Question

Show that the vector $ \hat i + \hat j + \hat k $ is equally inclined to the axes OX, OY and OZ.

Answer 1

Hint: To show that a given vector is equally inclined with OX, OY and OZ. For this we first find direction ratio of given vector, line OX, line OY and line OZ and then substituting values of direction ratios in formula to find angle between two lines with given direction ratio to find angles.
Formulas used: Angle between two lines having direction ratios given as: $ \cos \theta = \dfrac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {{a_1}^2 + {b_1}^2 + {c_1}^2} \sqrt {{a_2}^2 + {b_2}^2 + {c_2}^2} }} $ , where $ {a_1},{b_1},{c_1},\,{a_2},\,{b_2}\,\,and\,\,{c_2} $ are direction ratio of given lines

Complete step-by-step answer:
Given vector is $ \hat i + \hat j + \hat k $
To show that a given vector is equally inclined to axes OX, OY and OZ. We will find the angle that a given vector forms with OX, OY and OZ.
If all angles are the same then we can say that a given vector is equally inclined to axes OX, OY and OZ.
Dr’s of given vector are $ \left( {1,1,1} \right) $
Dr’s of x axis or line OX are $ \left( {1,0,0} \right) $
Let angle between above two lines be $ \theta $
Therefore, angle between two lines given as:
 $ \cos \theta = \dfrac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {{a_1}^2 + {b_1}^2 + {c_1}^2} \sqrt {{a_2}^2 + {b_2}^2 + {c_2}^2} }} $ , where $ {a_1},{b_1},{c_1},\,{a_2},\,{b_2}\,\,and\,\,{c_2} $ are direction ratio of given lines.
Substituting values in above formula. We have,
 $
  \cos \theta = \dfrac{{\left( 1 \right)\left( 1 \right) + \left( 1 \right)\left( 0 \right) + \left( 1 \right)\left( 0 \right)}}{{\sqrt {{{(1)}^2} + {{(1)}^2} + {{(1)}^2}} \sqrt {{{(1)}^2} + {{(0)}^2} + {{(0)}^2}} }} \\
   \Rightarrow \cos \theta = \dfrac{{1 + 0 + 0}}{{\sqrt {1 + 1 + 1} \sqrt {1 + 0 + 0} }} \\
   \Rightarrow \cos \theta = \dfrac{1}{{\sqrt 3 \sqrt 1 }} \\
   \Rightarrow \cos \theta = \dfrac{1}{{\sqrt 3 }} \;
  $
From above we see that angle between given vector and line OX is $ {\cos ^{ - 1}}\left( {\dfrac{1}{{\sqrt 3 }}} \right) $
Now, we will find the angle between given vector and line OY.
Drs of a given vector are $ \left( {1,1,1} \right) $ .
Dr’s of line OY or y axis are $ \left( {0,1,0} \right) $
The angle between the above two lines is $ \alpha $ .
Then angle between two lines given as:
  $ \cos \alpha = \dfrac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {{a_1}^2 + {b_1}^2 + {c_1}^2} \sqrt {{a_2}^2 + {b_2}^2 + {c_2}^2} }} $ , where $ {a_1},{b_1},{c_1},\,{a_2},\,{b_2}\,\,and\,\,{c_2} $ are direction ratio of given lines.
Substituting values in above formula. We have,
 $
  \cos \alpha = \dfrac{{\left( 1 \right)\left( 0 \right) + \left( 1 \right)\left( 1 \right) + \left( 1 \right)\left( 0 \right)}}{{\sqrt {{{(1)}^2} + {{(1)}^2} + {{(1)}^2}} \sqrt {{{(0)}^2} + {{(1)}^2} + {{(0)}^2}} }} \\
   \Rightarrow \cos \alpha = \dfrac{{0 + 1 + 0}}{{\sqrt {1 + 1 + 1} \sqrt { + 1 + 0} }} \\
   \Rightarrow \cos \alpha = \dfrac{1}{{\sqrt 3 \sqrt 1 }} \\
   \Rightarrow \cos \alpha = \dfrac{1}{{\sqrt 3 }} \;
  $
From above we see that angle between given vector and line OY is $ {\cos ^{ - 1}}\left( {\dfrac{1}{{\sqrt 3 }}} \right) $
Now, we will find the angle between the given vector and line OZ.
Dr’s of a given line are $ \left( {1,1,1} \right) $ .
Dr’s of line OZ or z axis are $ \left( {0,0,1} \right) $
Let angle between above two lines is $ \beta $
Therefore, angle between two lines given as:
 $ \cos \beta = \dfrac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {{a_1}^2 + {b_1}^2 + {c_1}^2} \sqrt {{a_2}^2 + {b_2}^2 + {c_2}^2} }} $ , where $ {a_1},{b_1},{c_1},\,{a_2},\,{b_2}\,\,and\,\,{c_2} $ are direction ratio of given lines.
Substituting values in above formula. We have,
 $
  \cos \beta = \dfrac{{\left( 1 \right)\left( 0 \right) + \left( 1 \right)\left( 0 \right) + \left( 1 \right)\left( 1 \right)}}{{\sqrt {{{(1)}^2} + {{(1)}^2} + {{(1)}^2}} \sqrt {{{(0)}^2} + {{(0)}^2} + {{(1)}^2}} }} \\
   \Rightarrow \cos \beta = \dfrac{{0 + 0 + 1}}{{\sqrt {1 + 1 + 1} \sqrt {0 + 0 + 1} }} \\
   \Rightarrow \cos \beta = \dfrac{1}{{\sqrt 3 \sqrt 1 }} \\
   \Rightarrow \cos \beta = \dfrac{1}{{\sqrt 3 }} \;
  $
From above we see that angle between given vector and line OZ is $ {\cos ^{ - 1}}\left( {\dfrac{1}{{\sqrt 3 }}} \right) $
Therefore, from above we see that given vector form an angle of $ {\cos ^{ - 1}}\left( {\dfrac{1}{{\sqrt 3 }}} \right) $ with lines OX, OY and OZ.
Hence, we can say that given vector is equally inclined with OX, OY and OZ.

Note: Angle between two lines can also be obtained by direction cosine of line. In this we first find the direction ratio of given line then using it we will find direction cosine of line and then using direction cosine in formula to find angle between lines and we will see that the result or angle obtained by using direction cosine or direction ratio method are same.