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 `hat i , hat j ,`and `hat k` respectively.
Now, consider the position vector `vec (OP)` of a point P(x, y, z) as in following Fig . Let `P_1` be the foot of the perpendicular from P on the plane XOY.
We, thus, see that `P_1P` 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) = z hatk `. Similarly, `vec (QP_1) = vec (OS) = y hat j` and `vec (OQ) = x hat i .`
Therefore, it follows that `vec (OP_1) = vec (OQ )+ vec (QP_1) = x hat i + y hat j`
and `vec (OP) = vec (OP_1) + vec (P_1P) = x hat i + y hat j + z hat k`
Hence, the position vector of P with reference to O is given by
`vec (OP) (or vec r) = x hat i + y hat j + zhat k`
This form of any vector is called its component form. Here, x, y and z are called as the scalar components of `vec r`, and `x hat i, y hat j` and `z hat k` are called the vector components of `vec r` along the respective axes. Sometimes x, y and z are also termed as rectangular components.
The length of the vector `vec r = x hat i + y hat j + z hat k ,` is readily determined by applying the Pythagoras theorem twice. We note that in the right angle triangle `OQP_1` in above 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) = 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
`|vec r| = |xhat i + y hat j +z hat k| = sqrt (x^2 + y^2 + z^2)`
If `vec a` and `vec b ` are any two vectors given in the component form `a_1hat i + a_2hat j + a_3hat k` and `b_1hat i + b_2hat j + b_3hat k` respectively , then
(i) the sum (or resultant) of the vectors `vec a "and" vec b` is given by
`vec a + vec b = (a_1 + b_1) hat i + (a_2 + b_2) hat j + (a_3 + b_3) hat k`
(ii) the difference of the vector `vec a` and `vec b` is given by
`vec a - vec b = (a_1 - b_1) hat i + (a_2 - b_2) hat j + (a_3 - b_3) hat k`
(iii) the vectors `vec a` and `vec b` are equal if and only if
`a_1 =b_1, a_2 = b_2 and a_3 = b_3`
(iv) the multiplication of vector `vec a` by any scalar λ is given by
`lambda vec a = (lambda a_1)hat i + (lambda a_2) hat j + (lambda a_3) hat k`
The addition of vectors and the multiplication of a vector by a scalar together give the following distributive laws:
Let `vec a` and `vec b` be any two vectors, and k and m be any scalars. Then
i) `k vec a + m vec a = (k+m)vec a`
ii) `k(m vec a) = (km) vec a`
iii) `k(vec a + vec b) = k vec a + k vec b`
Video link : https://youtu.be/hBlkiMDiBNI