Question

Find the unit vector in the direction of the vector $\vec a = 2\hat i + 3\hat j + \hat k$ .

Answer 1

Hint: A unit vector is described as a vector whose magnitude is one unit and having a particular direction. The unit vector of the given vector $\vec a$ will be directed in the same direction. The dot product of the reciprocal of the magnitude of the given vector $\left| {\vec a} \right|$ and the vector itself will produce the required unit vector

Formula Used:
1) The magnitude of a vector $\vec A$ is given by, $\left| {\vec A} \right| = \sqrt {{{\left( {{A_x}} \right)}^2} + {{\left( {{A_y}} \right)}^2} + {{\left( {{A_z}} \right)}^2}} $ where ${A_x}$ , ${A_y}$ and ${A_z}$ are respectively the x-component, y-component and z-component of $\vec A$ .
2) The unit vector in the direction of a vector $\vec A$ is given by, $\hat A = \dfrac{1}{{\left| {\vec A} \right|}} \cdot \vec A$ where $\left| {\vec A} \right|$ is the magnitude of the vector.

Complete step by step answer:
Step 1: List the given parameters.
A vector is represented as $\vec a = 2\hat i + 3\hat j + \hat k$ . A unit vector along this vector has to be determined.
The x-component of the given vector is ${a_x} = 2$ , its y-component is ${a_y} = 3$ and the z-component of is ${a_z} = 1$ .
Step 2: Express the relation for the magnitude of the given vector.
The magnitude of the given vector $\vec a = 2\hat i + 3\hat j + \hat k$ will be
$\left| {\vec a} \right| = \sqrt {{{\left( {{a_x}} \right)}^2} + {{\left( {{a_y}} \right)}^2} + {{\left( {{a_z}} \right)}^2}} $ ---------- (1)
where ${a_x}$ , ${a_y}$ and ${a_z}$ are respectively its x-component, y-component and z-component.
Substituting for ${a_x} = 2$ , ${a_y} = 3$ and ${a_z} = 1$ in equation (1) we get, $\left| {\vec a} \right| = \sqrt {{2^2} + {3^2} + {1^2}} = \sqrt {14} $
Thus the magnitude of the given vector is $\left| {\vec a} \right| = \sqrt {14} $ .
Step 3: Express the relation for a unit vector along the direction of the given vector.
The unit vector in the direction of the given vector $\vec a$ is given by, $\hat a = \dfrac{1}{{\left| {\vec a} \right|}} \cdot \vec a$ --------- (2)
where $\left| {\vec a} \right|$ is the magnitude of the vector.
Substituting for $\left| {\vec a} \right| = \sqrt {14} $ and $\vec a = 2\hat i + 3\hat j + \hat k$ in equation (2) we get, $\hat a = \dfrac{1}{{\sqrt {14} }} \cdot \left( {2\hat i + 3\hat j + \hat k} \right)$
The unit vector can be expressed as $\hat a = \dfrac{2}{{\sqrt {14} }}\hat i + \dfrac{3}{{\sqrt {14} }}\hat j + \dfrac{1}{{\sqrt {14} }}\hat k$ .

Note: Here, $\hat i$ , $\hat j$ and $\hat k$ are the unit vectors along the x-direction, y-direction and z-direction respectively. The magnitude of the given vector is a scalar quantity and it refers to the length of the vector. The dot product of a scalar quantity and a vector will be a vector. The dot product of the reciprocal of the magnitude of the vector and the vector itself is obtained by multiplying the magnitude with each component (x, y and z components) of the given vector.