♦ Product of Two Vectors

♦ Scalar (or dot) product of two vectors

♦ Two important properties of scalar product

♦ Projection of a vector on a line

♦ Scalar (or dot) product of two vectors

♦ Two important properties of scalar product

♦ Projection of a vector on a line

● 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.

`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`

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`

`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}`

`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}`

`\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`

`=>` 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

with magnitudes 1 and 2 respectively and when `vec a ⋅ vec b =1` .

Class 12 Chapter 10 Example 13

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

Class 12 Chapter 10 Example 14

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

, then show that the vectors `vec a + vec b` and `vec a − vec b` are perpendicular.

Class 12 Chapter 10 Example 15

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

Class 12 Chapter 10 Example 16

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

Class 12 Chapter 10 Example 17

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

Class 12 Chapter 10 Example 18

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

Class 12 Chapter 10 Example 19

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

Class 12 Chapter 10 Example 20

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

are collinear.

Class 12 Chapter 10 Example 21

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.