Question

To find a unit vector \(\left( a \right).\)

Short answer

The unit vector \({\rm{a}} = \frac{1}{{\sqrt 3 }}{\rm{i}} - \frac{1}{{\sqrt 3 }}{\rm{j}} - \frac{1}{{\sqrt 3 }}{\rm{k}}\) and \({\rm{a}} = - \frac{1}{{\sqrt 3 }}{\rm{i}} + \frac{{\rm{l}}}{{\sqrt 3 }}{\rm{j}} + \frac{1}{{\sqrt 3 }}{\rm{k}}\).

Step-by-step solution

Concept of three dimensional vector

Formula:

Consider three dimensional vector\({\bf{a}}\)as follows.

\({\rm{a}} = {a_1}{\rm{i}} + {a_2}{\rm{j}} + {a_3}{\rm{k}}\)

Calculation for a unit vector as\({\rm{i}} + {\rm{j}}\)

As a unit vector is orthogonal to \({\rm{i}} + {\rm{j}}\) that is, \(a \cdot (i + j) = 0\)

Substitute \({a_1}{\rm{i}} + {a_2}{\rm{j}} + {a_3}{\rm{k}}\) for \({\rm{a}}\).

Rearrange the equation.

\({a_1} = - {a_2}\)

Calculation for a unit vector as\({\rm{i}} + k\)

As a unit vector is orthogonal to \({\rm{i}} + {\rm{k}}\) that is, \({\rm{a}} \cdot ({\rm{i}} + {\rm{k}}) = 0\)

Substitute \({a_1}{\rm{i}} + {a_2}{\rm{j}} + {a_3}{\rm{k}}\) for \({\rm{a}}\).

Rearrange the equation.

\({a_1} = - {a_3}\)

Calculation of the magnitude of\(a\)

Compare the above equations of\({a_1}\).

$\({a_1} = - {a_2} = - {a_3}\)

Find magnitude of equation\(a\),

\(|{\rm{a}}| = \sqrt {a_1^2 + a_2^2 + a_3^2} \)

Substitute \(1\) for \(|{\rm{a}}|, - {a_1}\) for \({a_2}\) and \( - {a_1}\) for \({a_3}\).

\(\begin{aligned}{l}1 &= \sqrt {a_1^2 + {{\left( { - {a_1}} \right)}^2} + {{\left( { - {a_1}} \right)}^2}} \\1 &= \sqrt {a_1^2 + a_1^2 + a_1^2} \\1 &= \sqrt {3a_1^2} \\1 &= 3a_1^2\end{aligned}\)

Rearrange the equation.

\(\begin{aligned}{l}{a_1} &= \sqrt {\frac{1}{3}} \\{a_1} &= \pm \frac{1}{{\sqrt 3 }}\end{aligned}\)

In equation\({a_1}\),\( \pm \frac{1}{{\sqrt 3 }}\)for \({a_1}\). \({a_2} = \mp \frac{1}{{\sqrt 3 }}\)

In equation\({a_1}\) , substitute \( \pm \frac{1}{{\sqrt 3 }}\) for \({a_1}\).

\({a_3} = \mp \frac{1}{{\sqrt 3 }}\)

Substitution of the values of\({a_1},{a_2}\)

In equation of\(a\), substitute \(\frac{1}{{\sqrt 3 }}\) for \({a_1}, - \frac{1}{{\sqrt 3 }}\) for \({a_2}\) and \( - \frac{1}{{\sqrt 3 }}{a_2}\). \(a = \frac{1}{{\sqrt 3 }}{\rm{i}} - \frac{1}{{\sqrt 3 }}{\rm{j}} - \frac{1}{{\sqrt 3 }}{\rm{k}}\)

Thus, the value of unit vector \(a\) is \(\frac{1}{{\sqrt 3 }}{\rm{i}} - \frac{1}{{\sqrt 3 }}{\rm{j}} - \frac{1}{{\sqrt 3 }}{\rm{k}}\).

In equation \(a\),substitute \( - \frac{1}{{\sqrt 3 }}\) for \({a_1},\frac{1}{{\sqrt 3 }}\) for \({a_2}\) and \(\frac{1}{{\sqrt 3 }}{a_2}\). \(a = - \frac{1}{{\sqrt 3 }}{\rm{i}} + \frac{1}{{\sqrt 3 }}{\rm{j}} + \frac{1}{{\sqrt 3 }}{\rm{k}}\)

Thus, the value of unit vector \(a\) is \( - \frac{1}{{\sqrt 3 }}i + \frac{1}{{\sqrt 3 }}{\rm{j}} + \frac{1}{{\sqrt 3 }}{\rm{k}}\).