ML Aggarwal Class 12 Maths Solutions Section B Chapter 01 Vectors

1 Vectors

INTRODUCTION

In class XI, we studied scalars and vectors; various types of vectors - null vector, proper vector, unit vector, like and unlike vectors, equal vectors, collinear and non-collinear vectors, free and localised vectors, co-initial vectors, coplanar and non-coplanar vectors, negative of a vector; multiplication of a vector by a scalar, sum and difference of vectors; position vector of a point; triangle inequality; section formula; vectors in two and three dimensions; expressing a vector in terms of unit vectors along x, y and z axes (i.e. \( \hat{i}, \hat{j}, \hat{k} \)), and applications of vector algebra in geometry.

In the present chapter, we shall study:

• (i) Scalar (dot) product of vectors - its properties
• (ii) Vector (cross) product of vectors - its properties, area of a triangle, collinear vectors
• (iii) Scalar triple product - volume of a parallelopiped, coplanarity
• (iv) Proofs of formulae (using vectors) and applications of vectors in some geometrical problems

1.1 SCALAR (OR DOT) PRODUCT OF TWO VECTORS

Since vectors have directions, these cannot be multiplied like numbers. We shall see that there are two types of products of vectors - one of these yield a scalar quantity and the other a vector quantity. Moreover, as the origin of this branch of Mathematics lies in physical problems, the definitions of products of vectors should be so framed that are helpful in applications to physical sciences.

Angle between two vectors

Given two (non-zero) vectors \( \vec{a} \) and \( \vec{b} \), we can shift them parallel to themselves so that they intersect at a point. Then the angle, say θ, between them is called the angle between the vectors \( \vec{a} \) and \( \vec{b} \). Note that \( 0 \leq \theta \leq \pi \).

1.1.1 Scalar or dot product of two vectors

The scalar (or dot) product of two (non-zero) vectors \( \vec{a} \) and \( \vec{b} \), denoted by \( \vec{a} \cdot \vec{b} \) (read as \( \vec{a} \) dot \( \vec{b} \)), is defined as

\( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta = ab \cos \theta \),

where \( a = |\vec{a}|, b = |\vec{b}| \) and \( \theta \) (with \( 0 \leq \theta \leq \pi \)) is the angle between \( \vec{a} \) and \( \vec{b} \).

Remarks

1. If \( \vec{a} = \vec{0} \) or \( \vec{b} = \vec{0} \) or both \( \vec{a} = \vec{0} \) and \( \vec{b} = \vec{0} \), we define \( \vec{a} \cdot \vec{b} = 0 \).
2. The vectors \( \vec{a} \) and \( \vec{b} \) are known as factors of \( \vec{a} \cdot \vec{b} \).
3. The dot product of two vectors is a scalar quantity.
4. If \( \theta = 0 \), \( \vec{a} \cdot \vec{b} = ab \) (as \( \cos 0 = 1 \)).
5. If \( \theta = \pi \), \( \vec{a} \cdot \vec{b} = -ab \) (as \( \cos \pi = -1 \)).
   In particular, \( \vec{a} \cdot (-\vec{a}) = |\vec{a}| |-\vec{a}| \cos \pi \) (since the angle between \( \vec{a} \) and \( -\vec{a} \) is \( \pi \))
   \( = |\vec{a}| |-\vec{a}| (-1) \) (since \( \cos \pi = -1 \))
   \( = -a^2 \) (since \( |-\vec{a}| = a \))
6. Condition of perpendicularity of two vectors
   If \( \vec{a} \) and \( \vec{b} \) are perpendicular, the angle between them is \( \frac{\pi}{2} \) and we obtain \( \vec{a} \cdot \vec{b} = ab \cos \frac{\pi}{2} = 0 \).
   Conversely, if \( \vec{a} \cdot \vec{b} = 0 \) i.e. if \( ab \cos \theta = 0 \), then either \( a = 0 \) or \( b = 0 \) or \( \cos \theta = 0 \); it follows that either (or both) of the vectors is a zero vector or else they are perpendicular.
   Therefore, two non-zero (proper) vectors \( \vec{a} \) and \( \vec{b} \) are perpendicular iff \( \vec{a} \cdot \vec{b} = 0 \), which is the required condition of perpendicularity.
   Thus, \( \vec{a} \cdot \vec{b} = 0 \) iff \( \vec{a} = \vec{0} \) or \( \vec{b} = \vec{0} \) or \( \vec{a} \perp \vec{b} \).
7. Note that \( \vec{a} \cdot \vec{b} = \begin{cases} > 0 & \text{if } 0 \leq \theta < \frac{\pi}{2} \text{ i.e. if angle between vectors is acute} \\ 0 & \text{if } \theta = \frac{\pi}{2} \text{ i.e. if } \vec{a} \text{ and } \vec{b} \text{ are perpendicular} \\ < 0 & \text{if } \frac{\pi}{2} < \theta \leq \pi \text{ i.e. if angle between vectors is obtuse} \end{cases} \)
8. Square of a vector
   The scalar product of a vector \( \vec{a} \) with itself is called the square of the vector \( \vec{a} \), and is written as \( \vec{a} \cdot \vec{a} \) or \( (\vec{a})^2 \).
   \( (\vec{a})^2 = \vec{a} \cdot \vec{a} = aa \cos 0 = a \cdot a \cdot 1 = a^2 = |\vec{a}|^2 \).
   Thus, the square of a vector is equal to the square of its modulus.

The length of a vector can be found by using \( |\vec{a}| = \sqrt{\vec{a} \cdot \vec{a}} \) (since \( |\vec{a}|^2 = \vec{a} \cdot \vec{a} \))

9. Squares and scalar products of \( \hat{i}, \hat{j}, \hat{k} \)

Since \( \hat{i}, \hat{j}, \hat{k} \) are unit vectors along the coordinate axes i.e. along three mutually perpendicular lines, we have

\( \hat{i} \cdot \hat{i} = 1 \cdot 1 \cos 0 = 1 \).

Similarly, \( \hat{j} \cdot \hat{j} = 1, \hat{k} \cdot \hat{k} = 1 \).

Also \( \hat{i} \cdot \hat{j} = 1 \cdot 1 \cos 90° = 1 \cdot 1 \cdot 0 = 0 \).

Similarly, \( \hat{j} \cdot \hat{i} = 0, \hat{j} \cdot \hat{k} = 0, \hat{k} \cdot \hat{j} = 0, \hat{k} \cdot \hat{i} = 0, \hat{i} \cdot \hat{k} = 0 \).

1.1.2 Properties of scalar (or dot) product

Theorem 1. The scalar product is commutative i.e. \( \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} \).

Proof. If \( \vec{a} = \vec{0} \) or \( \vec{b} = \vec{0} \), then by definition, \( \vec{a} \cdot \vec{b} = 0 = \vec{b} \cdot \vec{a} \) so that \( \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} \).
If \( \vec{a}, \vec{b} \) are two non-zero vectors, let θ be the angle between \( \vec{a} \) and \( \vec{b} \) (with \( 0 \leq \theta \leq \pi \)). Then angle between \( \vec{b} \) and \( \vec{a} \) is also θ.
\( \therefore \vec{a} \cdot \vec{b} = ab \cos \theta = ba \cos \theta = \vec{b} \cdot \vec{a} \).
Thus, \( \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} \).

Theorem 2. If \( \vec{a}, \vec{b} \) are any vectors and m is any real number (scalar), then \( (m\vec{a}) \cdot \vec{b} = m(\vec{a} \cdot \vec{b}) = \vec{a} \cdot (m\vec{b}) \).

Corollary 1. \( \vec{a} \cdot (-\vec{b}) = -(\vec{a} \cdot \vec{b}) = (-\vec{a}) \cdot \vec{b} \).

Corollary 2. \( (-\vec{a}) \cdot (-\vec{b}) = \vec{a} \cdot \vec{b} \).

Theorem 3. The scalar product is distributive w.r.t. addition i.e. \( \vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} \).

Corollary 1. \( \vec{a} \cdot (\vec{b} + \vec{c} + \vec{d}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} + \vec{a} \cdot \vec{d} \) and, in general
\( \vec{a} \cdot (\vec{b} + \vec{c} + \vec{d} + \ldots) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} + \vec{a} \cdot \vec{d} + \ldots \)

Corollary 2. \( \vec{a} \cdot (\vec{b} - \vec{c}) = \vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{c} \).
We have \( \vec{a} \cdot (\vec{b} - \vec{c}) = \vec{a} \cdot (\vec{b} + (-\vec{c})) = \vec{a} \cdot \vec{b} + \vec{a} \cdot (-\vec{c}) = \vec{a} \cdot \vec{b} - (\vec{a} \cdot \vec{c}) = \vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{c} \).

Corollary 3. Scalar product of sums of vectors

We have \( (\vec{a} + \vec{b}) \cdot (\vec{c} + \vec{d}) = (\vec{a} + \vec{b}) \cdot \vec{c} + (\vec{a} + \vec{b}) \cdot \vec{d} = \vec{c} \cdot (\vec{a} + \vec{b}) + \vec{d} \cdot (\vec{a} + \vec{b}) \) (since the scalar product is commutative)
\( = \vec{c} \cdot \vec{a} + \vec{c} \cdot \vec{b} + \vec{d} \cdot \vec{a} + \vec{d} \cdot \vec{b} = \vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{c} + \vec{a} \cdot \vec{d} + \vec{b} \cdot \vec{d} \).

Corollary 4.

(i) \( (\vec{a} + \vec{b})^2 = (\vec{a})^2 + 2\vec{a} \cdot \vec{b} + (\vec{b})^2 \)

(ii) \( (\vec{a} - \vec{b})^2 = (\vec{a})^2 - 2\vec{a} \cdot \vec{b} + (\vec{b})^2 \)

(iii) \( (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = (\vec{a})^2 - (\vec{b})^2 \)

(iv) \( (\vec{a} + \vec{b})^2 + (\vec{a} - \vec{b})^2 = 2[(\vec{a})^2 + (\vec{b})^2] \)

(v) \( (\vec{a} + \vec{b})^2 - (\vec{a} - \vec{b})^2 = 4\vec{a} \cdot \vec{b} \).

Proof. (i) \( (\vec{a} + \vec{b})^2 = (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} = \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} \) (since \( \vec{b} \cdot \vec{a} = \vec{a} \cdot \vec{b} \))
\( = (\vec{a})^2 + 2\vec{a} \cdot \vec{b} + (\vec{b})^2 \).
Proofs of (ii), (iii), (iv) and (v) are left for the reader.

Corollary 5. If \( \vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} \), then either \( \vec{a} = \vec{0} \) or \( \vec{b} = \vec{c} \) or \( \vec{a} \) is perpendicular to \( (\vec{b} - \vec{c}) \).

Proof. \( \vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} \Rightarrow \vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{c} = 0 \Rightarrow \vec{a} \cdot (\vec{b} - \vec{c}) = 0 \Rightarrow \) either \( \vec{a} = \vec{0} \) or \( \vec{b} - \vec{c} = \vec{0} \) or \( \vec{a} \) is perpendicular to \( (\vec{b} - \vec{c}) \Rightarrow \) either \( \vec{a} = \vec{0} \) or \( \vec{b} = \vec{c} \) or \( \vec{a} \) is perpendicular to \( (\vec{b} - \vec{c}) \).

1.1.3 Scalar product of two vectors in terms of their rectangular components

Let \( \vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \) and \( \vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k} \),

then \( \vec{a} \cdot \vec{b} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \cdot (b_1\hat{i} + b_2\hat{j} + b_3\hat{k}) = a_1b_1\hat{i} \cdot \hat{i} + a_1b_2\hat{i} \cdot \hat{j} + a_1b_3\hat{i} \cdot \hat{k} + a_2b_1\hat{j} \cdot \hat{i} + a_2b_2\hat{j} \cdot \hat{j} + a_2b_3\hat{j} \cdot \hat{k} + a_3b_1\hat{k} \cdot \hat{i} + a_3b_2\hat{k} \cdot \hat{j} + a_3b_3\hat{k} \cdot \hat{k} = a_1b_1 + a_2b_2 + a_3b_3 \) (since \( (\hat{i})^2 = (\hat{j})^2 = (\hat{k})^2 = 1 \) and \( \hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0 \))

Thus, \( \vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3 \). Therefore, the scalar product of two vectors is equal to the sum of the products of their corresponding rectangular components.

Corollary. Let \( \vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}, \vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k} \), then \( \vec{a} \) and \( \vec{b} \) are perpendicular iff \( \vec{a} \cdot \vec{b} = 0 \) i.e. iff \( a_1b_1 + a_2b_2 + a_3b_3 = 0 \). Thus, \( \vec{a} \) and \( \vec{b} \) are perpendicular iff \( a_1b_1 + a_2b_2 + a_3b_3 = 0 \).

1.1.4 Angle between two vectors

Angle θ between two vectors \( \vec{a} \) and \( \vec{b} \) can be found by using \( \vec{a} \cdot \vec{b} = ab \cos \theta \Rightarrow \cos \theta = \frac{\vec{a} \cdot \vec{b}}{ab} = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \).

If \( \vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \) and \( \vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k} \), then angle θ between vectors \( \vec{a} \) and \( \vec{b} \) is given by

\( \cos \theta = \frac{a_1b_1 + a_2b_2 + a_3b_3}{\sqrt{a_1^2 + a_2^2 + a_3^2} \sqrt{b_1^2 + b_2^2 + b_3^2}} \).

1.1.5 Geometrical interpretation of scalar product

Let \( \overrightarrow{OA} \) and \( \overrightarrow{OB} \) represent vectors \( \vec{a} \) and \( \vec{b} \) respectively, then
\( a = |\vec{a}| = |\overrightarrow{OA}| = |OA| = OA \) and \( b = |\vec{b}| = |\overrightarrow{OB}| = |OB| = OB \).

Let \( \angle AOB = \theta \) and M, N be the feet of perpendiculars from A, B on OB, OA respectively.
From right angled \( \triangle OAM \),
\( \cos \theta = \frac{OM}{OA} \Rightarrow OM = OA \cos \theta \) ...(i)

Now, projection of \( \vec{a} \) on \( \vec{b} \) = OM = OA cos θ = a cos θ,
\( \therefore \vec{a} \cdot \vec{b} = ab \cos \theta = b(a \cos \theta) = b \cdot \text{projection of } \vec{a} \text{ on } \vec{b} \).

Similarly, projection of \( \vec{b} \) on \( \vec{a} \) = ON = OB cos θ = b cos θ,
\( \therefore \vec{a} \cdot \vec{b} = ab \cos \theta = a(b \cos \theta) = a \cdot \text{projection of } \vec{b} \text{ on } \vec{a} \).

Thus, \( \vec{a} \cdot \vec{b} \) can be defined as the product of the modulus of one vector and the projection of the other vector upon it.

Also, we find that projection of \( \vec{a} \) on \( \vec{b} \) = \( a \cos \theta = a \cdot \frac{\vec{a} \cdot \vec{b}}{ab} = \frac{\vec{a} \cdot \vec{b}}{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \).

Hence, the (scalar) projection of vector \( \vec{a} \) on vector \( \vec{b} \) = \( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \).

Similarly, the (scalar) projection of vector \( \vec{b} \) on vector \( \vec{a} \) = \( \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|} \).

1.1.6 Direction cosines of a given vector

Let \( \overrightarrow{AB} \) be any vector in space and \( \overrightarrow{OP} \) be a vector passing through origin O and parallel to the vector \( \overrightarrow{AB} \). If the vector \( \overrightarrow{OP} \) makes angles α, β and γ with the three rectangular axes respectively, then α, β and γ are called the direction angles of the vector \( \overrightarrow{AB} \) and cos α, cos β and cos γ are called direction cosines of the vector \( \overrightarrow{AB} \).

If α, β and γ are the direction angles of a (non-zero) vector \( \vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \), then its direction cosines are given as

\( \cos \alpha = \frac{\vec{a} \cdot \hat{i}}{|\vec{a}| |\hat{i}|} = \frac{\vec{a} \cdot \hat{i}}{|\vec{a}|}, \cos \beta = \frac{\vec{a} \cdot \hat{j}}{|\vec{a}|} \text{ and } \cos \gamma = \frac{\vec{a} \cdot \hat{k}}{|\vec{a}|} \)

i.e. \( \cos \alpha = \frac{a_1}{|\vec{a}|}, \cos \beta = \frac{a_2}{|\vec{a}|} \text{ and } \cos \gamma = \frac{a_3}{|\vec{a}|} \)

i.e. \( |\vec{a}| \cos \alpha = a_1, |\vec{a}| \cos \beta = a_2 \text{ and } |\vec{a}| \cos \gamma = a_3 \).

It follows that the scalar components \( a_1, a_2 \) and \( a_3 \) of the vector \( \vec{a} \) are the projections of \( \vec{a} \) along x-axis, y-axis and z-axis respectively.

ILLUSTRATIVE EXAMPLES

Example 1. If \( \vec{a} \) is a unit vector and \( (2\vec{a} + \vec{b}) \cdot (2\vec{a} - \vec{b}) = 2 \), then find \( |\vec{b}| \).
Solution: Since \( \vec{a} \) is a unit vector, \( |\vec{a}| = 1 \).
Given \( (2\vec{a} + \vec{b}) \cdot (2\vec{a} - \vec{b}) = 2 \Rightarrow 2\vec{a} \cdot 2\vec{a} - 2\vec{a} \cdot \vec{b} + \vec{b} \cdot 2\vec{a} - \vec{b} \cdot \vec{b} = 2 \Rightarrow 4|\vec{a}|^2 - |\vec{b}|^2 = 2 \) (since \( \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} \)) \Rightarrow 4 \cdot 1 - |\vec{b}|^2 = 2 \) (since \( |\vec{a}| = 1 \)) \Rightarrow |\vec{b}|^2 = 2 \Rightarrow |\vec{b}| = \sqrt{2} \) (as magnitude of a vector is non-negative)

Exam Tip: Apply the distributive property to expand the product, then use the fact that a unit vector's magnitude is 1. This is a key step that simplifies the problem.

Example 2. If \( (\vec{a} - \vec{b}) \cdot (\vec{a} + \vec{b}) = 8 \) and \( |\vec{a}| = 8|\vec{b}| \), find \( |\vec{a}| \) and \( |\vec{b}| \).
Solution: Given \( (\vec{a} - \vec{b}) \cdot (\vec{a} + \vec{b}) = 8 \Rightarrow \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} - \vec{b} \cdot \vec{a} - \vec{b} \cdot \vec{b} = 8 \Rightarrow |\vec{a}|^2 - |\vec{b}|^2 = 8 \) (since \( \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} \)) \Rightarrow 64|\vec{b}|^2 - |\vec{b}|^2 = 8 \) (since \( |\vec{a}| = 8|\vec{b}| \), given) \Rightarrow 63|\vec{b}|^2 = 8 \Rightarrow |\vec{b}| = \sqrt{\frac{8}{63}} \).
\( \therefore |\vec{a}| = 8|\vec{b}| = 8\sqrt{\frac{8}{63}} \).

Exam Tip: Use the algebraic identity \( (a-b)(a+b) = a^2 - b^2 \) in vector form. Substitute the given relationship between magnitudes to get a single-variable equation.

Example 3. If the angle between two vectors \( \vec{a} \) and \( \vec{b} \) of equal magnitude is 30° and their scalar product is \( 2\sqrt{3} \), find their magnitudes.
Solution: Given \( |\vec{a}| = |\vec{b}| \), θ (angle between \( \vec{a} \) and \( \vec{b} \)) = 30° and \( \vec{a} \cdot \vec{b} = 2\sqrt{3} \)
\( \Rightarrow |\vec{a}| |\vec{b}| \cos 30° = 2\sqrt{3} \Rightarrow |\vec{a}|^2 \cdot \frac{\sqrt{3}}{2} = 2\sqrt{3} \Rightarrow |\vec{a}|^2 = 4 \Rightarrow |\vec{a}| = 2 \)
Hence, \( |\vec{a}| = 2 = |\vec{b}| \).

Exam Tip: Start with the formula \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \). Use the equal magnitude condition to simplify quickly.

Example 4. Find the angle between two vectors \( \vec{a} \) and \( \vec{b} \) with magnitudes \( \sqrt{3} \) and 2 respectively and \( \vec{a} \cdot \vec{b} = \sqrt{6} \).
Solution: Let θ be the angle between the vectors \( \vec{a} \) and \( \vec{b} \).
Given \( \vec{a} \cdot \vec{b} = \sqrt{6} \Rightarrow |\vec{a}| |\vec{b}| \cos \theta = \sqrt{6} \Rightarrow \sqrt{3} \cdot 2 \cos \theta = \sqrt{6} \) (since \( |\vec{a}| = \sqrt{3} \) and \( |\vec{b}| = 2 \) given) \Rightarrow \cos \theta = \frac{\sqrt{6}}{2\sqrt{3}} = \frac{1}{\sqrt{2}} \Rightarrow \cos \theta = \cos 45° \Rightarrow \theta = 45° \).
Hence, the angle between the given vectors = 45°.

Exam Tip: Simplify fractions with radicals carefully. Check that your final cosine value lies between -1 and 1.

Example 5. Find the angle between two vectors \( \hat{i} - 2\hat{j} + 3\hat{k} \) and \( 3\hat{i} - 2\hat{j} + \hat{k} \).
Solution: Let \( \vec{a} = \hat{i} - 2\hat{j} + 3\hat{k} \) and \( \vec{b} = 3\hat{i} - 2\hat{j} + \hat{k} \).
Then \( \vec{a} \cdot \vec{b} = (\hat{i} - 2\hat{j} + 3\hat{k}) \cdot (3\hat{i} - 2\hat{j} + \hat{k}) = 1 \times 3 + (-2)(-2) + 3 \times 1 = 3 + 4 + 3 = 10 \)
\( |\vec{a}| = \sqrt{1^2 + (-2)^2 + 3^2} = \sqrt{14} \) and \( |\vec{b}| = \sqrt{3^2 + (-2)^2 + 1^2} = \sqrt{14} \).
Let θ be the angle between the given vectors, then θ is given by
\( \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{10}{\sqrt{14} \sqrt{14}} = \frac{10}{14} = \frac{5}{7} \Rightarrow \theta = \cos^{-1}\left(\frac{5}{7}\right) \).

Exam Tip: When finding magnitudes of vectors with components, compute the sum of squares of each component separately before taking the square root.

Example 6. If \( \vec{a} = \hat{i} + 2\hat{j} - 3\hat{k} \) and \( \vec{b} = 3\hat{i} - \hat{j} + 2\hat{k} \), show that the vectors \( \vec{a} + \vec{b} \) and \( \vec{a} - \vec{b} \) are perpendicular to each other.
Solution: Given \( \vec{a} = \hat{i} + 2\hat{j} - 3\hat{k} \) and \( \vec{b} = 3\hat{i} - \hat{j} + 2\hat{k} \),
\( \therefore \vec{a} + \vec{b} = (\hat{i} + 2\hat{j} - 3\hat{k}) + (3\hat{i} - \hat{j} + 2\hat{k}) = 4\hat{i} + \hat{j} - \hat{k} \) and
\( \vec{a} - \vec{b} = (\hat{i} + 2\hat{j} - 3\hat{k}) - (3\hat{i} - \hat{j} + 2\hat{k}) = -2\hat{i} + 3\hat{j} - 5\hat{k} \).
\( \therefore (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 4(-2) + 1 \times 3 + (-1)(-5) = -8 + 3 + 5 = 0 \).
Thus, the dot product of two non-zero vectors \( \vec{a} + \vec{b} \) and \( \vec{a} - \vec{b} \) is zero, therefore, the vectors \( \vec{a} + \vec{b} \) and \( \vec{a} - \vec{b} \) are perpendicular to each other.

Exam Tip: To prove perpendicularity, calculate the dot product and show it equals zero. This is the standard condition.

Example 7. If \( \vec{a} = 4\hat{i} + 2\hat{j} - \hat{k} \) and \( \vec{b} = 5\hat{i} + 2\hat{j} - 3\hat{k} \), find the angle between the vectors \( \vec{a} + \vec{b} \) and \( \vec{a} - \vec{b} \).
Solution: Given \( \vec{a} = 4\hat{i} + 2\hat{j} - \hat{k} \) and \( \vec{b} = 5\hat{i} + 2\hat{j} - 3\hat{k} \)
\( \therefore \vec{a} + \vec{b} = (4\hat{i} + 2\hat{j} - \hat{k}) + (5\hat{i} + 2\hat{j} - 3\hat{k}) = 9\hat{i} + 4\hat{j} - 4\hat{k} \) and
\( \vec{a} - \vec{b} = (4\hat{i} + 2\hat{j} - \hat{k}) - (5\hat{i} + 2\hat{j} - 3\hat{k}) = -\hat{i} + 0\hat{j} + 2\hat{k} \)
\( \therefore (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 9 \times (-1) + 4 \times 0 + (-4) \times 2 = -17 \)
\( |\vec{a} + \vec{b}| = \sqrt{9^2 + 4^2 + (-4)^2} = \sqrt{113} \) and \( |\vec{a} - \vec{b}| = \sqrt{(-1)^2 + 0^2 + 2^2} = \sqrt{5} \).
Let θ be the angle between the vectors \( \vec{a} + \vec{b} \) and \( \vec{a} - \vec{b} \) is given by
\( \cos \theta = \frac{(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b})}{|\vec{a} + \vec{b}| |\vec{a} - \vec{b}|} = \frac{-17}{\sqrt{113} \sqrt{5}} = \frac{-17}{\sqrt{565}} \Rightarrow \theta = \cos^{-1}\left(\frac{-17}{\sqrt{565}}\right) \).

Exam Tip: A negative dot product means the angle is obtuse (greater than 90°). Compute magnitudes accurately using the distance formula for vectors.

Example 8. Prove that the three vectors \( 3\hat{i} + \hat{j} + 2\hat{k}, \hat{i} - \hat{j} - \hat{k} \) and \( \hat{i} + 5\hat{j} - 4\hat{k} \) are at right angles to one another.
Solution: Let \( \vec{a} = 3\hat{i} + \hat{j} + 2\hat{k}, \vec{b} = \hat{i} - \hat{j} - \hat{k} \) and \( \vec{c} = \hat{i} + 5\hat{j} - 4\hat{k} \).
We note that all the three vectors are non-zero.
Now \( \vec{a} \cdot \vec{b} = (3\hat{i} + \hat{j} + 2\hat{k}) \cdot (\hat{i} - \hat{j} - \hat{k}) = (3)(1) + (1)(-1) + (2)(-1) = 0 \).
Thus, the dot product of two non-zero vectors \( \vec{a} \) and \( \vec{b} \) is zero, therefore, these vectors are perpendicular to each other.
Again \( \vec{b} \cdot \vec{c} = (\hat{i} - \hat{j} - \hat{k}) \cdot (\hat{i} + 5\hat{j} - 4\hat{k}) = (1)(1) + (-1)(5) + (-1)(-4) = 0 \)
and \( \vec{c} \cdot \vec{a} = (\hat{i} + 5\hat{j} - 4\hat{k}) \cdot (3\hat{i} + \hat{j} + 2\hat{k}) = (1)(3) + (5)(1) + (-4)(2) = 0 \).
As above, it follows that \( \vec{b}, \vec{c} \) are perpendicular and \( \vec{c}, \vec{a} \) are perpendicular. Hence, all the three given vectors are perpendicular to one another.

Exam Tip: To show mutual perpendicularity of three vectors, verify that the dot product of each pair is zero. Do all three calculations to be complete.

Example 9. Find the value of λ so that the two vectors \( 2\hat{i} + 3\hat{j} - \hat{k} \) and \( -4\hat{i} - 6\hat{j} + \lambda\hat{k} \) are (i) parallel (ii) perpendicular to each other.
Solution: Let \( \vec{a} = 2\hat{i} + 3\hat{j} - \hat{k}, \vec{b} = -4\hat{i} - 6\hat{j} + \lambda\hat{k} \).
(i) \( \vec{a} \) and \( \vec{b} \) are parallel to each other iff \( \frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3} \) i.e. iff \( \frac{2}{-4} = \frac{3}{-6} = \frac{-1}{\lambda} \Rightarrow \lambda = 2 \).
(ii) \( \vec{a} \) and \( \vec{b} \) are perpendicular to each other iff \( \vec{a} \cdot \vec{b} = 0 \)
i.e. if \( (2)(-4) + (3)(-6) + (-1)(\lambda) = 0 \Rightarrow \lambda = -8 - 18 = -26 \).

Exam Tip: For parallel vectors, all component ratios must be equal. For perpendicularity, set the dot product to zero.

Example 10. If \( \vec{a} = 3\hat{i} + 2\hat{j} + 9\hat{k} \) and \( \vec{b} = \hat{i} + \lambda\hat{j} + 3\hat{k} \), find the value of λ so that \( \vec{a} + \vec{b} \) is perpendicular to \( \vec{a} - \vec{b} \).
Solution: Given \( \vec{a} = 3\hat{i} + 2\hat{j} + 9\hat{k} \) and \( \vec{b} = \hat{i} + \lambda\hat{j} + 3\hat{k} \),
\( \therefore \vec{a} + \vec{b} = 4\hat{i} + (2 + \lambda)\hat{j} + 12\hat{k} \) and \( \vec{a} - \vec{b} = 2\hat{i} + (2 - \lambda)\hat{j} + 6\hat{k} \).
As the vector \( \vec{a} + \vec{b} \) is perpendicular to the vector \( \vec{a} - \vec{b} \),
we have \( (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 0 \Rightarrow 4 \times 2 + (2 + \lambda)(2 - \lambda) + 12 \times 6 = 0 \Rightarrow 8 + 4 - \lambda^2 + 72 = 0 \Rightarrow \lambda^2 = 84 \Rightarrow \lambda = \pm 2\sqrt{21} \).

Exam Tip: When expanding \( (2 + \lambda)(2 - \lambda) \), remember this is a difference of squares: \( 4 - \lambda^2 \).

Example 11. If \( \vec{a} \) and \( \vec{b} \) are unit vectors such that \( 2\vec{a} - 4\vec{b} \) and \( 10\vec{a} + 8\vec{b} \) are perpendicular to each other, find the angle between vectors \( \vec{a} \) and \( \vec{b} \). (I.S.C. 2005)
Solution: Since \( \vec{a} \) and \( \vec{b} \) are unit vectors, \( |\vec{a}| = 1 \) and \( |\vec{b}| = 1 \).
As the vectors \( 2\vec{a} - 4\vec{b} \) and \( 10\vec{a} + 8\vec{b} \) are perpendicular to each other,
\( (2\vec{a} - 4\vec{b}) \cdot (10\vec{a} + 8\vec{b}) = 0 \Rightarrow 20(\vec{a})^2 + 16\vec{a} \cdot \vec{b} - 40\vec{a} \cdot \vec{b} - 32(\vec{b})^2 = 0 \Rightarrow 20|\vec{a}|^2 - 24\vec{a} \cdot \vec{b} - 32|\vec{b}|^2 = 0 \Rightarrow 20 \cdot 1^2 - 24 \cdot 1 \cdot 1 \cdot \cos \theta - 32 \cdot 1^2 = 0 \) where θ is the angle between vectors \( \vec{a} \) and \( \vec{b} \) \Rightarrow 20 - 24 \cos \theta - 32 = 0 \Rightarrow -12 - 24 \cos \theta = 0 \Rightarrow \cos \theta = -\frac{1}{2} \Rightarrow \theta = 120° \).
Hence, the angle between vectors \( \vec{a} \) and \( \vec{b} \) is 120°.

Exam Tip: When vectors are unit vectors, use \( |\vec{a}|^2 = 1 \) directly to simplify. A negative cosine value signals an obtuse angle.

Example 12. Let \( \vec{a} = \hat{i} + 3\hat{j} + 7\hat{k} \) and \( \vec{b} = 7\hat{i} - \hat{j} + 8\hat{k} \), find:
(i) the projection of \( \vec{a} \) on \( \vec{b} \) (ii) projection of \( \vec{b} \) on \( \vec{a} \).
Solution: Given \( \vec{a} = \hat{i} + 3\hat{j} + 7\hat{k} \) and \( \vec{b} = 7\hat{i} - \hat{j} + 8\hat{k} \),
\( \therefore \vec{a} \cdot \vec{b} = 1 \times 7 + 3 \times (-1) + 7 \times 8 = 60 \).
(i) \( |\vec{b}| = \sqrt{7^2 + (-1)^2 + 8^2} = \sqrt{114} \).
The projection of \( \vec{a} \) on \( \vec{b} \) = \( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{60}{\sqrt{114}} \).
(ii) \( |\vec{a}| = \sqrt{1^2 + 3^2 + 7^2} = \sqrt{59} \).
The projection of \( \vec{b} \) on \( \vec{a} \) = \( \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|} = \frac{60}{\sqrt{59}} \).

Exam Tip: The projection formula is always (dot product) divided by (magnitude of the vector onto which you are projecting). Use the correct denominator.

Example 13. Find λ when the scalar projection of \( \vec{a} = \lambda\hat{i} + \hat{j} + 4\hat{k} \) on \( \vec{b} = 2\hat{i} + 6\hat{j} + 3\hat{k} \) is 4 units.
Solution: The scalar projection of \( \vec{a} \) on \( \vec{b} \) = \( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = 4 \) (given) ...(i)
Here \( \vec{a} \cdot \vec{b} = \lambda \times 2 + 1 \times 6 + 4 \times 3 = 2\lambda + 18 \)
and \( |\vec{b}| = \sqrt{2^2 + 6^2 + 3^2} = \sqrt{49} = 7 \).
From (i), we get \( \frac{2\lambda + 18}{7} = 4 \Rightarrow \lambda = 5 \).

Exam Tip: First calculate the dot product and the magnitude separately, then divide to find the projection.

Example 14. The vectors \( \vec{a} = 3\hat{i} + x\hat{j} - \hat{k} \) and \( \vec{b} = 2\hat{i} + \hat{j} + y\hat{k} \) are mutually perpendicular. Given that \( |\vec{a}| = |\vec{b}| \), find the values of x and y. (I.S.C. 2002)
Solution: Since \( \vec{a} \perp \vec{b}, \vec{a} \cdot \vec{b} = 0 \)
\( \Rightarrow 3 \times 2 + x \times 1 + (-1) \times y = 0 \Rightarrow y - x = 6 \) ...(i)
Given \( |\vec{a}| = |\vec{b}| \Rightarrow |\vec{a}|^2 = |\vec{b}|^2 \Rightarrow 3^2 + x^2 + (-1)^2 = 2^2 + 1^2 + y^2 \Rightarrow y^2 - x^2 = 5 \Rightarrow (y - x)(y + x) = 5 \Rightarrow 6(y + x) = 5 \) (using (i)) \Rightarrow y + x = \frac{5}{6} \) ...(ii)
Solving (i) and (ii), we get \( x = -\frac{31}{12} \) and \( y = \frac{41}{12} \).

Exam Tip: Use the perpendicularity condition to get one equation. Apply the equal magnitude condition to get a second equation. Solve the system of two equations to find both unknowns.

EXERCISE 1.1

1. Evaluate the scalar product \( (3\vec{a} - 5\vec{b}) \cdot (2\vec{a} + 7\vec{b}) \).

2. (i) If \( \vec{a} \) is a unit vector and \( (\vec{x} + \vec{a}) \cdot (\vec{x} - \vec{a}) = 12 \), find \( |\vec{x}| \).
(ii) If \( \vec{a} \) is a unit vector and \( (2\vec{a} + \vec{b}) \cdot (2\vec{a} - \vec{b}) = 2 \), then find \( |\vec{b}| \).

3. If \( (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 3 \) and \( |\vec{a}| = 2|\vec{b}| \), find \( |\vec{a}| \) and \( |\vec{b}| \).

4. (i) If \( |\vec{a}| = 3, |\vec{b}| = 4 \) and \( \vec{a} \cdot \vec{b} = 1 \), find \( (\vec{a} - \vec{b})^2 \).
(ii) If \( |\vec{a}| = 2, |\vec{b}| = 5 \) and \( \vec{a} \cdot \vec{b} = 8 \), find \( |\vec{a} - \vec{b}| \).
(iii) If \( |\vec{a}| = 2, |\vec{b}| = 3 \) and \( \vec{a} \cdot \vec{b} = -8 \), find \( |\vec{a} + 3\vec{b}| \).

5. If the angle between two vectors \( \vec{a} \) and \( \vec{b} \) of equal magnitude is 60° and their scalar product is \( \frac{1}{2} \), find their magnitudes.

6. (i) Find the angle between two vectors \( \vec{a} \) and \( \vec{b} \), given that \( |\vec{a}| = 3, |\vec{b}| = 4 \) and \( \vec{a} \cdot \vec{b} = 6 \).
(ii) Find the angle between vectors \( \vec{a} \) and \( \vec{b} \) such that \( |\vec{a}| = |\vec{b}| = 3 \) and \( \vec{a} \cdot \vec{b} = 1 \).
(iii) Find the angle between two vectors if they have same length \( \sqrt{2} \) and their scalar product is -1.

7. Find the angle between the vectors \( \vec{a} = \hat{i} + \hat{j} - \hat{k} \) and \( \vec{b} = \hat{i} - \hat{j} + \hat{k} \).

8. If \( \vec{a} = 5\hat{i} + 3\hat{j} + 4\hat{k} \) and \( \vec{b} = 6\hat{i} - 8\hat{j} \), then find
(i) \( |\vec{a}| \) (ii) \( |\vec{b}| \) (iii) \( |\vec{a} + \vec{b}| \) (iv) \( |\vec{a} - \vec{b}| \) (v) \( \vec{a} \cdot \vec{b} \) (vi) \( \vec{b} \cdot \vec{a} \) (vii) the angle θ between \( \vec{a} \) and \( \vec{b} \) (viii) the projection of \( \vec{a} \) on \( \vec{b} \) (ix) the projection of \( \vec{b} \) on \( \vec{a} \).

9. (i) Find \( (\vec{b} - \vec{a}) \cdot (3\vec{a} + \vec{b}) \) where \( \vec{a} = \hat{i} - 2\hat{j} + 5\hat{k}, \vec{b} = 2\hat{i} + \hat{j} - 3\hat{k} \).
(ii) Find \( (\vec{a} + 3\vec{b}) \cdot (2\vec{a} - \vec{b}) \) where \( \vec{a} = \hat{i} + \hat{j} + 2\hat{k}, \vec{b} = 3\hat{i} + 2\hat{j} - \hat{k} \).

10. (i) Find the projection of \( \vec{a} = 2\hat{i} + 3\hat{j} + 2\hat{k} \) on the vector \( \vec{b} = \hat{i} + 2\hat{j} + \hat{k} \)
(ii) Find the projection of \( \hat{i} - \hat{j} \) in the direction of \( \hat{i} + \hat{j} \).
(iii) Find λ when the projection of \( \hat{i} + \lambda\hat{j} + \hat{k} \) on \( \hat{i} + \hat{j} \) is \( \sqrt{2} \) units.