Question

A line in the three-dimensional space makes an angle \[\theta \left( {0 < \;\theta \; \leqslant \;\dfrac{\pi }{2}} \right)\;\] with both the x and y-axes. Then the set of all values of \[\theta \] is the interval
(A) \[\left( {0,\;\dfrac{\pi }{4}} \right]\]
(B) \[\left[ {\dfrac{\pi }{6},\;\dfrac{\pi }{3}} \right]\]
(C) \[\left[ {\dfrac{\pi }{4},\;\dfrac{\pi }{2}} \right]\]
(D) \[\left( {\dfrac{\pi }{3},\;\dfrac{\pi }{2}} \right]\;\]

Answer 1

Hint: Here first we will assume the third angle made by the line with z axis to be \[\alpha \] then we will use the concept of direction cosines of a line in three dimensional space i.e. the sum of squares of direction cosines of a line in three dimensional space is equal to 1 and then we will use the concept of domain of sine and cosine functions to get the value of \[\alpha \].

Complete step-by-step answer:
Let us assume the third angle made by the line with z axis to be \[\alpha \]
And it is given that the line in the three-dimensional space makes an angle \[\theta \left( {0 < \;\theta \; \leqslant \;\dfrac{\pi }{2}} \right)\;\] with both the x and y-axes.
Now we know that the sum of squares of direction cosines of a line in three dimensional space is equal to 1.
Hence applying this property we get:-
\[{\cos ^2}\theta + {\cos ^2}\theta + {\cos ^2}\alpha = 1\]
Solving it further we get:-
\[
  2{\cos ^2}\theta + {\cos ^2}\alpha = 1 \\
   \Rightarrow 2{\cos ^2}\theta = 1 - {\cos ^2}\alpha \\
 \]
Now we know that:-
\[
  {\cos ^2}\theta + {\sin ^2}\theta = 1 \\
   \Rightarrow {\sin ^2}\theta = 1 - {\cos ^2}\theta \\
 \]
Applying this identity in the above equation we get:-
\[2{\cos ^2}\theta = {\sin ^2}\alpha \]………………………………(1)
Now since we know that:-
\[ - 1 \leqslant \sin \alpha \leqslant 1\]
On squaring the sides we get:-
\[0 \leqslant {\sin ^2}\alpha \leqslant 1\] as square of any quantity is always greater than zero.
Now since from equation 1 \[2{\cos ^2}\theta = {\sin ^2}\alpha \]
Therefore,
\[0 \leqslant 2{\cos ^2}\theta \leqslant 1\]
Solving it further and dividing by 2 we get:-
\[\dfrac{0}{2} \leqslant \dfrac{{2{{\cos }^2}\theta }}{2} \leqslant \dfrac{1}{2}\]
Simplifying it further we get:-
\[0 \leqslant {\cos ^2}\theta \leqslant \dfrac{1}{2}\]
Now taking square root we get:-
\[\sqrt 0 \leqslant \sqrt {{{\cos }^2}\theta } \leqslant \sqrt {\dfrac{1}{2}} \]
Simplifying it further we get:-
\[\;0 \leqslant \left| {\cos \theta } \right| \leqslant \sqrt {\dfrac{1}{2}} \]
Now since modulus of any quantity is always greater than zero
Therefore the above expression can be written as:-
\[\left| {\cos \theta } \right| \leqslant \sqrt {\dfrac{1}{2}} \]
Now since it is given that \[0 < \;\theta \; \leqslant \;\dfrac{\pi }{2}\]
And cosine function is positive in this interval
Therefore,
\[\cos \theta \geqslant \dfrac{1}{{\sqrt 2 }}\]
Now since we know that,
\[\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}\]
Hence putting this value in above expression we get:-
\[
  \cos \theta \geqslant \cos \dfrac{\pi }{4} \\
   \Rightarrow \theta \geqslant \dfrac{\pi }{4} \\
 \]
Also, it is given that \[\theta \; \leqslant \;\dfrac{\pi }{2}\]
Therefore \[\theta \] lies in the interval \[\left[ {\dfrac{\pi }{4},\;\dfrac{\pi }{2}} \right]\]
Hence option C is correct.

Note: Students should note that cosine function is always positive in the interval \[\left[ {0,\;\dfrac{\pi }{2}} \right]\].
Also, the square and modulus of any quantity are greater than zero as they are always positive.
Students might make mistakes while changing and evaluating the inequalities so all the calculations should be done carefully. Consider the interval (0,4)- this means all the values in between 0 and 4, but 0 and 4 are not included. Consider the interval [0,4]- this means all the values in between 0 and 4, including 0 and 4.