♦ Addition of Vectors

♦ Properties of vector addition

♦ Multiplication of a Vector by a Scalar

♦ Components of a vector

♦ Vector joining two points

♦ Section formula

♦ Properties of vector addition

♦ Multiplication of a Vector by a Scalar

♦ Components of a vector

♦ Vector joining two points

♦ Section formula

`=>` A vector `vec(AB)` simply means the displacement from a point A to the point B. Now consider a situation that a girl moves from A to B and then from B to C (Fig).

`=>` The net displacement made by the girl from point A to the point C, is given by the vector AC and expressed as

`vec(AC) = vec(AB) + vec (BC)`

This is known as the triangle law of vector addition.

In general, if we have two vectors `veca` and `vecb` (Fig(i)), then to add them, they are positioned so that the initial point of one coincides with the terminal point of the other (Fig (ii)).

`=>` For example, in Fig (ii), we have shifted vector `vecb` without changing its magnitude and direction, so that it’s initial point coincides with the terminal point of `vec a`.

`=>` Then, the vector `veca + vecb`, represented by the third side AC of the triangle ABC, gives us the sum (or resultant) of the vectors `vec a` and `vec b` i.e., in triangle ABC (Fig (ii)), we have

`vec(AB) + vec (BC) = vec(AC)`

Now again, since `vec(AC) = - vec (CA)` , from the above equation, we have

`vec(AB) + vec (BC) + vec (CA) = vec ("AA") = 0`

`=>` This means that when the sides of a triangle are taken in order, it leads to zero resultant as the initial and terminal points get coincided (Fig(iii)).

`color{blue}{"Difference Of Vector"}`Now, construct a vector `vec (BC)` so that its magnitude is same as the vector `vec (BC)` , but the direction opposite to that of it (Fig(iii)), i.e.,

`vec(BC') = - vec (BC)`

`=>` Then, on applying triangle law from the Fig 10.8 (iii), we have

`color{red}{vec(AC') = vec (AB) + vec (BC') = vec (AB) + ( - vec(BC) ) = vec a - vec b}`

The vector `vec (AC)` is said to represent the difference of `vec a ` and `vec b`

`=>` If we have two vectors `vec a` and `vec b` represented by the two adjacent sides of a parallelogram in magnitude and direction (Fig), then their sum `vec a + vec b` is represented in magnitude and direction by the diagonal of the parallelogram through their common point. This is known as the parallelogram law of vector addition.

`color{red} "Note"` From Fig , using the triangle law, one may note that

`vec(OA) + vec (AC) = vec (OC)`

or ` vec(OA) + vec (OB) = vec (OC)` since `vec(AC) = vec (OB)`

which is parallelogram law.

`=>` The net displacement made by the girl from point A to the point C, is given by the vector AC and expressed as

`vec(AC) = vec(AB) + vec (BC)`

This is known as the triangle law of vector addition.

In general, if we have two vectors `veca` and `vecb` (Fig(i)), then to add them, they are positioned so that the initial point of one coincides with the terminal point of the other (Fig (ii)).

`=>` For example, in Fig (ii), we have shifted vector `vecb` without changing its magnitude and direction, so that it’s initial point coincides with the terminal point of `vec a`.

`=>` Then, the vector `veca + vecb`, represented by the third side AC of the triangle ABC, gives us the sum (or resultant) of the vectors `vec a` and `vec b` i.e., in triangle ABC (Fig (ii)), we have

`vec(AB) + vec (BC) = vec(AC)`

Now again, since `vec(AC) = - vec (CA)` , from the above equation, we have

`vec(AB) + vec (BC) + vec (CA) = vec ("AA") = 0`

`=>` This means that when the sides of a triangle are taken in order, it leads to zero resultant as the initial and terminal points get coincided (Fig(iii)).

`color{blue}{"Difference Of Vector"}`Now, construct a vector `vec (BC)` so that its magnitude is same as the vector `vec (BC)` , but the direction opposite to that of it (Fig(iii)), i.e.,

`vec(BC') = - vec (BC)`

`=>` Then, on applying triangle law from the Fig 10.8 (iii), we have

`color{red}{vec(AC') = vec (AB) + vec (BC') = vec (AB) + ( - vec(BC) ) = vec a - vec b}`

The vector `vec (AC)` is said to represent the difference of `vec a ` and `vec b`

`=>` If we have two vectors `vec a` and `vec b` represented by the two adjacent sides of a parallelogram in magnitude and direction (Fig), then their sum `vec a + vec b` is represented in magnitude and direction by the diagonal of the parallelogram through their common point. This is known as the parallelogram law of vector addition.

`color{red} "Note"` From Fig , using the triangle law, one may note that

`vec(OA) + vec (AC) = vec (OC)`

or ` vec(OA) + vec (OB) = vec (OC)` since `vec(AC) = vec (OB)`

which is parallelogram law.

`color{blue} "Property 1"` For any two vectors `vec a ` and `vecb`

`vec a + vec b = vec b + vec a` ....... (Commutative property)

`color {red} " Proof"` Consider the parallelogram ABCD (Fig). Let `vec (AB) = veca` and `vec (BC) = vec b` , then using the triangle law, from triangle ABC, we have

`vec(AC) = vec a + vec b`

`=>` Now, since the opposite sides of a parallelogram are equal and parallel, from Fig 10.10, we have, `vec(AD) = vec(BC) = vec b` and `vec(DC) = vec(AB) = vec a` .

Again using triangle law, from triangle ADC, we have

`vec(AC) = vec(AD) + vec (DC) = vec b + vec a`

Hence ` vec a + vec b = vec b + vec a`

`color{blue} "Property 2"` For any three vectors `veca, vecb` and `vecc`

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

`color{red} "Proof "` Let the vectors `veca, vecb` and `vecc` be represented by `vec(PQ), vec(QR)` and `vec(RS)`, respectively, as shown in Fig 10.11(i) and (ii).

Then `veca + vecb = vec(PQ) + vec(QR) = vec(PR)`

and `vecb + vecc = vec(QR) + vec(RS) = vec(QS)`

So `(veca +vec b) + vecc = vec(PR) + vec(RS) = vec(PS)`

and `veca + (vecb + vecc) = vec(PQ) + vec(QS) = vec(PS)`

Hence `(veca + vecb) + vecc= veca + (vecb + vecc)`

`color {blue} "Remark"` The associative property of vector addition enables us to write the sum of three vectors `veca ,vecb , vecc` as `veca + vecb + vecc ` without using brackets.

Note that for any vector `veca`, we have

`veca + vec0 = vec0 + veca = a`

Here, the zero vector `vec0` is called the additive identity for the vector addition.

`vec a + vec b = vec b + vec a` ....... (Commutative property)

`color {red} " Proof"` Consider the parallelogram ABCD (Fig). Let `vec (AB) = veca` and `vec (BC) = vec b` , then using the triangle law, from triangle ABC, we have

`vec(AC) = vec a + vec b`

`=>` Now, since the opposite sides of a parallelogram are equal and parallel, from Fig 10.10, we have, `vec(AD) = vec(BC) = vec b` and `vec(DC) = vec(AB) = vec a` .

Again using triangle law, from triangle ADC, we have

`vec(AC) = vec(AD) + vec (DC) = vec b + vec a`

Hence ` vec a + vec b = vec b + vec a`

`color{blue} "Property 2"` For any three vectors `veca, vecb` and `vecc`

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

`color{red} "Proof "` Let the vectors `veca, vecb` and `vecc` be represented by `vec(PQ), vec(QR)` and `vec(RS)`, respectively, as shown in Fig 10.11(i) and (ii).

Then `veca + vecb = vec(PQ) + vec(QR) = vec(PR)`

and `vecb + vecc = vec(QR) + vec(RS) = vec(QS)`

So `(veca +vec b) + vecc = vec(PR) + vec(RS) = vec(PS)`

and `veca + (vecb + vecc) = vec(PQ) + vec(QS) = vec(PS)`

Hence `(veca + vecb) + vecc= veca + (vecb + vecc)`

`color {blue} "Remark"` The associative property of vector addition enables us to write the sum of three vectors `veca ,vecb , vecc` as `veca + vecb + vecc ` without using brackets.

Note that for any vector `veca`, we have

`veca + vec0 = vec0 + veca = a`

Here, the zero vector `vec0` is called the additive identity for the vector addition.

`=>` Let `veca` be a given vector and `λ` a scalar.

`=>` Then the product of the vector `veca` by the scalar λ, denoted as λ`veca` , is called the multiplication of vector `veca` by the scalar `λ.`

`=>`Note that, λ`veca` is also a vector, collinear to the vector `veca` . The vector λ`veca` has the direction same (or opposite) to that of vector `veca` according as the value of λ is positive (or negative). Also, the magnitude of vector λ `veca` is |λ| times the magnitude of the vector `veca` , i.e.,

`| lamda veca | = | lamda | | vec a |`

A geometric visualisation of multiplication of a vector by a scalar is given in Fig

`=>` When `λ = –1,` then `λveca = −veca`, which is a vector having magnitude equal to the magnitude of `veca` and direction opposite to that of the direction of `veca` . The vector `– veca` is called the negative (or additive inverse) of vector `veca` and we always have

`veca + (-veca ) = ( - vec a ) + vec a = vec 0`

`=>` Also, if `lamda = 1/(|vec a |)`, provided `veca ≠ 0`, i.e. `veca` is not a null vector, then

`| lamda vec a | = | lamda | | vec a | = 1/ (| vec a | ) ( | vec a | ) = 1`

`=>` So, λ`veca` represents the unit vector in the direction of `veca` . We write it as

`"unit vector a" = 1/( | vec a| ) vec a`

`color{red} "Note"` For any scalar k , `kvec0 = vec 0`

`=>` Then the product of the vector `veca` by the scalar λ, denoted as λ`veca` , is called the multiplication of vector `veca` by the scalar `λ.`

`=>`Note that, λ`veca` is also a vector, collinear to the vector `veca` . The vector λ`veca` has the direction same (or opposite) to that of vector `veca` according as the value of λ is positive (or negative). Also, the magnitude of vector λ `veca` is |λ| times the magnitude of the vector `veca` , i.e.,

`| lamda veca | = | lamda | | vec a |`

A geometric visualisation of multiplication of a vector by a scalar is given in Fig

`=>` When `λ = –1,` then `λveca = −veca`, which is a vector having magnitude equal to the magnitude of `veca` and direction opposite to that of the direction of `veca` . The vector `– veca` is called the negative (or additive inverse) of vector `veca` and we always have

`veca + (-veca ) = ( - vec a ) + vec a = vec 0`

`=>` Also, if `lamda = 1/(|vec a |)`, provided `veca ≠ 0`, i.e. `veca` is not a null vector, then

`| lamda vec a | = | lamda | | vec a | = 1/ (| vec a | ) ( | vec a | ) = 1`

`=>` So, λ`veca` represents the unit vector in the direction of `veca` . We write it as

`"unit vector a" = 1/( | vec a| ) vec a`

`color{red} "Note"` For any scalar k , `kvec0 = vec 0`

`=>` Let us take the points `A(1, 0, 0), B(0, 1, 0)` and `C(0, 0, 1)` on the x-axis, y-axis and z-axis, respectively. Then, clearly

`|vec(OA)| =1 , |vec (OB)| = 1` and `|vec(OC)| = 1`

`=>` The vectors `vec(OA), vec(OB)` and `vec(OC)` , each having magnitude 1, are called unit vectors along the axes OX, OY and OZ, respectively, and denoted by `hati, hat j ` and `hat k` , respectively (Fig 10.13).

`=>` Now, consider the position vector `vec(OP)` of a point P(x, y, z) as in Fig.

`=>` Let `P_1` be the foot of the perpendicular from P on the plane XOY. We, thus, see that `P_1 P` is parallel to z-axis. As `hat i , hat j ` and `hat k` are the unit vectors along the x, y and z-axes, respectively, and by the definition of the coordinates of P, we have `vec(P_1P) = vec(OR) = zhatk`.

Similarly, `(QP_1) = vec(OS) = y hatj`

and `vec(OQ )= x hati `

.

`=>` Therefore, it follows that `vec(OP_1) = vec(OQ) + vec(QP_1) = x hat i + y hatj`

and `vec(OP) = vec(OP_1) + vec(P_1P) = xhati + yhatj + zhatk `

`=>` Hence, the position vector of P with reference to O is given by

`vec(OP)` (or `vecr` ) `= xhati + yhatj + zhatk`

`=>` This form of any vector is called its component form. Here, x, y and z are called as the scalar components of `vecr` , and `xhati, yhatj` and `zhatk` are called the vector components of `vecr` along the respective axes.

● The length of any vector `vecr = xhat i + y hat j + z hatk` , is readily determined by applying the Pythagoras theorem twice.

We note that in the right angle triangle OQP1 (Fig)

`|vec(OP_1) | = sqrt ( |vec(OQ)|^2 + | vec (QP_1)|^2 ) = sqrt (x^2 +y^2)`

and in the right angle triangle `OP_1P`, we have

`|vec(OP_1) | = sqrt(|vec(OP_1)|^2 + | vec(P_1P)|^2 ) = sqrt((x^2 + y^2 ) + z^2)`

`=>` Hence, the length of any vector `vec r = x hat i + y hat j + z hat k` is given by

`color{orange}{| vec r | = | x hati + y hatj + z hat k | = sqrt ( x^2 + y^2 + z^2)}`

`\color{green} ✍️` If `vec a` and `vec b` are any two vectors given in the component form `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 ` respectively, then

(i) the sum (or resultant) of the vectors `veca` and `vecb` is given by

`veca + vecb = (a_1 + b_1 )hat i + (a_2 + b_2 )hat j + (a_3 + b_3 ) hat k`

(ii) the difference of the vector `veca` and `vecb` is given by

`vec a − vec b = (a_1 − b_1 )hat i + (a_2 − b_2 ) hatj + (a_3 − b_3 ) hat k`

(iii) the vectors `veca` and `vecb` are equal if and only if

`a_1 = b_1, a_2 = b_2` and `a_3 = b_3`

(iv) the multiplication of vector `veca` by any scalar λ is given by

`λveca = (λa_1 ) hat i + (λa_2 ) hatj + (λa_3 )hat k`

The addition of vectors and the multiplication of a vector by a scalar together give the following distributive laws:

`\color{green} ✍️` Let `veca` and `vecb` be any two vectors, and k and m be any scalars. Then

`color{blue} {"(i)" k veca + mveca = (k + m)veca} `

`color{blue} {"(ii)" k(mveca) = (km)veca}`

`color{blue} {"(iii)" k(veca + vecb) = kveca + kvecb}`

`color{blue} "Remarks"`

(i) We can observe that whatever be the value of `λ,` the vector λ`veca` is always collinear to the vector `veca`.

`=>` If the vectors `veca` and `vec b` are given in the component form, i.e. `veca = a_1 hat i + a_2 hat j + a_3 hatk` and `vec b = b_1 hati + b_2 hatj + b_3 k`, then the two vectors are collinear if and only if

`b_1 hati + b_2 hat j + b_3 hat k = λ(a_1 hat i + a_2 hatj + a_3 hat k)`

`=> b_1 hat i + b_2 hat j + b_3 hat k = ( lamda a_1 ) hat i + ( lamda a_2 ) ha j + ( lamda a_3 ) hat k`

`=>` ` b_1 = λa_1 , b_2 = λa_2 , b_3 = λa_3`

`=> b_1/a_1 = b_2/a_2 = b_3/a_3 = lamda`

(ii) If `veca = a_1 hati + a_2 hatj + a_3 hat k` , then `a_1, a_2, a_3` are also called direction ratios of `vec a`.

(iii) In case if it is given that l, m, n are direction cosines of a vector, then `lhat i + mhatj + n hat k` `= (cosα)hat i + (cosβ) hatj + (cos γ)hat k` is the unit vector in the direction of that vector, where α, β and γ are the angles which the vector makes with `x, y` and `z` axes respectively.

`|vec(OA)| =1 , |vec (OB)| = 1` and `|vec(OC)| = 1`

`=>` The vectors `vec(OA), vec(OB)` and `vec(OC)` , each having magnitude 1, are called unit vectors along the axes OX, OY and OZ, respectively, and denoted by `hati, hat j ` and `hat k` , respectively (Fig 10.13).

`=>` Now, consider the position vector `vec(OP)` of a point P(x, y, z) as in Fig.

`=>` Let `P_1` be the foot of the perpendicular from P on the plane XOY. We, thus, see that `P_1 P` is parallel to z-axis. As `hat i , hat j ` and `hat k` are the unit vectors along the x, y and z-axes, respectively, and by the definition of the coordinates of P, we have `vec(P_1P) = vec(OR) = zhatk`.

Similarly, `(QP_1) = vec(OS) = y hatj`

and `vec(OQ )= x hati `

.

`=>` Therefore, it follows that `vec(OP_1) = vec(OQ) + vec(QP_1) = x hat i + y hatj`

and `vec(OP) = vec(OP_1) + vec(P_1P) = xhati + yhatj + zhatk `

`=>` Hence, the position vector of P with reference to O is given by

`vec(OP)` (or `vecr` ) `= xhati + yhatj + zhatk`

`=>` This form of any vector is called its component form. Here, x, y and z are called as the scalar components of `vecr` , and `xhati, yhatj` and `zhatk` are called the vector components of `vecr` along the respective axes.

● The length of any vector `vecr = xhat i + y hat j + z hatk` , is readily determined by applying the Pythagoras theorem twice.

We note that in the right angle triangle OQP1 (Fig)

`|vec(OP_1) | = sqrt ( |vec(OQ)|^2 + | vec (QP_1)|^2 ) = sqrt (x^2 +y^2)`

and in the right angle triangle `OP_1P`, we have

`|vec(OP_1) | = sqrt(|vec(OP_1)|^2 + | vec(P_1P)|^2 ) = sqrt((x^2 + y^2 ) + z^2)`

`=>` Hence, the length of any vector `vec r = x hat i + y hat j + z hat k` is given by

`color{orange}{| vec r | = | x hati + y hatj + z hat k | = sqrt ( x^2 + y^2 + z^2)}`

`\color{green} ✍️` If `vec a` and `vec b` are any two vectors given in the component form `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 ` respectively, then

(i) the sum (or resultant) of the vectors `veca` and `vecb` is given by

`veca + vecb = (a_1 + b_1 )hat i + (a_2 + b_2 )hat j + (a_3 + b_3 ) hat k`

(ii) the difference of the vector `veca` and `vecb` is given by

`vec a − vec b = (a_1 − b_1 )hat i + (a_2 − b_2 ) hatj + (a_3 − b_3 ) hat k`

(iii) the vectors `veca` and `vecb` are equal if and only if

`a_1 = b_1, a_2 = b_2` and `a_3 = b_3`

(iv) the multiplication of vector `veca` by any scalar λ is given by

`λveca = (λa_1 ) hat i + (λa_2 ) hatj + (λa_3 )hat k`

The addition of vectors and the multiplication of a vector by a scalar together give the following distributive laws:

`\color{green} ✍️` Let `veca` and `vecb` be any two vectors, and k and m be any scalars. Then

`color{blue} {"(i)" k veca + mveca = (k + m)veca} `

`color{blue} {"(ii)" k(mveca) = (km)veca}`

`color{blue} {"(iii)" k(veca + vecb) = kveca + kvecb}`

`color{blue} "Remarks"`

(i) We can observe that whatever be the value of `λ,` the vector λ`veca` is always collinear to the vector `veca`.

`=>` If the vectors `veca` and `vec b` are given in the component form, i.e. `veca = a_1 hat i + a_2 hat j + a_3 hatk` and `vec b = b_1 hati + b_2 hatj + b_3 k`, then the two vectors are collinear if and only if

`b_1 hati + b_2 hat j + b_3 hat k = λ(a_1 hat i + a_2 hatj + a_3 hat k)`

`=> b_1 hat i + b_2 hat j + b_3 hat k = ( lamda a_1 ) hat i + ( lamda a_2 ) ha j + ( lamda a_3 ) hat k`

`=>` ` b_1 = λa_1 , b_2 = λa_2 , b_3 = λa_3`

`=> b_1/a_1 = b_2/a_2 = b_3/a_3 = lamda`

(ii) If `veca = a_1 hati + a_2 hatj + a_3 hat k` , then `a_1, a_2, a_3` are also called direction ratios of `vec a`.

(iii) In case if it is given that l, m, n are direction cosines of a vector, then `lhat i + mhatj + n hat k` `= (cosα)hat i + (cosβ) hatj + (cos γ)hat k` is the unit vector in the direction of that vector, where α, β and γ are the angles which the vector makes with `x, y` and `z` axes respectively.

Q 3118767609

Find the values of `x, y` and `z` so that the vectors `vec a = x hat i + 2 hat j + z hat k` and

`vec b = 2 hat i + y hat j + hat k` are equal.

Class 12 Chapter 10 Example 4

`vec b = 2 hat i + y hat j + hat k` are equal.

Class 12 Chapter 10 Example 4

Note that two vectors are equal if and only if their corresponding components

are equal. Thus, the given vectors `vec a` and `vec b` will be equal if and only if

`x = 2, y = 2, z = 1`

Q 3118867700

Let `vec a = vec i + 2 hat j` and `vec b = 2 hat i + hat j` Is `| vec a | = | vec b |` ? Are the vectors and `vec a ` and `vec b`

equal?

Class 12 Chapter 10 Example 5

equal?

Class 12 Chapter 10 Example 5

We have `| vec a | = sqrt (1^2 + 2^2) = sqrt 5` and `| b | = sqrt(2^2 + 1^2) = sqrt 5`

So, `| vec a | = | vec b |`

. But, the two vectors are not equal since their corresponding components

are distinct.

Q 3128867701

Find unit vector in the direction of vector `vec a = 2 hat i + 3 hat j + hat k`

Class 12 Chapter 10 Example 6

Class 12 Chapter 10 Example 6

The unit vector in the direction of vector `vec a` is given by `vec a = 1/(|vec a|) vec a`

Now ` | vec a | = sqrt( 2^2 + 3^2 + 1^2) = sqrt(14)`

Therefore ` hat a = 1/sqrt(14) ( 2 hat i + 3 hat j + hat k) = 2/sqrt(14) hat i + 3/sqrt(14) hat j + 1/sqrt(14) hat k`

Q 3138867702

Find a vector in the direction of vector `vec a = hat i − 2 hat j` that has magnitude

`7` units.

Class 12 Chapter 10 Example 7

`7` units.

Class 12 Chapter 10 Example 7

The unit vector in the direction of the given vector `vec a` is

`hat a = 1/(|vec a|) vec a = 1/sqrt5 ( hat i - 2 hat j) = 1/sqrt5 hat i - 2/sqrt5 hat j`

Therefore, the vector having magnitude equal to `7` and in the direction of `vec a` is

` 7 hat a = 7 ( 1/sqrt5 hat i - 2/sqrt5 hat j) = 7/sqrt5 hat i - (14)/sqrt5 hat j`

Q 3148867703

Find the unit vector in the direction of the sum of the vectors,

`vec a = 2 hat i + 2 hat j – 5 hat k` and `vec b = 2 hat i + hat j + 3 hat k`

Class 12 Chapter 10 Example 8

`vec a = 2 hat i + 2 hat j – 5 hat k` and `vec b = 2 hat i + hat j + 3 hat k`

Class 12 Chapter 10 Example 8

The sum of the given vectors is

`vec a + vec b` ( `= vec c ` , say ) = ` 4 hat i + 3 hat j - 2 hat k`

and ` | vec c | = sqrt( 4^2 + 3^2 + (-2)^2) = sqrt(29)`

Thus, the required unit vector is

`hat c = 1/(|vec c|) vec c = 1/sqrt(29) ( 4 hat i + 3 hat j - 2 hat k) = 4/sqrt(29) hat i + 3/sqrt(29) hat j - 2/sqrt(29) hat k`

Q 3158867704

Write the direction ratio’s of the vector `vec a = hat i + hat j − 2 hat k` and hence calculate

its direction cosines.

Class 12 Chapter 10 Example 9

its direction cosines.

Class 12 Chapter 10 Example 9

Note that the direction ratio’s a, b, c of a vector `vec r = x hat i + y hat j + z hat k` are just

the respective components x, y and z of the vector. So, for the given vector, we have

a = 1, b = 1 and c = –2. Further, if l, m and n are the direction cosines of the given

vector, then

`l = a/(|vec r|) = 1/sqrt6 , m= b/(|vec r|) = 1/sqrt6 , n = c/(|vec r|) = (-2)/sqrt6` as ` | vec r| = sqrt 6`

Thus, the direction cosines are ` ( 1/sqrt6 , 1/sqrt6 , 2/sqrt6)`

`=>` If `P_1(x_1, y_1, z_1)` and `P_2(x_2, y_2, z_2)` are any two points, then the vector joining `P_1` and `P_2` is the vector `P_1P_2` (Fig).

`=>` Joining the points `P_1` and `P_2` with the origin O, and applying triangle law, from the triangle `OP_1P_2`, we have

`vec(OP_1) + vec(P_1P_2) = vec(OP_2)`

`=>` Using the properties of vector addition, the above equation becomes

`vec(P_1P_2) = vec(OP_2) − vec (OP_1)`

i.e `vec(P_1 P_2) = ( x_2 hat i + y_2 hatj + z_2 hat k ) - ( x_1 hat i + y_1 hatj + z_1 hatk )`

`= ( x_2 - x_1 ) hat i + ( y_2 - y_1 ) hat j + ( z_2 - z_1 ) hat k`

`=>` The magnitude of vector `vec(P_1P_2)` is given by

`color{orange}{vec(P_1P_2) = sqrt ( ( x_2 - x_1)^2 + (y_2 - y_1)^2 + ( z_2 - z_1 )^2 )}`

`=>` Joining the points `P_1` and `P_2` with the origin O, and applying triangle law, from the triangle `OP_1P_2`, we have

`vec(OP_1) + vec(P_1P_2) = vec(OP_2)`

`=>` Using the properties of vector addition, the above equation becomes

`vec(P_1P_2) = vec(OP_2) − vec (OP_1)`

i.e `vec(P_1 P_2) = ( x_2 hat i + y_2 hatj + z_2 hat k ) - ( x_1 hat i + y_1 hatj + z_1 hatk )`

`= ( x_2 - x_1 ) hat i + ( y_2 - y_1 ) hat j + ( z_2 - z_1 ) hat k`

`=>` The magnitude of vector `vec(P_1P_2)` is given by

`color{orange}{vec(P_1P_2) = sqrt ( ( x_2 - x_1)^2 + (y_2 - y_1)^2 + ( z_2 - z_1 )^2 )}`

Q 3168867705

Find the vector joining the points `P(2, 3, 0)` and `Q(– 1, – 2, – 4)` directed

from `P` to `Q`.

Class 12 Chapter 10 Example 10

from `P` to `Q`.

Class 12 Chapter 10 Example 10

Since the vector is to be directed from P to Q, clearly P is the initial point

and Q is the terminal point. So, the required vector joining P and Q is the vector `vec(PQ)`

given by

` vec(PQ) = (−1 − 2) hat i + (−2 − 3) hat j + (−4 − 0)hat k`

i.e. ` vec(PQ) = −3 hat i − 5 hat j − 4 hat k`.

`=>` Let P and Q be two points represented by the position vectors `vec(OP)` and `vec(OQ)`, respectively, with respect to the origin O. `=>` Then the line segment joining the points P and Q may be divided by a third point, say R, in two ways internally (Fig) and externally (Fig ).

`color{blue} "Case"` I When R divides PQ internally (Fig).

`=>` If R divides `vec(PQ)` such that m `vec (RQ) = n vec(PR)`,

`=>` where m and n are positive scalars, we say that the point R divides `vec(PQ)` internally in the ratio of m : n. Now from triangles ORQ and OPR, we have

`vec(RQ) = vec (OQ) - vec (OR) = vec b - vecr`

and ` vec (PR) = vec (OR) - vec (OP) = vec r - vec a`,

`=>` Therefore, we have `m (vec b − vec r ) = n (vec r − vec a )`

or ` vec r = ( m vec b + n vec a)/( m + n )` (on simplification)

`=>` Hence, the position vector of the point R which divides P and Q internally in the ratio of m : n is given by

`vec (OR) = ( m vec b + n vec a ) / ( m + n )`

`color{blue} "Case II"` When R divides PQ externally (Fig 10.17). We leave it to the reader as an exercise to verify that the position vector of the point R which divides the line segment `vec (PQ)` externally in the ratio

`m : n ("i.e" (PR)/(QR) = m /n )` is given by

`vec ( OR) = ( m vec b - n vec a ) / ( m - n )`

`"Concept : "` If `R` is the midpoint of `PQ` , then `m = n.` And therefore, from Case I, the midpoint R of `vec(PQ)`, will have its position vector as

`vec (OR) = ( vec a + vec b ) /2`

`color{blue} "Case"` I When R divides PQ internally (Fig).

`=>` If R divides `vec(PQ)` such that m `vec (RQ) = n vec(PR)`,

`=>` where m and n are positive scalars, we say that the point R divides `vec(PQ)` internally in the ratio of m : n. Now from triangles ORQ and OPR, we have

`vec(RQ) = vec (OQ) - vec (OR) = vec b - vecr`

and ` vec (PR) = vec (OR) - vec (OP) = vec r - vec a`,

`=>` Therefore, we have `m (vec b − vec r ) = n (vec r − vec a )`

or ` vec r = ( m vec b + n vec a)/( m + n )` (on simplification)

`=>` Hence, the position vector of the point R which divides P and Q internally in the ratio of m : n is given by

`vec (OR) = ( m vec b + n vec a ) / ( m + n )`

`color{blue} "Case II"` When R divides PQ externally (Fig 10.17). We leave it to the reader as an exercise to verify that the position vector of the point R which divides the line segment `vec (PQ)` externally in the ratio

`m : n ("i.e" (PR)/(QR) = m /n )` is given by

`vec ( OR) = ( m vec b - n vec a ) / ( m - n )`

`"Concept : "` If `R` is the midpoint of `PQ` , then `m = n.` And therefore, from Case I, the midpoint R of `vec(PQ)`, will have its position vector as

`vec (OR) = ( vec a + vec b ) /2`

Q 3178867706

Consider two points P and Q with position vectors `vec( OP) = 3 vec a − 2 vec b`

and `vec (OQ) = vec a + vec b` . Find the position vector of a point R which divides the line joining P and Q

in the ratio 2:1, (i) internally, and (ii) externally.

Class 12 Chapter 10 Example 11

and `vec (OQ) = vec a + vec b` . Find the position vector of a point R which divides the line joining P and Q

in the ratio 2:1, (i) internally, and (ii) externally.

Class 12 Chapter 10 Example 11

(i) The position vector of the point R dividing the join of P and Q internally in the

ratio `2:1` is

` vec(QR) = ( 2 ( veca + vec b) + ( 3 vec a - 2 vec b))/(2 +1) = (5 veca)/3`

(ii) The position vector of the point R dividing the join of P and Q externally in the

ratio `2:1` is

` vec(QR) = ( 2 ( veca + vec b) - ( 3 vec a - 2 vec b))/(2 - 1) = 4 vec b - vec a`

Q 3108867708

Show that the points `A (2 hat i − hat j + hat k), B ( hat i − 3 hat j − 5 hat k), C(3 hat i − 4 j − 4 hat k)` are

the vertices of a right angled triangle.

Class 12 Chapter 10 Example 12

the vertices of a right angled triangle.

Class 12 Chapter 10 Example 12

We have

`vec(AB) = (1− 2) hat i + (−3 +1) hat j + (−5 −1) hat k = − hat i − 2 hat j − 6 hat k`

` vec (BC) = (3 −1) hat i + (−4 + 3) hat j + (−4 + 5) hat k = 2 hat i − hat j + hat k`

and `vec (CA) = (2 − 3) hat i + (−1+ 4) hat j + (1+ 4) hat k = − hat i + 3 hat j + 5 hat k`

Further, note that

`| vec (AB) | ^2 = 41= 6 + 35 = | vec(BC) |^2 + | vec ( CA)|^2`

Hence, the triangle is a right angled triangle.