`=>` 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.
`=>` 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.