Consider three points $P = \left( { - \sin \left( {{\rm{\beta }} - {\rm{\alpha }}} \right),{\rm{\;}} - \cos {\rm{\beta }}} \right)$ ${\cal Q} = \left( {\cos \left( {{\rm{\beta }} - {\rm{\alpha }}} \right),{\rm{\;}}\sin {\rm{\beta }}} \right)$ And $R = \left( {\cos \left( {{\rm{\beta }} - {\rm{\alpha }} + {\rm{\theta }}} \right),{\rm{\;}}\sin ({\rm{\beta }} - {\rm{\theta }}} \right))$, Where $0 < \alpha ,\;\beta ,\;\theta < \frac{\pi }{4}$. Then,

$P$ lies on the line segment $R{\cal Q}$

${\cal Q}$ lies on the line segment $PR$

$R$ lies on the line segment ${\cal Q}P$

JEE MATHEMATICS MCQ: STRAIGHT LINES

1 Easy

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Analyzing +4 | -1

Consider three points

P = \left( { - \sin \left( {\mathrm{\beta } - \mathrm{\alpha }} \right),\mathrm{\;} - \cos \mathrm{\beta }} \right)

{\cal Q} = \left( {\cos \left( {\mathrm{\beta } - \mathrm{\alpha }} \right),\mathrm{\;}\sin \mathrm{\beta }} \right)

And

R = \left( {\cos \left( {\mathrm{\beta } - \mathrm{\alpha } + \mathrm{\theta }} \right),\mathrm{\;}\sin (\mathrm{\beta } - \mathrm{\theta }} \right))

,

Where

0 < \alpha ,\;\beta ,\;\theta < \frac{\pi }{4}

. Then,

View answer and explanation

Correct option: D

P,\;{\cal Q},\;R

are non-collinear

Option D (

P,\;{\cal Q},\;R

are non-collinear) is correct because the determinant formed by the coordinates of the points is non-zero, indicating non-collinearity.