Question

If a line makes angles ${90^\circ },{60^\circ }$ and $\theta $ with x, y, z axis respectively, where $\theta $ is acute, then find $\theta $.

Answer 1

Hint: The three angles mode by a line with the coordinates x, y and z can be taken as $\alpha ,\,\beta $ and $\gamma $ and to calculate $\theta $ we can use the direction cosine rule ${\cos ^2}\alpha + \,{\cos ^2}\beta + \,{\cos ^2}\gamma = 1$

Complete step by step solution
Given:
The angle mode with x-axis = $ {90^\circ } $
The angle mode with y-axis = $ {60^\circ } $
The angle mode with z-axis = $ \theta $
Also is an acute angle, so $ \theta < {90^\circ } $
Steps:
The direction cosines of a vector are the cosines of angles that the vector forms with the coordinate axis. The direction cosines (incomplete) set the directions of the vector.
Let the directions cosines for the line be l, m and n. The angle formed with x axis is taken as $\alpha $angle with y- axis is $\beta $and angle with z- axis as$\gamma $. Thus,
$l = \cos \alpha ,\,m = x\cos \beta ,\,n = \cos \gamma $
Now one property of the direction of cosines is that one sum of their square is equal to 1.
So,
${l^2} + {m^2} + {n^2} = 1$
Putting the values in the equation
${\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = 1$
$ \Rightarrow \,{\cos ^2}90 + {\cos ^2}60 + {\cos ^2}0 = 1$
$ \Rightarrow \,0 + {\left( {\dfrac{1}{2}} \right)^2} + {\cos ^2}0 = 1$
$ \Rightarrow \,{\cos ^2}\theta = \dfrac{3}{4}$
$ \Rightarrow \cos \theta = \pm \dfrac{{\sqrt 3 }}{2}$
$ \Rightarrow \theta = {30^\circ } $

Note:
Students must have proper knowledge about the direction cosines and their relation. Also to find the direction cosines of a vector, say $\overrightarrow {\alpha ,} $ we need to divide the corresponding coordinate of the vector by the length of the vector, The coordinates of the unit vector is equal to its direction cosines.