# 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.

**Product Of Vectors**

**Question 1**

If **a** = 6**i** +2 **j** +3 **k**, **b** = 2**i** – 9 **j+** 6**k**, then find the angle between the vectors **a** and **b**

**Sol: **

Given vectors are **a** = 6**i** +2 **j** +3 **k**, **b** = 2**i** – 9 **j+** 6**k**

If θ is the angle between the vectors **a** and **b**, then cos θ =

**a **.**b = **(6**i** +2 **j** +3 **k**). (2**i** – 9 **j+** 6**k**) = 6(2) + 2 (– 9) + 3(6)

** = **12 – 18 + 18 = 12

= 7

= 11

**Question 2**

If **a** = **i** +2 **j** –3 **k**, **b** = 3**i** – **j+** 2**k**, then show that **a** + **b **and **a** – **b** are perpendicular to each other.

**Sol: **

Given vectors are **a** = **i** +2 **j** –3 **k**, **b** = 3**i** – **j+** 2**k**

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

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

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

= – 8 + 3 + 5

= 0

∴ **a** + **b **and **a** – **b** is perpendicular to each other.

**Question 3**

If **a** and **b **be non-zero, non-collinear vectors. If , then find the angle between **a** and **b**

**Sol: **

Squaring on both sides

(**a** + **b**) (**a** + **b**) = (**a – b**) (**a – b**)

**a**^{2} + 2 **a**. **b** + **b**^{2} = **a**^{2} – 2 **a**.**b** + **b**^{2}

** **⟹ 4 **a**.**b** = 0

**a**.**b** = 0

∴ the angle between **a** and **b ** is 90^{0}

**Question 4**

If = 11, = 23 and = 30, then find the angle between the vectors** a** and **b** and also find

**Sol: **

(11)^{2} **– **2 ×11×23 cos θ + (23)^{2} = 900

121 **– **506 cos θ + 529 = 900

650 **– **506 cos θ = 900

= (11)^{2} **+ **2 ×11×23 cos θ + (23)^{2}

= 121 + 2 ×11×23 × + 529

= 400

**Question 5**

If **a** = **i** – **j** – **k** and **b** = 2**i** – 3**j** + **k**, then find the projection vector of **b** on **a** and its magnitude.

**Sol: **

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

** **a.b = (**i** – **j** –** k**). (2**i** – 3**j** + **k**) = 2 + 3 – 1 = 4

** **The projection vector of **b** on **a = **

** The magnitude** of the projection vector = =

**Question 6**

If the vectors λ **i** – 3**j** + 5**k** and 2λ **i** – λ **j** – **k **are perpendicular to each other, then find λ

**Sol:**

let **a **= λ **i** – 3**j** + 5**k**, **b** = 2λ **i** – λ **j** – **k**

Given, that **a **and **b **are perpendicular to each other

⟹ **a**.**b** = 0

(λ **i** – 3**j** + 5**k**). (2λ **i** – λ **j** – **k**) = 0

2 λ^{2} + 3 λ – 5 = 0

2 λ^{2} + 5 λ – 2 λ – 5 = 0

λ (2 λ + 5) – 1 (2 λ + 5) = 0

(2 λ + 5) ((λ – 1) = 0

λ = 1 or λ = -5/2

**Question 7**

Find the Cartesian equation of the plane passing through the point (– 2, 1, 3) and perpendicular to the vector 3**i** + **j** + 5**k**

**Sol:**

let P (x, y, z) be any point on the plane

⟹ **OP** = x**i** + y**j** + z**k**

**OA** = – **2i** +**j** +3**k**

**AP** = **OP** – **OA** = (x**i** + y**j** + z**k**) – (– **2i** +**j** +3**k**)

** AP **= (x + 2) **i **+ (y – 1) **j** + (z – 3) **k**

** AP** is perpendicular to the vector 3**i** + **j** + 5**k **

⟹ 3 (x + 2) + (y – 1) + 5(z – 3) = 0

⟹ 3x + 6 + y – 1 + 5z – 15 = 0

∴ 3x + y + 5z – 10 = 0 is the required Cartesian equation of the plane

**Question 8**

Find the angle between the planes 2x – 3y – 6z = 5 and 6x + 2y – 9z = 4

**Sol:**

Given plane equations are: 2x – 3y – 6z = 5,6x + 2y – 9z = 4

Vector equations of the above planes are: **r**. (2**i** – 3**j** – 6**k**) = 5 and **r**. (6**i** + 2**j** – 9**k**) = 4

⟹ **n _{1}** = 2

**i**– 3

**j**– 6

**k**and

**n**= 6

_{2}**i**+ 2

**j**– 9

**k**

If θ is the angle between the planes **r**. **n _{1}** = d

_{1}and

**r**.

**n**= d

_{2}_{2}, then

**Question 9**

**a** = 2**i** – **j** + **k**, **b** = **i** – 3**j** – 5**k**. Find the vector **c** such that **a**, **b**, and **c** form the sides of a triangle.

**Sol:**

Given **a** = 2**i** – **j** + **k**, **b** = **i** – 3**j** – 5**k**

** **If a, **b**, and **c** form the sides of a triangle, then **a** + **b** + **c** = 0

⟹ **a** + **b **= – c

⟹ **c** = – (**a** + **b**)

= – [(2**i** – **j** + **k)** +( **i** – 3**j** – 5**k**)]

= – (3**i** –4 **j** –4**k**)

∴ **c** = – 3**i** +4 **j** + 4**k**

**Question 10**

Find the equation of the plane through the point (3, –2, 1) and perpendicular to the vector (4, 7, –4).

**Sol:**

Let **a** = 3**i** – 2**j** +** k **and **n = **4**i **+ 7**j **– 4**k **

** **The equation of the plane passing through point A(**a**) and perpendicular to the vector **n **is (**r** – **a**). **n** = 0

⟹ [**r** – (3**i** – 2**j** +** k**)]. (4**i **+ 7**j **– 4**k**) = 0

⟹ **r**. (4**i **+ 7**j **– 4**k**)– [(3**i** – 2**j** +** k**). (4**i **+ 7**j **– 4**k**)] = 0

**r**. (4**i **+ 7**j **– 4**k**)– (12 – 14 – 4) = 0

**r**. (4**i **+ 7**j **– 4**k**)– 6 = 0

**r**. (4**i **+ 7**j **– 4**k**) = 6

**Question 11**

Find the unit vector parallel to the XOY-plane and perpendicular to the vector 4**i** – 3**j** + **k**

**Sol: **

The vector which is parallel to the XOY-plane is of the form x**i** + y**j **

The vector which is parallel to the XOY-plane and perpendicular to 4**i** – 3**j** + **k**

is 3**i** + 4**j **

∴ The unit vector parallel to the XOY-plane and perpendicular to the vector 4**i** – 3**j** + **k **=

**Question 12**

If **a** + **b** + **c** = 0, = 3, = 5 and = 7, then find the angle between **a** and **b**

**Sol:**

Given, **a** + **b** + **c** = 0, = 3, = 5 and = 7

**a** + **b** = – **c**

3^{2} + 5^{2} + 2 cos θ = 7^{2}

9 + 25 + 2.3.5 cos θ = 49

34 + 30cos θ = 49

30cos θ = 49 – 34

30cos θ = 15

cos θ = 15/30 = 1/2

∴ θ = π/3

**Question 13**

If **a** = 2**i** – 3**j** + 5**k**, **b** = – **i** + 4**j** + 2**k**, then find **a** × **b** and unit vector perpendicular to both **a** and **b**

**Sol:**

Given,** a** = 2**i** – 3**j** + 5**k**, **b** = – **i** + 4**j** + 2**k **

** = i **(–6 – 20) – **j** (4 + 5) + **k** (8 – 3)

** = **–26**i – **9**j **+ 5**k**

The unit vector perpendicular to both **a** and **b** =

**Question 14**

If **a** = **i** + **j** + 2**k** and **b** = 3**i** + 5**j** – **k** are two sides of a triangle, then find its area.

**Sol:**

Given, **a** = **i** + 2**j** + 3**k** and **b** = 3**i** + 5**j** – **k**

If** a, b **are two sides of a triangle, then area of the triangle =

** = i **(–2 – 15) – **j** (–1 – 9) + **k** (5 – 6)

** = **–17**i + **10** j **– **k**

**Question 15**

Find the area of the parallelogram for which the vectors **a** = 2**i** – 3**j **and **b** = 3**i** – **k **are adjacent sides.

**Sol: **

Given, **a** = 2**i** – 3**j **and **b** = 3**i** – **k **are adjacent sides of a parallelogram

The area of the parallelogram whose vectors **a** , **b** are adjacent sides =

**= i **(3 – 0) – **j** (–2 – 0) + **k** (0 + 9)

** ** **=**3** i +**2 **j** +9 **k**

∴ The area of the parallelogram =

**Question 16**

Let **a**, **b** be two non-collinear unit vectors. If α = **a** – (**a** . **b**) **b** and β = **a** × **b**, then show that

**Sol:**

= 1 – cos^{2} θ

= sin^{2} θ

= 1 + cos^{2} θ – 2cos^{2} θ

= 1– cos^{2} θ

= sin^{2} θ

**Question 17**

**Sol: **

Let **a** = x**i** + y**j** + z**k**

** = i** ( 0 – 0) – **j **(0 – z) + **k** (0 – y)

** = **– y**k **+ z**j**

Similarly, ** ** = x^{2} + z^{2}** **and** ** = y^{2} + x^{2}** **

** + ****+** ** = y**^{2} + z^{2} + x^{2} + z^{2} + y^{2} + x^{2}** **

= 2(x^{2} + y^{2} +z^{2})

**Question 18**

If = 2, = 3 and (**p**, **q**) = , then find

**Sol:**

= 2 × 3 sin

= 2 × 3×1/2

= 3

**Question 19**

If 4**i** + **j** + p**k **is parallel to the vector **i** + 2**j** + 3**k**, then find p.

**Sol:**

Given 4**i** + **j** + p**k **is parallel to the vector **i** + 2**j** + 3**k**

⇒ p = 12

**Question 20**

Compute **a**× (**b** + **c**) + **b**× (**c** + **a**) + **c**× (**a** + **b**)

**Sol:**

**a**× (**b** + **c**) + **b**× (**c** + **a**) + **c**× (**a** + **b**)

**= a**× **b** +** a**× **c **+ **b**× **c** +** b**× **a **+** c **× **a** +** c **× **b**

**= a**× **b** +** a**× **c **+ **b**× **c** –**a **× **b **–** a** × **c **–** b** ×**c**

**= 0**

**Question 21**

Compute 2**j**× (3**i – **4**k**) + (**i** + 2**j**) × **k**

**Sol:**

2**j**× (3**i – **4**k**) + (**i** + 2**j**) × **k **

**= **6(**j** × **i**) – 8(**j** × **k**) + (**i **× **k**) + 2(**j** × **k**)

= **– **6**k – **8**i** **– j **+ 2**i**

= **– **6**i –j** **–**6** j**

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