Question

Let \[\alpha \] be the angle between the lines whose direction cosines satisfy the equations \[l + m - n = 0\] and \[{l^2} + {m^2} - {n^2} = 0\]. Then what is the value of \[\sin^{4}\alpha + \cos^{4}\alpha \] ?
A. \[\dfrac{3}{4}\]
B. \[\dfrac{1}{2}\]
C. \[\dfrac{5}{8}\]
D. \[\dfrac{3}{8}\]

Answer 1

Hint: First, Simplify the given equations and find the values of \[l\], and \[m\]. Substitute the values in the equation and find the values of other variables. Since the direction cosines satisfy the equations, so the values of the variables are the direction ratios. After that, use the formula of the angle between the lines and find the value of \[\cos\alpha \]. Then use the trigonometric identity \[\sin^{2}A + \cos^{2}A = 1\] and calculate the value of \[\sin^{2}\alpha \]. In the end, add the fourth power of both values to get the required answer

Formula Used: Angle between the lines with direction ratios \[\left( {{a_1},{b_1},{c_1}} \right)\] and \[\left( {{a_2},{b_2},{c_2}} \right)\] is: \[\cos\theta = \left| {\dfrac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {{a_1}^2 + {b_1}^2 + {c_1}^2} \sqrt {{a_2}^2 + {b_2}^2 + {c_2}^2} }}} \right|\]
\[\sin^{2}A + \cos^{2}A = 1\]

Complete step by step solution:
Given:
The direction cosines of two lines satisfy the equations \[l + m - n = 0\] and \[{l^2} + {m^2} - {n^2} = 0\].
The angle between the lines is \[\alpha \].
Let’s solve the given equations.
\[l + m - n = 0\] \[.....\left( 1 \right)\]
\[ \Rightarrow \]\[l + m = n\]
Take square on the both sides.
\[{l^2} + {m^2} + 2lm = {n^2}\]
\[ \Rightarrow \]\[{l^2} + {m^2} = {n^2} - 2lm\]
Substitute the value of \[{l^2} + {m^2}\] in the another given equation.
\[{n^2} - 2lm - {n^2} = 0\]
\[ \Rightarrow \]\[ - 2lm = 0\]
\[ \Rightarrow \]\[lm = 0\]
\[ \Rightarrow \]\[l = 0\] or \[m = 0\]

Now substitute the values \[l = 0\] and \[m = 0\] in the equation \[\left( 1 \right)\].
Case 1: \[l = 0\]
Then equation \[\left( 1 \right)\] becomes,
\[m = n\]
We know that if \[l, m, n\] are direction cosines, then \[{l^2} + {m^2} + {n^2} = 1\].
Substitute the values in the above equation. We get
\[0 + {m^2} + {m^2} = 1\]
\[ \Rightarrow \]\[2{m^2} = 1\]
\[ \Rightarrow \]\[{m^2} = \dfrac{1}{2}\]
\[ \Rightarrow \]\[m = \pm \dfrac{1}{{\sqrt 2 }}\]
Thus, \[n = \pm \dfrac{1}{{\sqrt 2 }}\]
Therefore,
\[\left( {l,m,n} \right) = \left( {0,\dfrac{1}{{\sqrt 2 }},\dfrac{1}{{\sqrt 2 }}} \right)\] or \[\left( {l,m,n} \right) = \left( {0, - \dfrac{1}{{\sqrt 2 }}, - \dfrac{1}{{\sqrt 2 }}} \right)\]

Case 2: \[m = 0\]
Then equation \[\left( 1 \right)\] becomes,
\[l = n\]
We know that if \[l, m, n\] are direction cosines, then \[{l^2} + {m^2} + {n^2} = 1\].
Substitute the values in the above equation. We get
\[{l^2} + 0 + {l^2} = 1\]
\[ \Rightarrow \]\[2{l^2} = 1\]
\[ \Rightarrow \]\[{l^2} = \dfrac{1}{2}\]
\[ \Rightarrow \]\[l = \pm \dfrac{1}{{\sqrt 2 }}\]
Thus, \[n = \pm \dfrac{1}{{\sqrt 2 }}\]
Therefore,
\[\left( {l,m,n} \right) = \left( {\dfrac{1}{{\sqrt 2 }},0,\dfrac{1}{{\sqrt 2 }}} \right)\] or \[\left( {l,m,n} \right) = \left( { - \dfrac{1}{{\sqrt 2 }},0, - \dfrac{1}{{\sqrt 2 }}} \right)\]

Let’s consider \[\left( {0,\dfrac{1}{{\sqrt 2 }},\dfrac{1}{{\sqrt 2 }}} \right)\] and \[\left( {\dfrac{1}{{\sqrt 2 }},0,\dfrac{1}{{\sqrt 2 }}} \right)\] are the direction cosines of the two lines.
Apply the formula of the angle between the lines.
 \[\cos\alpha = \left| {\dfrac{{\left( 0 \right)\left( {\dfrac{1}{{\sqrt 2 }}} \right) + \left( {\dfrac{1}{{\sqrt 2 }}} \right)\left( 0 \right) + \left( {\dfrac{1}{{\sqrt 2 }}} \right)\left( {\dfrac{1}{{\sqrt 2 }}} \right)}}{{\sqrt {{{\left( 0 \right)}^2} + {{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^2} + {{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^2}} \sqrt {{{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^2} + {{\left( 0 \right)}^2} + {{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^2}} }}} \right|\]
\[ \Rightarrow \]\[\cos\alpha = \left| {\dfrac{{\dfrac{1}{2}}}{{\sqrt {\dfrac{1}{2} + \dfrac{1}{2}} \sqrt {\dfrac{1}{2} + \dfrac{1}{2}} }}} \right|\]
\[ \Rightarrow \]\[\cos\alpha = \left| {\dfrac{{\dfrac{1}{2}}}{{\sqrt 1 \sqrt 1 }}} \right|\]
\[ \Rightarrow \]\[\cos\alpha = \dfrac{1}{2}\]
Take the fourth power on both sides.
\[\cos^{4}\alpha = \dfrac{1}{{16}}\] \[.....\left( 2 \right)\]

Now apply the formula \[\sin^{2}A + \cos^{2}A = 1\].
\[\sin^{2}\alpha = 1 - \cos^{2}\alpha \]
Substitute \[\cos\alpha = \dfrac{1}{2}\] in the above equation.
\[\sin^{2}\alpha = 1 - {\left( {\dfrac{1}{2}} \right)^2}\]
\[ \Rightarrow \]\[\sin^{2}\alpha = 1 - \dfrac{1}{4}\]
\[ \Rightarrow \]\[\sin^{2}\alpha = \dfrac{3}{4}\]
Take square on both sides.
\[\sin^{4}\alpha = \dfrac{9}{{16}}\] \[.....\left( 3 \right)\]

Now add the equations \[\left( 2 \right)\] and \[\left( 3 \right)\].
\[\sin^{4}\alpha + \cos^{4}\alpha = \dfrac{9}{{16}} + \dfrac{1}{{16}}\]
\[ \Rightarrow \]\[\sin^{4}\alpha + \cos^{4}\alpha = \dfrac{{10}}{{16}}\]
\[ \Rightarrow \]\[\sin^{4}\alpha + \cos^{4}\alpha = \dfrac{5}{8}\]

Hence the correct option is C.

Note: Students often get confused about the formula of the angle between the lines.
If the slopes of the lines are given, then we can use the formula \[\tan\theta = \left| {\dfrac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}}} \right|\] .
If the direction cosines are given, then we can use the formula \[\cos\theta = \left| {\dfrac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {{a_1}^2 + {b_1}^2 + {c_1}^2} \sqrt {{a_2}^2 + {b_2}^2 + {c_2}^2} }}} \right|\].