The component of vector $2i + 3j + 2k$ perpendicular to $i + j + k$ is:
A) $\dfrac{5}{3}\left( {i - 2j + k} \right)$
B) $\dfrac{1}{3}\left( {5i + j - 2k} \right)$
C) $\dfrac{{\left( {7i - 10j + 7k} \right)}}{3}$
D) $\dfrac{{5i - 8j + 5k}}{3}$

Hint: A vector quantity is such a quantity that has both magnitude as well as direction as opposed to a scalar quantity which only has a magnitude. For performing calculations with vector quantities a separate branch of mathematics known as vector algebra was formed. Vector algebra deals with the algebraic operations like addition, subtraction, multiplication etc. of vector quantities.

Complete step by step answer:
  Letus consider that we have been provided with two vectors a and b such that,
$\vec a = 2i + 3j + 2k$
$\vec b = \;i + {\text{j}} + {\text{k}}$
We know that the component of vector a perpendicular to vector b can be obtained by the following expression.
$\vec c = \vec a - \dfrac{{\vec a \cdot \vec b}}{{{{\left| {\vec b} \right|}^2}}} \times \vec b$ …….(1)
Where, vector c is the component of vector a perpendicular to the vector b.
The magnitude of vector a is,
$\left| {\vec a} \right| = \sqrt {{2^2} + {3^2} + {2^2}} = \sqrt {17} $
The magnitude of vector b is,
$\left| {\vec b} \right| = \sqrt {{1^2} + {1^2} + {1^2}} = \sqrt 3 $.....(2)
The scalar or dot product of vectors a & b is given by,
$\vec a \cdot \vec b = 2(1) - 3(1) + 2(1) = 1$......(3)
Now, putting all the values from equations (2) & (3) in equation (1) we get,
$\vec c = 2i + 3j + 2k - \dfrac{1}{{{{\left( {\sqrt 3 } \right)}^2}}} \times \left( {i + {\text{j}} + {\text{k}}} \right)$
$\vec c = \dfrac{5}{3}\left( {i - 2j + k} \right)$
i.e. $\dfrac{5}{3}\left( {i - 2j + k} \right)$ is the vector which is the component of vector a and also perpendicular to vector b.

Hence option (A) is the correct answer option.

Note: For a vector quantity q $\vec q = ai + bj + ck$ a, b and c are the magnitudes of the quantity along x, y and z directions respectively. i is the unit vector along x - direction, j is the unit vector along y - direction, k is the unit vector along z - direction. So if a ${\vec q}$ is a force vector and it is given in Newton, then it means that a Newton of force is applied in x - direction, b Newton Of force is applied in y - direction and c Newton of force is acting in y - direction.