Question

Find unit vector in direction of $2\hat i + 3\hat j + 4\hat k$ ?

Answer 1

You can solve this question rather easily by trying to recall what a unit vector is. It is a vector that has the same direction as the given vector, but a magnitude of 1, hence, earning the name of “Unit” vector. To find the unit vector, divide the vector by its magnitude.

Complete step by step answer:
We will proceed with the solution exactly as told in the hint section of the solution to the question. First, we’ll look at what a unit vector is, then we will use the definition to find the unit vector of the given vector.
The unit vector is a vector that has a magnitude of 1, thus earning it the name of the “unit” vector. Mathematically, we can write that:
$\hat v = \dfrac{{\vec v}}{{\left| {\vec v} \right|}}$
Where $\vec v$ is a vector,
$\left| {\vec v} \right|$ is the magnitude of the given vector and,
$\hat v$ is the unit vector.
To find a unit vector, all we have to do is to divide the given vector by its magnitude, and this will reach us to our answer.
The vector given to us in the question is:
$\vec a = 2\hat i + 3\hat j + 4\hat k$
Its magnitude can be found out as:
$\Rightarrow \left| {\vec a} \right| = \sqrt {{2^2} + {3^2} + {4^2}}$
$\Rightarrow \left| {\vec a} \right| = \sqrt {29}$
Now, using the formula,
$\hat v = \dfrac{{\vec v}}{{\left| {\vec v} \right|}}$
We can write,
$\Rightarrow \hat a = \dfrac{{\vec a}}{{\left| {\vec a} \right|}}$
We have already found out the value of magnitude of the given vector, substituting the values here, we get,
$\Rightarrow \hat a = \dfrac{1}{{\sqrt {29} }}\left( {2\hat i + 3\hat j + 4\hat k} \right)$

$\therefore$ The value of the unit vector in the direction of $2\hat i + 3\hat j + 4\hat k$ is $\dfrac{1}{{\sqrt {29} }}\left( {2\hat i + 3\hat j + 4\hat k} \right)$.

Note:
Many students get confused and think of a unit vector as a vector whose all three direction ratios are 1, like $\hat i + \hat j + \hat k$, which is a completely wrong assumption as the magnitude of this vector is actually $\sqrt 3$ and not 1.