Worksheets Class 12 Mathematics Three Dimensional Geometry

Worksheets for Class 12

Students should refer to Worksheets Class 12 Mathematics Three Dimensional Geometry Chapter 11 provided below with important questions and answers. These important questions with solutions for Chapter 11 Three Dimensional Geometry have been prepared by expert teachers for Class 12 Mathematics based on the expected pattern of questions in the class 12 exams. We have provided Worksheets for Class 12 Mathematics for all chapters on our website. You should carefully learn all the important examinations questions provided below as they will help you to get better marks in your class tests and exams.

Three Dimensional Geometry Worksheets Class 12 Mathematics

Question. If centroid of tetrahedron OABC, where A, B, C are given by (a, 2, 3), (1, b, 2) and (2, 1, c) respectively be (1, 2, –1), then distance of P(a, b, c) from origin is equal to:
(a) √107
(b) √14
(c) √(107 /14)
(d) None of these

Answer

A

Question. Distance of the point (x1, y1, z1) from the line (x – x2)/l = (y – y2)/m = (z – z2)/n, where l, m and n are the direction cosines of line is:
(image 14)
Ans : A

Answer

A

Question. The angle between the lines x/1 = y/0 = z/-1 and x/3 = y/4 = z/5 is :
(a) cos-1(1/5)
(b) cos-1(1/3)
(c) cos-1(1/2)
(d) cos-1(1/4)

Answer

A

Question. A line which makes angle 60° with y-axis and z-axis, then the angle which it makes with x-axis is:
(a) 45°
(b) 60°
(c) 75°
(d) 30°

Answer

A

Question. The angle between the pair of lines with direction ratios (1, 1, 2) and (√3 − 1,−√3 − 1,4) is:
(a) 30°
(b) 45°
(c) 60°
(d) 90°

Answer

C

Question. If direction ratios of two lines are 5, −12,13 and −3, 4, 5 then the angle between them is:
(a) cos-1(1/65)
(b) cos-1(2/65)
(c) cos-1(3/65)
(d) π/2

Answer

A

Question. A line passes through the points (6, –7, –1) and (2,–3, 1). The direction cosines of line, so directed that the angle made by it with the positive direction of x-axis is acute, are:
(a) 2/3, -2/3, -1/3
(b) -2/3, 2/3, 1/3
(c) 2/3, 2/3, 1/3
(d) 2/3, -2/3, 1/3

Answer

A

Question. If the direction cosines of a line are (1/c, 1/c, 1/c), then:
(a) c > 0
(b) c = ± √3
(c) 0 < c < 1
(d) c > 2

Answer

B

Question. If r is a vector of magnitude 21 and has (d)r.’s 2, –3, 6. Then r is equal to:
(a) 6i − 9j+18k
(b) 6i + 9j+18k
(c) 6i − 9j−18k
(d) 6i + 9j−18k

Answer

A

Question. The projection of a line on co-ordinate axes are 2, 3, 6. Then the length of the line is:
(a) 7
(b) 5
(c) 1
(d)11

Answer

B

Question. If the direction ratio of two lines are given by 3lm − 4ln + mn = 0 and l + 2m + 3n = 0 , then the angle between the lines is:
(a) π/2
(b) π/3
(c) π/4
(d) π/6

Answer

A

Question. If a line makes angles α, β, γ, δ with four diagonals of a cube, then the value of sin2α + sin2β + sin2γ + sin2δ is:
(a) 4/3
(b) 1
(c) 8/3
(d) 7/3

Answer

C

Question. The (d)(c)’s of the line 6x − 2 = 3y + 1= 2z − 2 are:
(a) 1/√3, 1/√3, 1/√3
(b) 1/√14, 2/√14, 3/√14
(c) 1, 2, 3
(d) None of these

Answer

B

Question. The angle between the lines whose direction cosines are proportional to (1, 2, 1) and (2, –3, 6) is:
(a) cos-1(2/7√6)
(b) cos-1(1/7√6)
(c) cos-1(3/7√6)
(d) cos-1(5/7√6)

Answer

A

Question. The distance of the point (4, 3, 5) from the y-axis is:
(a) √34
(b) √5
(c) √41
(d) √15

Answer

C

Question. If the x-co-ordinate of a point P on the join of Q (2, 2, 1) and R (5, 1, –2) is 4, then its z-co-ordinate is:
(a) 2
(b) 1
(c) –1
(d) –2

Answer

C

Question. The vector equation of line through the point A(3, 4, –7) and B(1, –1, 6) is
(a) r = (3i + 4j− 7k) +λ (i − j+ 6k)
(b) r = (i − j+ 6k) +λ (3i + 4j− 7k)
(c) r = (3i + 4j− 7k) +λ (−2i −5j+13k)
(d) r = (i − j+ 6k) +λ (4i + 3j−k)

Answer

C

Question. If the lines (x-1)/-3 = (y-2)/2k = (z-3)/2 and (x-1)/3k = (y-5)/1 = (z-6)/5 are at right angles, then k = ?
(a) –10
(b) 10/7
(c) –10/7
(d) –7/10

Answer

A

Question. The angle between the lines whose direction cosines satisfy the equations l + m + n = 0, l2 + m2 − n2 = 0, is given by:
(a) 2π/3
(b) π/6
(c) 5π/6
(d) π/3

Answer

D

Question. The angle between two lines (x+1)/2 = (y+3)/2 = (z – 4)/-1 and (x – 4)/1 = (y+4)/2 = (z+1)/2 is :
(a) cos-1(1/9)
(b) cos-1(2/9)
(c) cos-1(3/9)
(d) cos-1(4/9)

Answer

D

Question. The equation of a line 4x – 4y – z + 11 = 0 = x + 2y – z – 1 can be put as: (Image 72)   

Answer

A,B

Question. The equation of the line passing through (1, 2, 3) and parallel to the planes x − y + 2z = 5 and 3x + y + z = 6 , is: (Image 60)     

Answer

A

Question. The line x+3/3 = y–2/–2 = z+1/1 and the plane 4 x + 5y + 3z − 5 = 0 intersect at a point:
(a) (3, 1, –2)
(b) (3, – 2, 1)
(c) (2, –1, 3)
(d) (–1, –2, –3)

Answer

D

Question. If a plane passes through the point (1,1,1) and is perpendicular to the line x–1/3 = y–1/0 = z-1/4 then its perpendicular distance from the origin is:
(a) 3/4
(b) 4/3
(c) 7/3
(d) 1

Answer

C

Question. A plane which passes through the point (3, 2, 0) and the line x–3/1 = y–6/5 = z-4/4 is:
(a) x − y + z =1
(b) x + y + z = 5
(c) x + 2y − z = 0
(d) 2x − y + z = 5

Answer

A

Question. The equation of a sphere which passes through (1,0,0), (0,1,0) and (0,0,1) and whose centre lies on the curve 4xy= 1is:       
(a) x2 + y2 + z2 – x – y – z = 0
(b) x2 + y2 + z2 + x+y + z –2 = 0
(c) x2 + y2 + z2 + x + y + z = 0
(d) x2 + y2 + z2 – x – y – z–2 = 0

Answer

A,B

Question. The angle between the line x/2 = y/3 = z/4 and the plane 3x + 2y − 3z = 4 is
(a) 45°
(b) 0°
(c) cos–1(24/√29√22)
(d) 90°

Answer

B

Question. The plane ax + by + cz =1meets the co-ordinate axes in A, B and (c) The centroid of the triangle is:
(a) (3a,3b,3c)
(b) (a/3 , b/3 , c/3)
(c) (3/a,3/b,3/c)
(d) (1/3a, 1/3b, 1/3c)

Answer

D

Question. If P be the point (2, 6, 3), then the equation of the plane through P at right angle to OP, O being the origin, is:
(a) 2x + 6y + 3z = 7
(b) 2x − 6y + 3z = 7
(c) 2x + 6y − 3z = 49
(d) 2x + 6y + 3z = 49

Answer

D

Question. If the straight lines x–1/2 = y+1/K = z/2 and x+1/5 = y+1/2 = z/k are coplanar, then the plane(s) containing these two lines is/are:       
(a) y + 2 z = –1
(b) y + z = –1
(c) y – z = –1
(d) y – 2z = –1

Answer

B,C

Question. If y(x) satisfies the differential equation y′− y tan x=2xsec x and y(0), then: (Image 66)       

Answer

B,D

Question. A line l passing through the origin is perpendicular to the lines l1 : (3+ t)iˆ + (−1+ 2t) ˆj + (4 + 2t)kˆ,−∞ < t < ∞ 
l2 : (3+ 2s)iˆ + (3+ 2s) ˆj + (2 + s)kˆ,−∞ < s < ∞
Then, the coordinate(s) of the point(s) on l2 at a distance of √17 from the point of intersection of l and l1 is (are)     
(a) (7/3 , 7/3 , 7/3)
(b) (–1,–1,0)
(c) (1,1,1)
(d)  (7/9 , 7/9 , 8/9)

Answer

B,C

Question. The lines x–2/1 = y–3/1 = z–4/–k and x–1/k = y–4/2 = z–5/1 are coplanar if       
(a) k = 0
(b) k = – 1
(c) k = – 3
(d) k = 3

Answer

A,C

Assertion and Reason
Note: Read the Assertion (A) and Reason (R) carefully to mark the correct option out of the options given below:
(a) If both assertion and reason are true and the reason is the correct explanation of the assertion.
(b) If both assertion and reason are true but reason is not the correct explanation of the assertion.
(c) If assertion is true but reason is false.
(d) If the assertion and reason both are false.
e. If assertion is false but reason is true.

Question. Assertion: The direction cosines of the line 6x – 2 = 3y +1 = 2z – 2     
are same as the direction cosines of the normal to the plane 2x + 3y + z = 14
Reason: The direction angles of a normal to the plane are π/4 , π/4 , π/2 and the length of the perpendicular form the origin on the plane is √2, equation of the plane is x + y = 2.

Answer

D

Question. Assertion: If the distance of the point P(1, –2, 1) from the plane x + 2y – 2z = α
where α > 0, is 5, then the foot of the perpendicular from P to the plane is (8/3, 4/3, –7/3) 
Reason: A line through P(1, –2, 1)and perpendicular to the plane x + 2y – 2z = α intersects the plane at Q. If PQ =5 then α= 10.       

Answer

A

Question. Assertion: The point A(1,0,7) is the mirror image of the point (1, 6, 3) in the line x/1 = y–1/2 = z–2/3.     
Reason: The line x/1 = y–1/2 = z–2/3 bisects the line segment joining A(1,0,7)and B(1,6,3).

Answer

B

Question. Assertion: The point A(3,1,6) is the mirror image of the point B(1,3,4) in the plane x – y + z = 5. 
Reason: The plane x – y + z = 5 bisects the line segment joining A(3,1,6) and B(1,3,4).

Answer

B

Question. Assertion: The distance between the line r=2iˆ+2ˆj+3kˆ+λ (iˆ−ˆj + 4kˆ) and the plane r.(iˆ + 5 ˆj + kˆ) = 5 is 10/3√3.         
Reason: If a line is parallel to a plane, then the distance between the line and the plane is equal to the length of the perpendicular form any point on the line to the plane.

Answer

A

Question. (Image 79)     
Assertion: L1 and L2 are coplanar and the equation of the plane containing them is 5x + 2y − 3z −8 = 0
Reason: L1 and L2 intersect at a point.

Answer

B

Question. Vertices of a triangle ABC are A(1,1,0), B(1,0,1) and C(0,1,1)       
Assertion: The radius of the circum circle of the triangle ABC is √(2/3).
Reason: The centre of the circum circle of the triangle ABC lies on the plane x + y + z – 2 = 0

Answer

B

Question. Consider the line L : x/1 = y/2 = z/3 and the plane π : x + y + z = 0 
Assertion: If P is a point on L at a distance √14 from the origin O and N is the foot of the perpendicular from P to the plane π, then ON = √2
Reason: If R is a point on L such that the perpendicular distance of R from the plane π is √3 then the coordinates of R are (1, 2, 3).         

Answer

C

Question. Consider the plane 3x – 6y – 2z = 15 and 2x + y – 2z = 5.
Assertion: The parametric equations of the line of intersection of the given planes are x = 3 + 14t, y =1 + 2t, z15t; t being the parameter       
Reason: the vector 14iˆ + 2 ˆj +15kˆ is parallel to the line of intersection of the given planes.

Answer

D

Question. (Image 81)       
Assertion: The unit vector perpendicular to both L1 and L2 is –i–7j+5k/5√3
Reason: The distance of the point (1, 1, 1) from the plane passing through the point (–1, –2, –1) and whose normal is perpendicular to both the lines L1 and L2 is 23/5√3.

Answer

C

Paragraph –I

Consider the lines (Image 560)

Question. The unit vector perpendicular to both L1 and L2 is: (Image 85)         

Answer

B

Question. The shortest distance between L1 and L2 is:         
(a) 0 unit
(b) 17 /√3 unit
(c) 41/ 5√3 unit
(d) 17 / 5√3 unit

Answer

D

Question. The distance of the point (1, 1, 1) from the plane passing through the point (–1, –2, –1) and whose normal is perpendicular to both the lines L1 and L2 is:       
(a) 2 / √75 unit
(b) 7 / √75 unit
(c) 13/ √75 unit
(d) 23/ √75 unit

Answer

C

Paragraph –II

Let two planes P1 :2x – y + z = 2 and P3 : x + 2y – z = 3

Question. The equation of the plane through the intersection of P1 and P2 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

Answer

B

Question. Equation of the plane which passes through the point (–1,3,2) and is perpendicular to each the planes P1 and P2 is:         
(a) x + 3y – 5z + 2 = 0
(b) x + 3y + 5z – 18 = 0
(c) x – 3y – 5z + 20 = 0
(d) x – 3y + 5z = 0

Answer

C

Question. The equation of the acute angle bisector of planes P1 and P2 is:       
(a) x − 3y + 2z +1 = 0
(b) 3x + y − 5 = 0
(c) x + 3y − 2z +1 = 0
(d) 3x + z + 7 = 0

Answer

A

Question. The equation of the bisector of angle of the planes P1 and P2 which not containing origin is: 
(a) x – 3y + 2z + 1 = 0
(b) x + 3y = 5
(c) x + 3y + 2z + 2 = 0
(d) 3x + y = 5

Answer

D

Question. The image of plane P1 in the plane mirror P2 is:       
(a) x + 7y – 4x + 5 = 0
(b) 3x + 4y – 5z + 9 = 0
(c) 7x – y + 4z – 9 = 0
(d) None of the above

Answer

C

Match the Column

Question. Consider the following linear equations ax + by + cz = 0, bx + cy + az = 0, cx + ay + bz = 0 (Image 93)     
(a) A→3, B→1, C→4, D→2
(b) A→3, B→2, C→1, D→4
(c) A→1, B→3, C→2, D→4
(d) A→4, B-1, C→3, D→2

Answer

B

Question. Consider the lines (Image 94) and the planes P1 : 7x + y + 2z = 3, P : 3x + 5y − 6z = 4. Let ax + by + cz = d the equation of the plane passing through the point of intersection of lines L1 and L2 and perpendicular to planes P1 and P2. Match Column I with Column II and select the correct answer using the code given below the lists: (Image 94)       
(a) A→3, B→2, C→4, D→2
(b) A→2, B→4, C→3, D→1
(c) A→1, B→3, C→2, D→4
(d) A→4, B→1, C→3, D→2

Answer

A

Question. Match the statement of Column with those in Column II: (Image 95)         
(a) A→3,5 B→2,4 C→1,4
(b) A→3,4 B→1,4 C→2,4
(c) A→3,2 B→2,3 C→1,4
(d) A→3,5 B→4,3 C→1,2

Answer

A

Question. A variable plane is at a constant distance p form the origin and meets the axes in A, B and (c) If the locus of the centroid of the tetrahedron OABC is x–2 + y–2 + z–2 =λp–2 then the value of 160λ must be       

Answer

2560

Question. The lines x+4/3 = y+6/5 = z–1/–2 and 3x – 2y + z + 5 = 0 = 2x + 3y + 4z–k are coplanar for k is equal to       

Answer

2

Question. The shortest distance between the z-axis and the lines x + y + 2z − 3 = 0, 2x + 3y + 4z − 4 = 0must be       

Answer

2

Question. If the volume of tetrahedron formed by planes whose equations are y + z = 0, z + x = 0, x + y = 0 and x + y + z = 1is λ cubic unit then the value of 729λ must be 100. If the angle of intersection of the sphere x + y + z − 2x −4 y − 6z +10 = 0 with the sphere, the extremities of whose diameter are (1, 2, –3) and (5,0,1) is cos–1(λ), then the value of 9999|λ| must be       

Answer

486

Question. If the angle of intersection of the sphere x2 + y2 + z2 − 2x −4 y − 6z +10 = 0 with the sphere, the extremities of whose diameter are (1, 2, –3) and (5,0,1) is cos–1(λ), then the value of 9999|λ| must be     

Answer

6666