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If $x \cos \theta+y \sin \theta=2$ is perpendicular to the line $x-y=3$, then what is one of the value of $\theta ?$
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Verified Answer
The correct answer is:
$\pi / 4$
Consider a line
$x \cos \theta+y \sin \theta=2$
$\Rightarrow y \sin \theta=-x \cos \theta+2$
$\Rightarrow y=-x \frac{\cos \theta}{\sin \theta}+\frac{2}{\sin \theta}$
$\Rightarrow \mathrm{y}=-x \cot \theta+2 \operatorname{cosec} \theta$
On comparing this equation with $y=m x+c$ we get slope of line $x \cos \theta+y \sin \theta=2$ is $-\cot \theta$
Also, we have a line $x-y=3$ $\Rightarrow y=x-3$
slope of line $x-y=3$ is 1 . Since, both the lines are perpendicular to each other Product of their slopes $=-1$ $\Rightarrow(-\cot \theta)(1)=-1$
$\Rightarrow \cot \theta=1=\cot \frac{\pi}{4}$
$\Rightarrow \theta=\frac{\pi}{4}$
$x \cos \theta+y \sin \theta=2$
$\Rightarrow y \sin \theta=-x \cos \theta+2$
$\Rightarrow y=-x \frac{\cos \theta}{\sin \theta}+\frac{2}{\sin \theta}$
$\Rightarrow \mathrm{y}=-x \cot \theta+2 \operatorname{cosec} \theta$
On comparing this equation with $y=m x+c$ we get slope of line $x \cos \theta+y \sin \theta=2$ is $-\cot \theta$
Also, we have a line $x-y=3$ $\Rightarrow y=x-3$
slope of line $x-y=3$ is 1 . Since, both the lines are perpendicular to each other Product of their slopes $=-1$ $\Rightarrow(-\cot \theta)(1)=-1$
$\Rightarrow \cot \theta=1=\cot \frac{\pi}{4}$
$\Rightarrow \theta=\frac{\pi}{4}$
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