Welcome To Basics In Maths

Addition of Vectors (Qns.& Ans) V.S.A.Q.’S

1612493167414

These solutions designed by the ‘Basics in Maths’ team. These notes to do help the intermediate First-year Maths students.

Inter Maths – 1A two marks questions and solutions are very useful in IPE examinations.


Addition of Vectors

QUESTION 1

Find the unit vector in the direction of Addition of Vectors 17

Sol: Given vector isAddition of Vectors 17

 The unit vector in the direction of a vector Addition of Vectors 9 is given byAddition of Vectors 15 

Addition of Vectors 16

QUESTION 2

Find a vector in the direction of a where Addition of Vectors 19that has a magnitude of 7 units.

Sol: Given vector is Addition of Vectors 19

 The unit vector in the direction of a vector Addition of Vectors 9 is Addition of Vectors 20

 The vector having the magnitude 7 and in the direction of  is

  Addition of Vectors 21

QUESTION 3

Find the unit vector in the direction of the sum of the vectors, a = 2i + 2j – 5k and b = 2i + j + 3k

Sol Given vectors are a = 2i + 2j – 5k and b = 2i + j + 3k

a + b = (2i + 2j – 5k) + (2i + j + 3k) = 4i + 3j – 2k

Addition of Vectors 22

Addition of Vectors 23      

 QUESTION 4

Write the direction cosines of the vector Addition of Vectors 24

Sol: Given vector is Addition of Vectors 24

Addition of Vectors 25

    ∴ Direction cosines are Addition of Vectors 27

QUESTION 5

Show that the points whose position vectors are – 2a + 3b + 5c, a + 2b + 3c, 7 ac are collinear when a, b, c are non-collinear vectors

Sol: Let OA = – 2a + 3b + 5c, OB = a + 2b + 3c, OC = 7 acA

B = OB – OA = a + 2b + 3c (– 2a + 3b + 5c)

         AB = 3a b – 2c

 AC = OC – OA = 7 ac(– 2a + 3b + 5c)

 AC = 9a – 3b – 6c = 3(3a b – 2c)

          AC = 3 AB

   A, B and C are collinear

 QUESTION 6

ABCD is a parallelogram if L and M are middle points of BC and CD. Then find (i) AL and AM in terms of AB and AD (ii) 𝛌, if AM = 𝛌 AD – LM

Sol: Given, ABCD is a parallelogram and L and M are middle points of BC and CD

Addition of Vectors 1

(i) Take A as the origin

 M is the midpoint of CD

  AM =Addition of Vectors 2

          = AD + ½ AB (∵ AB = DC)

 L is the midpoint of BC

  AL =Addition of Vectors 3

      = AB + ½ AD ((∵ BC = AD)

(ii) AM = 𝛌 AD – LM

AM + LM= 𝛌 AD 

AD + ½ AB + AD + ½ AB – (AB + ½ AD) = 𝛌 AD 

AD + ½ AB + AD + ½ AB –  AB – ½ AD = 𝛌 AD 

3/2 AD = 𝛌 AD 

𝛌 = 3/2

QUESTION 7

If G is the centroid of the triangle ABC, then show that OG = Addition of Vectors 4  whenAddition of Vectors 9,Addition of Vectors 10 Addition of Vectors 42 are the position vectors of the vertices of triangle ABC.

Sol: OA = a, OB = b, OC = c and OD = d

Addition of Vectors 6

 D is the midpoint of BC

OD = Addition of Vectors 7

G divides median AD in the ratio 2: 1

OG =Addition of Vectors 8

OG =Addition of Vectors 4

QUESTION 8

If Addition of Vectors 9= Addition of Vectors 11, Addition of Vectors 10 =  Addition of Vectors 12  are collinear vectors, then find m and n.

Sol: Given Addition of Vectors 9 , Addition of Vectors 10 are collinear vectors

Addition of Vectors 9 = λAddition of Vectors 10

Addition of Vectors 13

 Equating like vectors

 2 = 4 λ; 5 = m λ; 1 = n λ

 λ =Addition of Vectors 14

 5 = Addition of Vectors 14m ⟹ m =10

1 = Addition of Vectors 14n ⟹ n = 2

∴ m = 10, n = 2

QUESTION 9

Let If  Addition of Vectors 28, Addition of Vectors 29. Find the unit vector in the direction of a + b.

Sol: Given vectors are Addition of Vectors 28 and  Addition of Vectors 29

     a + b = Addition of Vectors 33

    The unit vector in the direction of a + b = Addition of Vectors 30  

       = Addition of Vectors 31

        =  Addition of Vectors 32

QUESTION 10

If the vectors – 3i + 4j + λk and μi + 8j + 6k. are collinear vectors, then find λ and μ.

Sol: let a = – 3i + 4j + λk, b = μi + 8j + 6k

     ⟹   a = tb

 – 3i + 4j + λk = t (μi + 8j + 6k)

 – 3i + 4j + λk = μt i + 8t j + 6t k

Equating like vectors

– 3 = μt; 4 = 8t, λ = 6t

4 = 8t

 t =Addition of Vectors 14

– 3 =Addition of Vectors 14 μ ⟹μ=– 6

λ = Addition of Vectors 14 6 ⟹ λ = 3

∴ μ=– 6, λ = 3

QUESTION 11

ABCD is a pentagon. If the sum of the vectors AB, AE, BC, DC, ED and AC is 𝛌 AC then find the value of 𝛌

Sol: Given, ABCD is a pentagon

        AB + AE + BC + + DC + ED + AC = 𝛌 AC

         (AB + BC) + (AE + DC + ED) + AC = 𝛌 AC

         AC + AC + AC = 𝛌 AC

         3 AC = 𝛌 AC

         𝛌 = 3

QUESTION 12

If the position vectors of the points A, B and C are – 2i + jk and –4i + 2j + 2k and 6i – 3j – 13k respectively and AB = 𝛌 AC, then find the value of 𝛌

Sol: Given, OA = – 2i + jk , OB = –4i + 2j + 2k and OC  = 6i – 3j – 13k

  AB = OB – OA = –4i + 2j + 2k – (– 2i + jk)

          = –4i + 2j + 2k +2ij + k

        = –2i + j + 3k

 AC = OC – OA = 6i – 3j – 13k – (– 2i + jk)

       = 6i – 3j – 13k +2ij + k

     = 8i –4 j –12k

     = – 4 (2i + j + 3k)

AC = – 4 AB

 Given AB = 𝛌 AC

    𝛌 = – 1/4

QUESTION 13

If OA = i + j +k, AB = 3i – 2j + k, BC = i + 2j – 2k, CD = 2i + j +3k then find the vector OD

Sol: Given OA = i + j +k, AB = 3i – 2j + k, BC = i + 2j – 2k, CD = 2i + j +3k

         OD = OA + AB + BC + CD

                = i + j +k + 3i – 2j + k + i + 2j – 2k + 2i + j +3k

                = 7i + 2j +4k

QUESTION 14

Let a = 2i +4 j –5 k, b = i + j+ k, c = j +2 k, then find the unit vector in the opposite direction of a + b + c

Sol:  Given, a = 2i +4 j –5 k, b = i + j+ k, c = j +2 k

         a + b + c = 2i +4 j –5 k + i + j+ k + j +2 k

                           = 3i +6j –2k

 The unit vector in the opposite direction of a + b + c is Addition of Vectors 34

 ⟹ Addition of Vectors 35

      =Addition of Vectors 36

QUESTION 15

Is the triangle formed by the vectors 3i +5j +2k, 2i –3j –5k, 5i – 2j +3k

Sol: Let a =3i +5j +2k, b = 2i –3j –5k, c = 5i – 2j +3k

  Addition of Vectors 37

 ∴ Given vectors form an equilateral triangle.

QUESTION 16

Using the vector equation of the straight line passing through two points, prove that the points whose position vectors are a, b,  (3a – 2b) are collinear.

Sol: the vector equation of the straight line passing through two points a, b is

         r = (1 – t) a+ t b

         3a – 2b = (1 – t) a+ t b

          Equating like vectors

          1 – t = 3 and t = – 2

∴ Given points are collinear.

QUESTION 17

OABC is a parallelogram If OA = a and OC = c, then find the vector equation of the side BC

Sol: Given, OABC is a parallelogram and OA = a, OC = c

Addition of Vectors 38

 The vector equation of BC is a line which is passing through C(c) and parallel to OA

 ⟹ the vector equation of BC is r = c + t a

QUESTION 18

If a, b, c are the position vectors of the vertices A, B, and C respectively of triangle ABC, then find the vector equation of the median through the vertex A

Sol: Given OA = a, OB = b, OC = c

Addition of Vectors 39

 D is mid of BC

  OD =Addition of Vectors 40 = Addition of Vectors 41

 The equation of AD is

  r = (1 – t) a + t ( Addition of Vectors 41 )

QUESTION 19

Find the vector equation of the line passing through the point 2i +3j +k and parallel to the vector 4i – 2j + 3k

Sol: Let a =2i +3j +k, b = 4i – 2j + 3k  

         The vector equation of the line passing through a and parallel to the vector b is r = a + tb

       r = 2i +3j +k + t (4i – 2j + 3k)

          = (2 + 4t) i + (3 – 2t) j + (1 + 3t) k

QUESTION 20

Find the vector equation of the plane passing through the points i – 2j + 5k, 2j –k and – 3i + 5j

Sol: The vector equation of the line passing through a, b and cis r = (1 – t – s) a + tb + sc

      r = (1 – t – s) (i – 2j + 5k) + t (2j –k) + s (– 3i + 5j)

              = (1 – t – 4s) i + (– 2 – 3t + 7s) j + (5 – 6t – 5s) k


Visit My Youtube Channel:  Click  on below  logo

My Youtube channel Logo

 

Leave a Comment

Your email address will not be published. Required fields are marked *

error: Content is protected !!