Students and Teachers Forum

1. Which transformation does the matrix   0 -1-1   0 represent? Using the matrix, find the image of the point A(6, -2).
 

Here, 0-1-10 represents the reflection in y = -x.  Now,      Object in matrix form = 6-2 and M = 0-1-10 We have,  Image =  M × Object                  = .....

2. What transformation does the matrix 0-1-10 denote? Using the given matrix, find the image of the point A(5, -7).
 

Here, Given transformation matrix = 0-1-10 Let P(x, y) be an object.  Object in matrix form xy      = 0-1-10 × xy    = 0 × x-1 ×y-1 × x+0 × y    = -y-x. Thus, (x, y) .....

3. Find the 2 × 2 matrix which the unit square 01100011 transform to a parallelogram 03410132..
 

Here, given unit square = Object = 01100011 and  Parallelogram = Image = 03410132 Let the required 2 × 2 transformation matrix be M =  abcd. We have,       Transformation matrix × object = Image  So, .....

4. Find the transformation matrix which represents the reflection on the line y = -x. 
 

Let A(x, y) be a point and A'(-y, -x) be the image after reflection on the line y = -x.  So, Object → Image  (x, y) → (y, -x)  or, - y = 0 × x + (-1) × y  or, -x = (-1) × x + 0 × .....

5. Which transformation does the matrix -1001     represent? Find it. 

Here, Given transformation matrix = M = -1001     Let A(x, y) be aobject. We know,         Image = M × object                     = -1001 xy   .....

6. Define the point of inversion.

In a circle with centre O & radius r, if P be the point other then the inverse of P w.r.t. the circle is a point P' on line OP such that OP x OP' = r2, then the image P' is called the inversion pont of .....

7. K(2, 5), L(-1, 3), and M (4, 1) are the vertices of a triangle KLM. Find the coordinates of the images of triangle KLM under the rotation of positive 90°about origin followed by enlargement E [(0, 0), 2]. Present the object and images in the same graph paper.

Here,         Vertices of ∆ KLM are; K(2, 5),  L(-1, 3) and  M (4, 1). We know that,  Object                                 .....

8. State the single transformation equivalent to the combination reflections on the X- axis and Y-axis respectively. Using this single transformation find the coordinates of the vertices of the image of ∆ PQR having vertices P(4,3), Q(1,1), and R(5,-1,). Also, draw the object and image on the same graph.

 Here,          Vertices of triangle PQR are P(4, 3),  Q(1, 1),  and  R(5, -1). We have, Object                               .....

9. The point P(2, -3) is reflected on the X-axis. The image P' thus obtained is rotation through +90° about origin. Find the coordinates of the final image P'' and P'.

Here, We have formula, Object (x, y) Reflection x-axis  -   ———————— .....

10. L (2, 2) O (6, 2) V (7, 4) and E (3, 4) are the vertices of a parallelogram LOVE. Find the co- ordinates of the vertices of the images of parallelogram LOVE under the rotation of positive 90° about origin followed by enlargement E [(0, 0), 3]. Represent the object and images on the same graph paper.

Here,  Given vertices are. L (2, 2), O (6, 2), V (7, 4), E (3, 4) Now, Object                                               .....

11. State the single transformation equivalent to the combination of reflections on the X- axis and Y-axis respectively. Using this single transformation find the coordinates of the vertices of the image of ∆ABC having vertices A(2,3), B(3,4), C(1,-2,) . Also, draw the object and image on the same graph.

 Here,           Vertices of a triangle are A (2,3),  B (3,4),  C (1,-2,). We know that, Object                                 .....

12. C(2, 5), A(-1, 3) and T(4, 1) are  the vertices of the  triangle CAT. Find the coordinates of the images of triangle CAT under the rotation of positive 90° about origin followed by enlargement E [(0,0),  2]. Represent the object and images in the same graph paper.

Here, vertices of ∆ CAT are;           C(2,5) , A(-1,3) and T(4,1). We know that, Object                                     .....

13. Triangle PQR having vertices P(2, 1), Q(5, 3) and R(7, -1) is reflected on the y-axis, the image so formed is enlarged by E[(0, 0),2]. Write the co-ordinates of the vertices of the images thus obtained and present the ∆PQR and its images in the same graph paper. 

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14. P(1, 2), Q(2, 1) and R(4, 3) are the vertices of the triangle PQR. Find the coordinates of the images of the triangle PQR under the reflection on the line y = x followed by the enlargement E [(0, 0), 2]. Present the ∆PQR and its images in the same graph paper.

Here, Vertices of ∆PQR; P(1, 2), Q(2, 1) and R(4, 3)     We know that, Object                                             .....

15. Translate the ∆PQR vertices are P (2, 4), Q(-1, 2) and R (5, -1) by translation vector T = - 23 and then reflect the image so obtained in the line x + y = 0. Write the coordinates of the vertices of images and draw the object and images on the same graph paper.

Here,         Vertices of  ∆PQR are; P (2, 4), Q(-1, 2) and R (5, -1). We know that, Object                                     .....

16. The vertices of ∆ABC are A (2, 0), B(3, 1) and C (1, 1). ∆ABC is translated by the vector 2-3 . The image so formed is enlarged by E [(0, 0), 2]. Write the coordinates of the vertices of images thus obtained and represent the object and the images on the same graph paper. 

Here, the vertices of ∆ABC are;  A (2, 0), B(3, 1)  and C (1, 1). We know that, Object                                           .....

17. A  ∆ABC with vertices A(1, 2) B(4, -1) and C(2, 5) is reflected successively in the line y = x and the X- axis. Find the stating
co- ordinates and graphically represent the image under these transformations. State also the single transformation given by the combinations of these transformations.

Here,          Vertices of triangle ABC  are A(1, 2)  B(4, -1)  and  C(2, 5). We know that, Object                               .....

18. In the given figure, ∡RPQ = 90° and S is the mid-point of RQ. Prove by vector method that: PS = QS = RS.

Let, ABC be a right angled triangle whose right angle is at A.  Let, P be the middle point of the hypotenuse BC. Let, A be the origin .....

19.  If a→ = 4, b→ = 5 and a→.  b→ = 10 then find the angle between a→ and b→.  

Here,  a→ = 4,  b→ = 5  and a→. b→ = 10 Let θ be angle between a→ and b→ then,  cos θ = a→. b→ a→b→   = 104 ×5   = 12 or, cos θ  = .....

20. If 3i → + 6j→ and 5i → + 2j → are the position vectors of the points P and Q respectively. Find the position vectors of the point M which divides the line PQ internally in the ratio 3:2. 

Here,  OP → = 3i → + 6j→ and OQ→ = 5i → + 2j→ Ratio = 3 : 2 = m : n. We have,  OM → = mb →+ na → m+n         = 3OQ →+ 2OP .....

21. In the given figure, PQ = PR and QS = RS, then prove by vector method: PS ⊥ QR. 

Let PQR be an isosceles triangle with PQ = QR and PS is median of ∆PQR. PS→ = 12( PQ→ + PR→) [Midpoint theorem] QR→ = (PR→ ) - PQ→) [Triangle law of vector subtraction] PS→. QR→ = 12( PQ→ + PR→)) .....

22. If a→ = 4, b→ = 5 and a→. b  → = 10 then find the angle between a→ and b→.

Here, a→ = 4, b→ = 5. and a→. b→ = 10 Let θ be angle between a→ and b→ then,      cos θ   =  a→. b→ a→b→             .....

23. If p→. q→ = 18√3, p→ = 6 and q→ = 6, find the angle between p→ and q→ .

Here, We have given, p→ . q→ = 18√3 p→ = 6 and q→ = 6, Let θ be angle between p→ and q→ then, cos θ = p→. q→ p→q→ = 1836 ×6 = 3 2 or, cos θ = cos30° or, θ .....

24. If a→ = 4i→ + 2j→ and b→ = -i→ + 2j→, then find the angle between a→ and b→.

Here, a→ = 4i→ + 2j→ and b→ = -i→ + 2j→, So, a→. b→ = (4i→ + 2j→).( -i→ + 2j→)             = -4 + 4 ∴ a→. b→ = 0 If a→. b→ = .....

25. The position vectors of A and B are 9i→ + 7j→ and i→ - 3j→ respectively. If M is the mid-point of AB, find the position vector of M. 

Here,         let O be the origin then,        OA         → = 9i→ + 7j→ And         OB      → = i→ - .....

26. If the position vectors of M and N are 9i→ + 3j→ and 3i→ + 5j→ respectively then find the position vector of a point P such that MP→ = PN→.

Here,  Let O be the centre OM→ = 9i→ + 3j→ And ON→ = 3i→ + 5j→ We have given,         MP→ = PN→ or, OP→ - OM→ = ON→ - OP→ or, 2OP→ = OM→ + .....

27. The position vector of A and B are 2i→ + 7j→ and 7i→ - 3j→. Find the position vector of a point which divides AB internally in the ratio of 2:3. 

Here,     let O be the origin then,      OA→ = 2i→ + 7j→ and OB→ = 7i→ + j→      And the given ratio = m:n                  .....

28. If the position of the midpoint of the line segment EF is 4i→ + 7j→ and the position vector of a point E is -i→ - 4j→, find the position vector of the point F. 

Here, let O be the origin and P be the mid-point of EF. So,  OP→ = 4i→ + 7j→ OE→ = -3i→ - 4j→ Since P is the mid-point of EF.                 So, OP→ = OE →+ .....

29. If 10i→ - 7j→ and i→ + 10j→ are perpendicular to each other find the value of a. 

Here,   Given perpendicular vectors are:     10i→ - 7j→ &  ai→ + 10j→ . Using the condition of perpendicularity     (10i→ - 7j→). (ai→ + 10j→) = .....

30. Prove by vector method that the line joining the mid-point of the base and the vertex of an isosceles triangle is perpendicular to the base.

Let PQR be an isosceles triangle with PQ = PR and PS is median of ∆PQR. (i) PS→ = 12(PQ→ + PR→)[Mid-point theorem] (ii)QR→ = (PR→ - PQ→)[Triangle law of vector subtraction] (iii)

31. If a→ = 5√3, b→ = 6, and θ = 30°, find the value of a→ . b→.

Here, a→ = 5√3, b→ = 6, and θ = 30°. We know that, cos θ = a→. b→ 53 ×6 or, 3 2 = a→. b→ 53 ×6 or, a→. b→ = 45 Thus, the value of a→. b→ is .....

32. If the position vectors of the vertices A, B and C of ∆ABC are (3i→ + 5j→), (5i→ - j→) and (i→ + 8j→) respectively, find the position vectors of the centroid G of the triangle. 

Let O be the origin then, OA→ = 3i→ + 5j→ OB→ = 5i→ - j→ OC→ = i→ + 8j→ . Let G be the centroid then,  OG→ = 13(OA→ + OB→ + OC→)        = 13(3i→ + .....

33. If p→ + q→ + r→ = 0, p→ = 6, q→ = 7 and r→ = √127, find the angle between p→ and q→.

Here, p→ + q→ + r→ = 0,.................(i) p→ = 6, q→ = 7 and r→ = √127 Now,       p      → + q→ = -r→                [ .....

34. In the figure, if PA → = 14PQ →,
Prove that a → = 14(3p→ + q→): 

Here,  PA → = 14PQ → (By the triangle law of vector subtraction.) or, OA → - OP → = 14(OQ → - OP →) or, a → - p → = 14(q → - p →) or, 4a → - 4p → = q → - p → or, 4a → = .....

35. If OA→ = a→ , OB→ = b→ and the point M divides the line segment AB internally in the ratio of m1:m2, then prove that:
       OM       →  =  m1b→+ m2a→ m1+ m2

Let O be the origin OA→ = a→ , OB→ = b→ and M be a point on AB which divides AB in the ratio of m1:m2. From the figure, AMMB = m1 m2 or, AM→ MB→ = m1 m2 or, m2(OM→ - OA→ ) = m1(OB→ - OM→ ) or, .....

36. Prove vectorically that the diagonals of a rectangle are equal.

Here, Let ABCD be a rectangle and  AC and BD are its diagonals. (i) AC2 = (AC→)2 = (AB→ + BC→)2 = (DC→ + BC→)2 [∴ AB = DC]       = DC→2 + 2DC→. BC→ + BC→2    .....

37. If p→  = 2i - 2j & q→  = 4j , what is the value of p→ .q→ ?

Here,  p→  = 2i - 2j & q→  = 4j = 0i + 4j , Now,  p→ .q→ = 2⨯0 + (-2)(4)                       = - .....

38. If a→  = (3, 0) & b→  = (4, -3), what is the value of a→ .b→ ?

Here,  a→  = (3, 0) & b→  = (4, -3), Now,  a→ .b→ = (3, 0).(4, -3)                       = 3⨯4 + 0(-3)         .....

39. If j→  is unit vector along the y-axis, what is the value of j2→ ?

We know,  j→  = (0, 1). So, j2 = → j→ .j→  = (0, 1).(0, 1) = 0 + 1 = .....

40. If i→  be the unit vector along the x-axis, what is the value of i→ 2 ?

We know,  a→  = (1, 0). So, i2 = → i→ .i→  = (1, 0).(1, 0) = 1 + 0 = .....

41. If a→  = (2, 0) & b→  = (0, 4), what is the value of a→ .b→ ?

Here,  a→ . b→ = (2, 0).(0, 4)                = 2 x 0 + 0 x 4              = 0 + 0              = .....

42. If the position vectors of two points A and B are 32 and –52 respectively then find position vector of the mid-point C of AB. 

Let O be the origin then, OA→ = 32 and OB→ = -54 Let C be the mid-point of AB.  Then,  OC→ = 12(OA→ + OB→)       = 1232+ -54       = 123-52+4       = 12-26   .....

43. If a→.b→=0, what is the angles between vectors a→&b→?

If a→.b→=0, then the angle between a→&b→ is .....

44. If a→.b→= |a|→|b|→, what is the angle between a→&b→?

The angle between a→&b →is 0o or .....

45. In the given figure, PQ = QR and RS = SP, then prove by vector method: PR ⊥ QS.

Here, Given PQ = QR = RS = SP To prove: PR ⊥ QS Proof: 1.  PR→ = PS→ + SR→[Triangle law of vector addition]    2. QS       → = QP→ + PS→[Triangle law of vector .....

46. If a→ + 2b→ and 5a→ - 4b→ are perpendicular to each other and a→ and b→ are unit vectors, find the angle between a→ and b→.

Here, a→ and b→ are unit vectors, so a→ = and b→ = 1 Since, (a→ + 2b→) and (5a→ - 4b→) are perpendicular to each other. So, (a→ + 2b→). (5a→ – 4b→) = 0 or, 5a2 - 4a→. b→ + .....

47. If a→ = -5i→ + 3j→ and b→ = pi→ + (p + 2) j→ are perpendicular to each other, find the value of p.

Here, a→ = -5i→ + 3j→ and b→ = pi→ + (p + 2) j→ Since, a→ ⊥ b→, So a→ . b→ = 0 i.e. (-5i→ + 3j→) . [pi→ + (p + 2) j→] = 0 or, - 5 × p + 3 (p + 2 ) = 0 or, -5p + 3p + .....

48. If b→ = 6, a→. b→ = 12 and angle between a→ and b→ (θ ) = 60°, find the value of a→.

Here, b→ = 6, a→. b→ = 12 and θ = 60°, We know that, cos θ = a→. b→ a→b→ cos60° = 12a→ ×6 12 = 2a→ Thus, the value of a→ is .....

49. If the position vectors of two points A and B are 34 and 45 respectively then find the position vector of the mid-point K of AB.

Here,  Let O be the origin then,  OA→ = 34 And OB→ = 45 We know that,  OK→ = OA→+ OB→2       = 1234+ 45       = 123+44+5       = 1279       = .....

50. In the given parallelogram SR→ = 4i→ - 2j→ and PR→ = 6i→ + 5j→, find QR→.

Here,         SR          → = 4i→ - 2j→ and PR→ = 6i→ + 5j→ . We know that,  QR→ = PS→ = PR→ + RS→       = 6i→ + .....