Question Number 1851 by 112358 last updated on 14/Oct/15
![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.](Q1851.png)
$${A}\:{plane}\:{has}\:{equation}\:{x}−{z}=\mathrm{4}\sqrt{\mathrm{3}}. \\ $$$${The}\:{line}\:{l}\:{has}\:{vector}\:{equation} \\ $$$$\boldsymbol{{r}}=\lambda\left[\left({cos}\theta+\sqrt{\mathrm{3}}\right)\boldsymbol{{i}}+\left(\sqrt{\mathrm{2}}{sin}\theta\right)\boldsymbol{{j}}+\left({cos}\theta−\sqrt{\mathrm{3}}\right)\boldsymbol{{k}}\right] \\ $$$${where}\:\lambda\:{is}\:{a}\:{scalar}\:{parameter}. \\ $$$${If}\:{l}\:{meets}\:{the}\:{plane}\:{at}\:{P},\:{show}\:{that}, \\ $$$${as}\:\theta\:{varies},\:{P}\:\:{describes}\:{a}\:{circle}.\: \\ $$
Answered by 123456 last updated on 15/Oct/15
![r=λ[(cos θ+(√3))i+(√2)sin θ j+(cos θ−(√3))k] x−z=4(√3) λ(cos θ+(√3))−λ(cos θ−(√3))=4(√3) 2λ(√3)=4(√3) λ=2 ..... rotation of axis [(x),(z) ]= [((cos 45°),(−sin 45°)),((sin 45°),(cos 45°)) ] [(x_r ),(z_r ) ] [((cos 45°),(sin 45°)),((−sin 45°),(cos 45°)) ] [(x),(z) ]= [(x_r ),(z_r ) ] ..... [(x),(y),(z) ]= [((λ(cos θ+(√3)))),((λ(√2)sin θ)),((λ(cos θ−(√3)))) ] [(x_r ),(y_r ),(z_r ) ]= [((λ(√2)cos θ)),((λ(√2)sin θ)),((−λ(√6))) ] ...... x_r ^2 +y_r ^2 =2λ^2 (equation of circle)](Q1853.png)
$$\boldsymbol{{r}}=\lambda\left[\left(\mathrm{cos}\:\theta+\sqrt{\mathrm{3}}\right)\boldsymbol{{i}}+\sqrt{\mathrm{2}}\mathrm{sin}\:\theta\:\boldsymbol{{j}}+\left(\mathrm{cos}\:\theta−\sqrt{\mathrm{3}}\right)\boldsymbol{{k}}\right] \\ $$$${x}−{z}=\mathrm{4}\sqrt{\mathrm{3}} \\ $$$$\lambda\left(\mathrm{cos}\:\theta+\sqrt{\mathrm{3}}\right)−\lambda\left(\mathrm{cos}\:\theta−\sqrt{\mathrm{3}}\right)=\mathrm{4}\sqrt{\mathrm{3}} \\ $$$$\mathrm{2}\lambda\sqrt{\mathrm{3}}=\mathrm{4}\sqrt{\mathrm{3}} \\ $$$$\lambda=\mathrm{2} \\ $$$$..... \\ $$$$\mathrm{rotation}\:\mathrm{of}\:\mathrm{axis} \\ $$$$\begin{bmatrix}{{x}}\\{{z}}\end{bmatrix}=\begin{bmatrix}{\mathrm{cos}\:\mathrm{45}°}&{−\mathrm{sin}\:\mathrm{45}°}\\{\mathrm{sin}\:\mathrm{45}°}&{\mathrm{cos}\:\mathrm{45}°}\end{bmatrix}\begin{bmatrix}{{x}_{{r}} }\\{{z}_{{r}} }\end{bmatrix} \\ $$$$\begin{bmatrix}{\mathrm{cos}\:\mathrm{45}°}&{\mathrm{sin}\:\mathrm{45}°}\\{−\mathrm{sin}\:\mathrm{45}°}&{\mathrm{cos}\:\mathrm{45}°}\end{bmatrix}\begin{bmatrix}{{x}}\\{{z}}\end{bmatrix}=\begin{bmatrix}{{x}_{{r}} }\\{{z}_{{r}} }\end{bmatrix} \\ $$$$..... \\ $$$$\begin{bmatrix}{{x}}\\{{y}}\\{{z}}\end{bmatrix}=\begin{bmatrix}{\lambda\left(\mathrm{cos}\:\theta+\sqrt{\mathrm{3}}\right)}\\{\lambda\sqrt{\mathrm{2}}\mathrm{sin}\:\theta}\\{\lambda\left(\mathrm{cos}\:\theta−\sqrt{\mathrm{3}}\right)}\end{bmatrix} \\ $$$$\begin{bmatrix}{{x}_{{r}} }\\{{y}_{{r}} }\\{{z}_{{r}} }\end{bmatrix}=\begin{bmatrix}{\lambda\sqrt{\mathrm{2}}\mathrm{cos}\:\theta}\\{\lambda\sqrt{\mathrm{2}}\mathrm{sin}\:\theta}\\{−\lambda\sqrt{\mathrm{6}}}\end{bmatrix} \\ $$$$...... \\ $$$${x}_{{r}} ^{\mathrm{2}} +{y}_{{r}} ^{\mathrm{2}} =\mathrm{2}\lambda^{\mathrm{2}} \:\:\:\:\:\:\left(\mathrm{equation}\:\mathrm{of}\:\mathrm{circle}\right) \\ $$