Mathematics - Vector Question with Solution | TestHub

Sol. Let $\mathrm{P}(\cos \theta ,\sin \theta ,0)$ be a point on the circle then its distance d from the

$$\text{ plane }=d=\left|\frac{3-\cos \theta -\sin \theta }{\sqrt{3}}\right|=\left|\frac{3-\sqrt{2}\sin \left(\theta +\frac{\pi }{4}\right)}{\sqrt{3}}\right|$$

for maxima, $\theta =\frac{-3\pi }{4}$

$$\begin{array}{rl}& \text{ and for minima, }\theta =\frac{\pi }{4}\\ & \Rightarrow \mathrm{A}\left(-\frac{1}{\sqrt{2}},\frac{-1}{\sqrt{2}},0\right)\text{ and }\\ & \Rightarrow \left|{\mathrm{x}}_1+{\mathrm{y}}_1-{\mathrm{x}}_2-{\mathrm{y}}_2\right|=2\sqrt{2}\end{array}$$