Question

Find the unit vector in the direction of the vector \[a = 2i + 3j + k\].

Answer 1

Hint:
We know that unit vector in the direction of given vector is given by \[\dfrac{{vector}}{{magnitude{\text{ }}of{\text{ }}vector}} = \dfrac{{\vec a}}{{\left| {\vec a} \right|}}\], where magnitude of vector \[\vec a = xi + yj + zk\] is given by \[\sqrt {{x^2} + {y^2} + {z^2}} \]. So let’s solve it!

Complete step by step solution:
Given vector is \[a = 2i + 3j + k\].
Comparing it with the general vector \[\vec a = xi + yj + zk\] we get x=2, y=3 and z=1. So magnitude of this vector is given by
\[ \Rightarrow \sqrt {{x^2} + {y^2} + {z^2}} \]
\[ \Rightarrow \sqrt {{2^2} + {3^2} + {1^2}} \]
Taking the squares,
\[ \Rightarrow \sqrt {4 + 9 + 1} \]
Adding them we get,
\[ \Rightarrow \sqrt {14} \]
This is the magnitude of the given vector. Now the unit vector is given by,
\[\dfrac{{vector}}{{magnitude{\text{ }}of{\text{ }}vector}} = \dfrac{{\vec a}}{{\left| {\vec a} \right|}}\]
\[ \Rightarrow \dfrac{{2i + 3j + k}}{{\sqrt {14} }}\]
\[\hat a \Rightarrow \dfrac{2}{{\sqrt {14} }}\hat i + \dfrac{3}{{\sqrt {14} }}\hat j + \dfrac{1}{{\sqrt {14} }}\hat k\]

This is the unit vector in the direction of given vector \[\hat a \Rightarrow \dfrac{2}{{\sqrt {14} }}\hat i + \dfrac{3}{{\sqrt {14} }}\hat j + \dfrac{1}{{\sqrt {14} }}\hat k\]

Note:
Here when we find the modulus of a given vector of which the value is always positive. Also when we separate the direction vectors simplify the fraction as much as we can. The cap symbol on the vector shows that it is a unit vector. Unit vectors I, j and k are in the direction of the x, y and z axis respectively.