Mathematics Scalar (or dot) product of two vectors
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### Topic Covered

♦ Product of Two Vectors
♦ Scalar (or dot) product of two vectors
♦ Two important properties of scalar product
♦ Projection of a vector on a line

### Product of Two Vectors

● Multiplication of two vectors is also defined in two ways, namely, scalar (or dot) product where the result is a scalar, and vector (or cross) product where the result is a vector.

### Scalar (or dot) product of two vectors

color{blue} "Definition" The scalar product of two nonzero vectors veca and vecb , denoted by vec a ⋅ vecb , is

defined as color{blue}{veca ⋅ vecb = | vec a | | vecb | cosθ},

where, θ is the angle between vec a and vec b, 0 ≤ θ ≤ π (Fig).

\color{green} ✍️If either vec a = vec0 or vecb = vec0, then θ is not defined, and in this case,

we define vec a ⋅ vec b = 0

color{red}{"Point To Consider :"}

1. vec a ⋅ vec b is a real number.

2. Let a and b be two nonzero vectors, then color{blue}{vec a ⋅ vec b = 0} if and only if veca and vec b are color{blue}{"perpendicular"} to each other. i.e.

veca * vec b = 0 => vec a bot vec b

3. If θ = 0, then veca ⋅ vec b = |vec a | | vecb |

In particular, veca ⋅ veca =| veca |^2 , as θ in this case is 0.

4. If θ = π, then vec a ⋅ vecb = −| veca | |vec b |

In particular, veca ⋅ (−veca) = −| veca |^2 , as θ in this case is π.

5. For hat i , hat j  and hat k, we have

hat i ⋅ hat i = hatj ⋅ hatj = hat k ⋅ hat k =1,
 hati ⋅ hatj = hat j ⋅ hat k = hat k ⋅ hat i = 0

6. The angle between two nonzero vectors veca and vec b is given by

cos theta = ( vec a * vec b)/( | vec a | | vec b | ) or color{blue}theta = cos^-1 ( (vec a * vec b )/(|vec a | | vec b | ) )}

7. The scalar product is commutative. i.e.

vec a * vec b = vec b * vec a

### Properties of scalar product

color {blue} "Property 1" (Distributivity of scalar product over addition) Let veca, vecb and vec c be any three vectors, then

vec a * ( vec b + vec c ) = vec a * vec b + vec a * vec c

color {blue} "Property 2" Let vec a and vec b be any two vectors, and λ be any scalar. Then

(λ vec a) ⋅ vec b = (λ vec a) ⋅ vec b = λ(vec a ⋅ vec b) = vec a ⋅ (λ vec b)

=> If two vectors vec a and  vec b are given in component form as a_1 hat i + a_2 hat j + a_3 hat k and  b_1 hat i + b_2 hat j + b_3 hat k , then their scalar product is given as

vec a ⋅ vec b = (a_1 hati + a_2 hatj + a_3 hat k)⋅(b_1 hat i + b_2 hat j + b_3 hat k)

= a_1 hat i ⋅ (b_1 hat i + b_2 hatj + b_3 hat k) + a_2 hatj ⋅ (b_1 hat i + b_2 hatj + b_3 hat k) + a_3 hat k ⋅ (b _1 hati + b_2 hatj + b_3 hat k)

= a_1 b_1 ( hat i ⋅ hat i) + a_1 b_2 (hat i ⋅ hatj) + a_1 b_3 (hat i ⋅ hat k) + a_2 b _1( hat j ⋅ hat i) + a_2 b_2 ( hat j ⋅ hatj) + a_2 b_3 ( hat j ⋅ hatk)
+ a_3 b_1 (hat k ⋅ hat i) + a b (hat k ⋅ hat j) + a_3 b_3 (hat k ⋅ hat k) (Using the above Properties 1 and 2)

= a_1b_1 + a_2b_2 + a_3b_3 (Using Observation 5)

Thus  color{blue}{vec a ⋅ vec b = a_11b_1 + a_2b_2 + a_3b_3}

### Projection of a vector on a line

\color{green} ✍️ A vector vec(AB) makes an angle θ with a given directed line l (say), in the anticlockwise direction (Fig).

=> Then the projection of vec(AB) on l is a vector vec p (say) with magnitude |vec ( AB)| cosθ, and the direction of vec p being the same (or opposite) to that of the line l, depending upon whether cos θ is positive or negative.

The vector vec p is called the projection vector, and its magnitude |vec p| is simply called as the projection of the vector AB on the directed line l.

For example, in each of the following figures (Fig (i) to (iv)), projection vector of vec (AB) along the line l is vector vec (AC).

\color{green} {✍️ "Observations"}

1. If hat p is the unit vector along a line l, then the projection of a vector vec a on the line l is given by vec a ⋅ hat p .

2 . Projection of a vector vec a on other vector vec b , is given by

color{orange}{ vec a * ( vec b/ (| vec b | )) , "or" \ \ 1/ (| vec b |) ( vec a * vec b)}

3. If θ = 0, then the projection vector of vec (AB) will be vec(AB) itself and if θ = π, then the projection vector of vec (AB) will be vec (BA)

4. If theta = pi/2  or  theta = (3 pi ) /2  , then the projection vector of vec (AB) will be zero vector.

color {blue }"Remarks" If α, β and γ are the direction angles of vector vec a = a_1 hat i + a_2 hat j + a_3 hat k , then its direction cosines may be given as

cos alpha = ( vec a * hat i )/ ( | vec a | | hat i | ) = a_1/ (| vec a | ), cos beta = a_2 /( | vec a | ) , and cos gamma = a_3 / (| vec a | )

"Note :" |vec a | cosα, |vec a |cosβ and |vec a| cosγ are respectively the projections of vec a  along OX, OY and OZ. i.e., the scalar components a_1, a_2 and a_3 of the vector vec a, are precisely the projections of veca along x-axis, y-axis and z-axis, respectively. Further, if vec a is a unit vector, then it may be expressed in terms of its direction cosines as vec a = cosα hat i + cosβ hat j + cos γ hat k
Q 3118867709

Find the angle between two vectors vec a and vec b
with magnitudes 1 and 2 respectively and when vec a ⋅ vec b =1 .
Class 12 Chapter 10 Example 13
Solution:

Given  vec a ⋅ vec b =1 , |vec a| = 1 and  |vec b | = 2 . We have

 theta = cos^(-1) ( (vec a ⋅ vec b)/( |vec a| |vec b|) ) = cos^(-1) (1/2) = pi/3
Q 3118067800

Find angle ‘θ’ between the vectors vec a = hat i + hat j − hat k and vec b = hat i − hat j + hat k .
Class 12 Chapter 10 Example 14
Solution:

The angle θ between two vectors vec a and vec b is given by

 cosθ = ( vec a vec b)/( |vec a | | vec b|)

Now  vec a vec b = ( hat i + hat j − hat k) . ( hat i - hat j + hat k) = 1−1−1= −1.

Therefore, we have  cos θ = (-1)/3

hence the required angle is  θ = cos^(-1) ( - 1/3)
Q 3128067801

If vec a = 5 hat i − hat j − 3 hat k and vec b = hat i + 3 hat j − 5 hat k
, then show that the vectors vec a + vec b and vec a − vec b are perpendicular.
Class 12 Chapter 10 Example 15
Solution:

We know that two nonzero vectors are perpendicular if their scalar product
is zero.
Here vec a +vec b = (5 hat i − hat j − 3 hat k) + ( hat i + 3 hat j − 5 hat k) = 6 hat i + 2 hat j −8 hat k

and  vec a − vec b = (5 hat i − hat j − 3 hat k) − ( hat i + 3 hat j − 5 hat k) = 4 hat i − 4 hat j + 2 hat k

So (vec a + vec b) ⋅ (vec a − vec b) = (6 hat i + 2 hat j −8 hat k) ⋅ (4 hat i − 4 hat j + 2 hat k) = 24 − 8 − 16 = 0.

Hence vec a +vec b and vec a − vec b

are perpendicular vectors.
Q 3148067803

Find the projection of the vector vec a = 2 hat i + 3 hat j + 2 hat k on the vector vec b = hat i + 2 hat j + hat k.
Class 12 Chapter 10 Example 16
Solution:

The projection of vector vec a on the vector vec b is given by

 1/(|vec b |) ( vec a . vec b) = (2 xx 1 + 3 xx 2 + 2 xx 1)/sqrt( (1)^2 + (2)^2 + (1)^2) = (10)/sqrt6 = 5/6 sqrt6
Q 3158067804

Find | vec a − vec b | , if two vectors and vec a  and  vec b are such that | vec a| = 2 , | vec b | = 3 and vec a ⋅vec b = 4 .
Class 12 Chapter 10 Example 17
Solution:

We have

| vec a − vec b |^2 = ( vec a − vec b) .(vec a − vec b)

 = vec a . vec a - vec a . vec b - vec b . vec a + vec b . vec b

= |vec a |^2 −2(vec a ⋅vec b) + | vec b |^2

= (2)^2 − 2(4) + (3)^2

Therefore | vec a − vec b | = sqrt 5
Q 3168067805

If vec a is a unit vector and (vec x − vec a) ⋅ ( vec x + vec a) = 8 , then find | vec x |

Class 12 Chapter 10 Example 18
Solution:

Since vec a is a unit vector, | vec a | = 1 . Also,

(vec x − vec a) ⋅ ( vec x + vec a) = 8

or vec x ⋅ vec x + vec x ⋅vec a − vec a ⋅ vec x − vec a ⋅ vec a = 8

or | vec x |^2 − 1 = 8 i.e. | vec x |^2 = 9

Therefore |vec x| = 3 (as magnitude of a vector is non negative).
Q 3178067806

For any two vectors vec a and vec b , we always have  | vec a ⋅ vec b | <= | vec | | vec b| (Cauchy- Schwartz inequality).
Class 12 Chapter 10 Example 19
Solution:

The inequality holds trivially when either vec a = vec 0 or vec b = vec 0 . Actually, in such a
situation we have  | vec a ⋅ vec b | = 0 = | vec a | | vec b | . So, let us assume that  | vec a | ≠ 0 ≠ | vec b|  .
Then, we have

 ( | vec a , vec b | )/( | vec a| . | vec b | ) = |cosθ | ≤ 1

Therefore  | vec a , vec b | <= | vec a | | vec b |
Q 3188067807

For any two vectors vec a and  vec b, we always  | vec a + vec b | <= | vec a | + | vec b | (triangle inequality).
Class 12 Chapter 10 Example 20
Solution:

The inequality holds trivially in case either
vec a = 0 or vec b = 0 (How?). So, let | vec a | != vec 0 != | vec b|. Then,
|vec a + vec b |^2 = 2 (vec a + vec b) = (vec a + vec b) ⋅ (vec a + vec b)

= vec a ⋅ vec a + vec a ⋅ vec b +vec b ⋅vec a + vec b ⋅ vec b

= | vec a ^2 +2 vec a ⋅vec b+ | vec b |^2

≤ | vec a |^ 2 + 2 | vec a ⋅vec b | + | vec b |^2

≤ | vec a |^2 + 2 | vec a || vec b | + | vec b |^2

= (| vec a | + | vec b | )^2

Hence |vec a + vec b |^2 <= | vec a | + | vec b |
Q 3108067808

Show that the points A (−2 hat i + 3 hat j + 5 hat k) , B( hat i + 2 hat j + 3 hat k) and C(7 hat i − hat k)
are collinear.
Class 12 Chapter 10 Example 21
Solution:

We have
vec(AB) = (1+ 2) hat i + (2 − 3)hat j + (3 − 5) hat k = 3 hat i − hat j − 2 hat k ,

 vec(BC) = (7 −1) hat i + (0 − 2) hat j + (−1− 3) hat k = 6 hat i − 2 hat j − 4 hat k ,

 vec(AC) = (7 + 2) hat i + (0 − 3) hat j + (−1− 5) hat k = 9 hat i − 3 hat j − 6 hat k

 | vec( AB) | = sqrt(14) , | vec( BC) | = 2 sqrt(14) and | vec (AC) | = 3 sqrt(14)

Therefore | vec(AC) | = | vec (AB) | + | vec( BC) |

Hence the points A, B and C are collinear.