Question

If \[f\left( \alpha \right) = x\cos \alpha + y\sin \alpha - p\left( \alpha \right)\] , then the lines \[f\left( \alpha \right) = 0\] and \[f\left( \beta \right) = 0\] are perpendicular to each other, if
\[\left( 1 \right){\text{ }}\alpha = \beta \]
\[\left( 2 \right){\text{ }}\alpha + \beta = \dfrac{\pi }{2}\]
\[\left( 3 \right){\text{ }}\left| {\left( {\alpha - \beta } \right)} \right| = \dfrac{\pi }{2}\]
\[\left( 4 \right){\text{ }}\alpha \pm \beta = \dfrac{\pi }{2}\]

Answer 1

Hint: Find the equation of line \[f\left( \beta \right)\] by replacing alpha by beta in equation of \[f\left( \alpha \right)\] .Then find the slopes of both the equations of lines. Then use the property that the product of slopes of two perpendicular lines is \[ - 1\] . Put the values of slopes in the equation, \[{m_1}{m_2} = - 1\] . Solve it further to get the answer.

Complete step-by-step answer:
It is given to us that the lines \[f\left( \alpha \right) = 0\] and \[f\left( \beta \right) = 0\] are perpendicular to each other. It is also given to us that
\[f\left( \alpha \right) = x\cos \alpha + y\sin \alpha - p\left( \alpha \right)\] ----------- (i)
Therefore to find \[f\left( \beta \right)\] replace \[\alpha \] by \[\beta \] in the equation (i) . Hence, we get
\[f\left( \beta \right) = x\cos \beta + y\sin \beta - p\left( \beta \right)\] ----------- (ii)
So we know that perpendicular lines are the lines that intersect at right angles. Also two lines are perpendicular if and only if the product of their slopes is -1. In other words, the slope of a line that is perpendicular to a given line is the negative reciprocal of that slope.
And it is given to us that the lines \[f\left( \alpha \right) = 0\] and \[f\left( \beta \right) = 0\] are perpendicular to each other. So,
Let slope of line \[f\left( \alpha \right) = x\cos \alpha + y\sin \alpha - p\left( \alpha \right)\] is \[{m_1}\]
And slope of line \[f\left( \beta \right) = x\cos \beta + y\sin \beta - p\left( \beta \right)\] is \[{m_2}\]
Now as these both lines are perpendicular therefore, \[{m_2} = - \dfrac{1}{{{m_1}}}\] .
The standard form of a line is given below,
\[ax + by - c = 0\] ------ (iii)
And slope of this line is \[m = - \dfrac{a}{b}\]
On comparing equation (i) with equation (iii), the slope of line \[f\left( \alpha \right)\] will be
\[{m_1} = - \dfrac{{\cos \alpha }}{{\sin \alpha }}\] -------- (iv)
Similarly on comparing equation (ii) with equation (iii), the slope of line \[f\left( \beta \right)\] will be
\[{m_2} = - \dfrac{{\cos \beta }}{{\sin \beta }}\] -------- (v)
We know that the product of slopes of two perpendicular lines is \[ - 1\] . Therefore,
\[ \Rightarrow {m_1}{m_2} = - 1\]
Now substitute the values of \[{m_1}\] and \[{m_2}\] from the equations (iv) and (v) in the above equation,
\[ \Rightarrow \left( { - \dfrac{{\cos \alpha }}{{\sin \alpha }}} \right).\left( { - \dfrac{{\cos \beta }}{{\sin \beta }}} \right) = - 1\]
As we know that \[' - ' \times ' - ' = ' + '\] therefore the above equation becomes
\[ \Rightarrow \left( {\dfrac{{\cos \alpha }}{{\sin \alpha }}} \right).\left( {\dfrac{{\cos \beta }}{{\sin \beta }}} \right) = - 1\]
On multiplying the terms in the numerator and denominator we get,
\[ \Rightarrow \dfrac{{\cos \alpha \cos \beta }}{{\sin \alpha \sin \beta }} = - 1\]
On shifting the term in the denominator to the right hand side we get,
\[ \Rightarrow \cos \alpha \cos \beta = - \sin \alpha \sin \beta \]
Now shift the right hand side term to the left hand side,
\[ \Rightarrow \cos \alpha \cos \beta + \sin \alpha \sin \beta = 0\]
Here by using formula, \[\cos \left( {\alpha - \beta } \right) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \] the above equation becomes
\[ \Rightarrow \cos \left( {\alpha - \beta } \right) = 0\]
As we are assuming that \[\alpha - \beta \] is positive, therefore the above equation can be written as
\[ \Rightarrow \cos \left| {\alpha - \beta } \right| = 0\]
On shifting cos to the right hand side we get,
\[ \Rightarrow \left| {\alpha - \beta } \right| = {\cos ^{ - 1}}\left( 0 \right)\]
We know that the value of \[\cos \dfrac{\pi }{2} = 0\] therefore,
\[ \Rightarrow \left| {\alpha - \beta } \right| = {\cos ^{ - 1}}\left( {\cos \dfrac{\pi }{2}} \right)\]
Here cos inverse eliminates cos by which the above equation becomes
\[ \Rightarrow \left| {\alpha - \beta } \right| = \dfrac{\pi }{2}\]
Hence the correct option is \[\left( 3 \right){\text{ }}\left| {\left( {\alpha - \beta } \right)} \right| = \dfrac{\pi }{2}\]
So, the correct answer is “Option B”.

Note: Remember that the result of two negative terms will be positive. Also remember that cos inverse always eliminates cos. Keep in mind that the product of the slopes of two perpendicular lines is \[ - 1\] . Check all the steps of the solution and what we have done in every step.