## Addition of Vectors

These solutions were designed by the ‘Basics in Maths’ team. These notes to do help 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

The unit vector in the direction of a vector is given by

**QUESTION 2**

Find a vector in the direction of **a where** that has a magnitude of 7 units.

The unit vector in the direction of a vector is

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

**QUESTION 3**

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

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

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

** ****QUESTION 4**

Write the direction cosines of the vector

**QUESTION 5**

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

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

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

** AB = **3**a **–** b **– 2**c**

** AC **= **OC – OA **= 7** a** – **c** – **(**– 2**a **+ 3**b **+ 5**c)**

** AC = **9**a **– 3**b **– 6**c =** 3**(**3**a **–** b **– 2**c)**

** 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

(i) Take A as the origin

M is the midpoint of CD

** = AD **+ ½** AB **(∵ AB = DC)

L is the midpoint of BC

** = 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 = ** when, are the position vectors of the vertices of triangle ABC.

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

** **D is the midpoint of BC

G divides median AD in the ratio 2: 1

**QUESTION 8**

If = , = are collinear vectors, then find m and n.

**Sol:** Given , are collinear vectors

Equating like vectors

2 = 4 λ; 5 = m λ; 1 = n λ

∴ m = 10, n = 2

**QUESTION 9**

Let If , . Find the unit vector in the direction of **a + b.**

** **The unit vector in the direction of **a + b **=

**QUESTION 10**

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

**Sol:** let **a **= – 3**i** + 4**j** + λ**k**, **b** = μ**i** + 8**j** + 6**k**

** ⟹ **** a **= t**b**

– 3**i** + 4**j** + λ**k** = t (μ**i** + 8**j** + 6**k)**

– 3**i** + 4**j** + λ**k** = μt **i + **8t **j + **6t **k**

Equating like vectors

– 3 = μt; 4 = 8t, λ = 6t

4 = 8t

∴ 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 – 2**i** + **j** – **k** and –4**i** + 2**j** + 2**k **and 6**i** – 3**j** – 13**k **respectively and** AB = **𝛌** AC, **then find the value of 𝛌

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

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

= –4**i** + 2**j** + 2**k **+2**i** – **j** + **k**

= –2**i** + **j** + 3**k **

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

= 6**i** – 3**j** – 13**k **+2**i** – **j** + **k**

= 8**i** –4 **j** –12**k**

** = **– 4 (2**i** + **j** + 3**k)**

**AC = **– 4** AB**

Given** AB = **𝛌** AC**

** **𝛌 = – 1/4

**QUESTION 13**

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

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

** OD = OA **+** AB **+** BC **+** CD **

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

** = **7**i** + 2**j** +4**k**

**QUESTION 14**

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

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

** = **3**i** +6**j** –2**k**

** **The unit vector in the opposite direction of **a **+ **b** + **c **is

**QUESTION 15**

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

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

∴ 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,** (3**a** – 2**b**) are collinear.

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

**r** = (1 – t) **a**+ t **b **

** **3**a – **2**b = (**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**

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

D is mid of BC

The equation of AD is

**QUESTION 19**

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

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

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

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

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

**QUESTION 20**

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

Sol: The vector equation of the line passing through **a,** **b and c**is **r** = (1 – t – s) a + t**b **+ s**c**

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

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