A plane has equation x−z=4(√3). The line l has vector equation r=λ[(cosθ+(√3))i+((√2)sinθ)j+(cosθ−(√3))k] where λ is a scalar parameter. If l meets the plane at P, show that, as θ varies, P describes a circle.


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Question Number 1851 by 112358 last updated on 14/Oct/15

$A p l a n e h a s e q u a t i o n x - z = 4 \sqrt{3} .$ $T h e l i n e l h a s v e c t o r e q u a t i o n$ $r = λ [(c o s θ + \sqrt{3}) i + (\sqrt{2} s i n θ) j + (c o s θ - \sqrt{3}) k]$ $w h e r e λ i s a s c a l a r p a r a m e t e r .$ $I f l m e e t s t h e p l a n e a t P, s h o w t h a t,$ $a s θ v a r i e s, P d e s c r i b e s a c i r c l e .$
	Answered by 123456 last updated on 15/Oct/15

	$r = λ [(\cos θ + \sqrt{3}) i + \sqrt{2} \sin θ j + (\cos θ - \sqrt{3}) k]$ $x - z = 4 \sqrt{3}$ $λ (\cos θ + \sqrt{3}) - λ (\cos θ - \sqrt{3}) = 4 \sqrt{3}$ $2 λ \sqrt{3} = 4 \sqrt{3}$ $λ = 2$ $. . . . .$ $rotation of axis$ $[\begin{matrix} x \\ z \end{matrix}] = [\begin{matrix} \cos 45 ° & - \sin 45 ° \\ \sin 45 ° & \cos 45 ° \end{matrix}] [\begin{matrix} x_{r} \\ z_{r} \end{matrix}]$ $[\begin{matrix} \cos 45 ° & \sin 45 ° \\ - \sin 45 ° & \cos 45 ° \end{matrix}] [\begin{matrix} x \\ z \end{matrix}] = [\begin{matrix} x_{r} \\ z_{r} \end{matrix}]$ $. . . . .$ $[\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} λ (\cos θ + \sqrt{3}) \\ λ \sqrt{2} \sin θ \\ λ (\cos θ - \sqrt{3}) \end{matrix}]$ $[\begin{matrix} x_{r} \\ y_{r} \\ z_{r} \end{matrix}] = [\begin{matrix} λ \sqrt{2} \cos θ \\ λ \sqrt{2} \sin θ \\ - λ \sqrt{6} \end{matrix}]$ $. . . . . .$ $x_{r}^{2} + y_{r}^{2} = 2 λ^{2} (equation of circle)$
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