Answer
Verified
387.6k+ views
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.
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.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Using the following information to help you answer class 12 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Select the correct plural noun from the given singular class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
The sum of three consecutive multiples of 11 is 363 class 7 maths CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How many squares are there in a chess board A 1296 class 11 maths CBSE