The angle between the lines x cos- -y sin- -a and x sin- -y cos- -a is

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Intercept Form of a Line

The angle bet...

Question

The angle between the lines $x \cos α + y \sin α = a$ and $x \sin β - y \cos β = a$ is

$β - α$

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$π + β - α$

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$\frac{π}{2} + β - α$

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$\frac{π}{2} - β + α$

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Solution

The correct option is D
$\frac{π}{2} - β + α$

Explanation for correct answer:
Finding the angles between the lines
Given, $x \cos α + y \sin α = a$ and $x \sin β - y \cos β = a$ are two lines.
For the general equation of the line, slope is $m = \frac{- a}{b}$
Finding the slope of $x \cos α + y \sin α = a$
$\Rightarrow m_{1} = \frac{- \cos (α)}{\sin (α)}$
And finding the slope of the $x \sin β - y \cos β = a$
$\Rightarrow m_{2} = \frac{- \sin (β)}{- \cos (β)} = \frac{\sin (β)}{\cos (β)}$
Then the angle between two lines is given by,
$\begin{array}{l} tanθ = | \frac{m_{1} - m_{2}}{1 + (m_{1} . m_{2})} | \end{array}$
$\begin{array}{l} \tan θ = |\frac{\frac{- \cos α}{\sin α} - \frac{\sin β}{c o s β}}{1 + (\frac{- \cos α}{\sin α}) (\frac{\sin β}{c o s β})}| \\ = |\frac{\frac{- \cos α c o s β - \sin β \sin α}{\sin α c o s β}}{\frac{\sin α c o s β - \sin β c o s α}{\sin α c o s β}}| \\ = |\frac{- [\cos α c o s β + \sin β \sin α]}{\sin α c o s β - c o s α \sin β}| \end{array}$
We know
$\sin (α - β) = \sin α c o s β - c o s α \sin β$
$\cos (α - β) = \cos α c o s β + \sin β \sin α$
Substituting the trigonometric identities, we get
$\begin{array}{rcl} \tan (θ) & = & |\frac{- \cos (α - β)}{\sin (α - β)}| \\ = & \frac{\cos (α - β)}{\sin (α - β)} \\ = & \cot (α - β) \\ \tan (θ) & = & \tan (\frac{π}{2} + (α - β)) \\ θ & = & \frac{π}{2} - β + α \end{array}$
Therefore, the angle between the lines is $\frac{π}{2} - β + α$
Hence, option (D) is the correct answer.

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