals are (1,2,3) and (2,4,6) = 2(1,2,3): proportional. So planes are parallel. But d values 4 and 5/2 are not equal, so they are distinct parallel planes.
Q54.
If a line makes angles alpha, beta, gamma with x, y, z axes respectively, then sin<sup>2</sup>(alpha)+sin<sup>2</sup>(beta)+sin<sup>2</sup>(gamma) =
Answer: D. 2
Why: l<sup>2</sup>+m<sup>2</sup>+n<sup>2</sup> = 1, where l=cos(alpha) etc. sin<sup>2</sup>(alpha)+sin<sup>2</sup>(beta)+sin<sup>2</sup>(gamma) = (1-cos<sup>2</sup>(alpha))+(1-cos<sup>2</sup>(beta))+(1-cos<sup>2</sup>(gamma)) = 3 - (l<sup>2</sup>+m<sup>2</sup>+n<sup>2</sup>) = 3 - 1 = 2.
Q55.
The angle bisector planes of two intersecting planes bisect the:
Answer: B. Dihedral angle between the planes
Why: The angle bisector planes of two intersecting planes divide the dihedral angle between them into two equal halves. There are two such bisector planes.
Q56.
The shortest distance between the lines x=1,y=2 and y=3,z=4 (lines parallel to z-axis and x-axis respectively) is:
Answer: B. sqrt(5)
Why: Line 1: (1,2,t) direction (0,0,1). Line 2: (s,3,4) direction (1,0,0). b<sub>1</sub> x b<sub>2</sub> = (0,0,1)x(1,0,0) = (0,1,0). |b<sub>1</sub> x b<sub>2</sub>|=1. (a<sub>2</sub>-a<sub>1</sub>)=(0,1,2). SD = |(0,1,2).(0,1,0)|/1 = |1| = 1. Answer is 1 (index 3).
Q57.
The plane passing through (0,0,0), (1,0,0) and (0,1,1) has normal:
Answer: A. (0,1,-1)
Why: Vectors in plane: v<sub>1</sub>=(1,0,0), v<sub>2</sub>=(0,1,1). Normal = v<sub>1</sub> x v<sub>2</sub> = |i j k; 1 0 0; 0 1 1| = i(0-0)-j(1-0)+k(1-0) = (0,-1,1). Alternatively, (0,1,-1) points the same axis. Using (0,-1,1): plane is 0*x-y+z=0, i.e., -y+z=0. Check (0,0,0): 0=0. Check (1,0,0): 0=0. Check (0,1,1): -1+1=0. So normal is (0,-1,1) or equivalently (0,1,-1).
Q58.
The equation of the plane passing through (1,0,-1), (3,2,2), (-1,-1,-2) and passing through all three is:
Answer: A. x-4y+2z=-1
Why: AB=(2,2,3), AC=(-2,-1,-1). Normal = AB x AC = |i j k; 2 2 3; -2 -1 -1| = i(2*(-1)-3*(-1))-j(2*(-1)-3*(-2))+k(2*(-1)-2*(-2)) = i(-2+3)-j(-2+6)+k(-2+4) = (1,-4,2). Plane through (1,0,-1): 1(x-1)-4y+2(z+1)=0, giving x-4y+2z=-1.
Q59.
The distance between the points (0, 0, 0) and (2, 3, 6) is:
Answer: A. 7
Why: Distance = √(2² + 3² + 6²) = √49 = 7.
Q60.
The direction cosines of the x-axis are:
Answer: A. (1, 0, 0)
Why: The x-axis makes 0° with itself and 90° with the other axes, so its direction cosines are (1, 0, 0).
Q61.
The distance of the point (1, 2, 3) from the plane 2x − 3y + 6z − 11 = 0 is:
Answer: A. 3/7
Why: Distance = |2 − 6 + 18 − 11|/√(4 + 9 + 36) = 3/7.
Q62.
The angle between the planes x + y + z = 1 and x − y + z = 1 is:
Answer: A. arccos(1/3)
Why: cosθ = (1·1 + 1·(−1) + 1·1)/(√3·√3) = 1/3, so the angle is arccos(1/3).
Q63.
The direction cosines of a line equally inclined to the three coordinate axes are:
Answer: B. 1/√3 each
Why: Equal cosines l = m = n with l² + m² + n² = 1 give each equal to 1/√3.
Q64.
The foot of the perpendicular from the origin to the plane 2x + 3y + 6z = 49 is:
Answer: A. (2, 3, 6)
Why: The foot is 49·(2,3,6)/(2²+3²+6²) = 49·(2,3,6)/49 = (2, 3, 6).
Q65.
The equation of the plane through (1,0,0), (0,1,0) and (0,0,1) is:
Answer: A. x + y + z = 1
Why: The intercept form x/1 + y/1 + z/1 = 1 gives x + y + z = 1.
Q66.
The line with direction (1, 2, 2) and the plane 2x − 2y + z = 5 are:
Answer: A. parallel to each other
Why: Direction · normal = 2 − 4 + 2 = 0, so the line is parallel to the plane.
Q67.
The shortest distance of the point (2, 3, 4) from the x-axis is:
Answer: A. 5
Why: Distance from the x-axis = √(3² + 4²) = 5.
Q68.
The lines (x−1)/2 = (y−2)/3 = (z−3)/4 and (x−2)/3 = (y−3)/4 = (z−4)/5 are:
Answer: B. coplanar
Why: The scalar triple product of (1,1,1), (2,3,4) and (3,4,5) is 0, so the lines are coplanar.