# Mathematics-I (3110014)

BE | Semester-1   Winter-2019 | 17-01-2020

## Q3) (c)

#### Find the maximum and minimum distance from the point $\left(1,2,2\right)$ to the sphere ${\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}=36$.

$\mathrm{D}=\sqrt{{\left(\mathrm{x}-1\right)}^{2}+{\left(\mathrm{y}-2\right)}^{2}+{\left(\mathrm{z}-2\right)}^{2}}$

$\mathrm{f}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)+\mathrm{\lambda }·\mathrm{\phi }\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=0$

$⟹2\left(\mathrm{x}-1\right)+\mathrm{\lambda }\cdot \left(2\mathrm{x}\right)=0$

$⟹\mathrm{\lambda }\cdot \left(2\mathrm{x}\right)=-2\left(\mathrm{x}-1\right)$

$⟹\mathrm{\lambda }=\frac{1-\mathrm{x}}{\mathrm{x}}=\frac{1}{\mathrm{x}}-1$

$⟹2\left(\mathrm{y}-2\right)+\mathrm{\lambda }\cdot \left(2\mathrm{y}\right)=0$

$⟹\mathrm{\lambda }\cdot \left(2\mathrm{y}\right)=-2\left(\mathrm{y}-2\right)$

$⟹\mathrm{\lambda }=\frac{2-\mathrm{y}}{\mathrm{y}}=\frac{2}{\mathrm{y}}-1$

$⟹2\left(\mathrm{z}-2\right)+\mathrm{\lambda }\cdot \left(2\mathrm{z}\right)=0$

$⟹\mathrm{\lambda }\cdot \left(2\mathrm{z}\right)=-2\left(\mathrm{z}-2\right)$

$⟹\mathrm{\lambda }=\frac{2-\mathrm{z}}{\mathrm{z}}=\frac{2}{\mathrm{z}}-1$

$\frac{1}{\mathrm{x}}-1=\frac{2}{\mathrm{y}}-1$

$⟹\frac{1}{\mathrm{x}}=\frac{2}{\mathrm{y}}$

$⟹\mathrm{y}=2\mathrm{x}$

$\frac{1}{\mathrm{x}}-1=\frac{2}{\mathrm{z}}-1$

$⟹\frac{1}{\mathrm{x}}=\frac{2}{\mathrm{z}}$

$⟹\mathrm{z}=2\mathrm{x}$

${\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}=36$

$⇒{\mathrm{x}}^{2}+4{\mathrm{x}}^{2}+4{\mathrm{x}}^{2}=36$

$⇒9{\mathrm{x}}^{2}=36$

$⇒{\mathrm{x}}^{2}=4$

$⇒\mathrm{x}=±2$

$\mathrm{Now},$

$\mathrm{y}=2\mathrm{x}=2\left(±2\right)=±4$

$\mathrm{z}=2\mathrm{x}=2\left(±2\right)=±4$

$\mathrm{D}=\sqrt{{\left(\mathrm{x}-1\right)}^{2}+{\left(\mathrm{y}-2\right)}^{2}+{\left(\mathrm{z}-2\right)}^{2}}=\sqrt{{\left(2-1\right)}^{2}+{\left(4-2\right)}^{2}+{\left(4-2\right)}^{2}}=3$

$\mathrm{D}=\sqrt{{\left(\mathrm{x}-1\right)}^{2}+{\left(\mathrm{y}-2\right)}^{2}+{\left(\mathrm{z}-2\right)}^{2}}=\sqrt{{\left(-2-1\right)}^{2}+{\left(-4-2\right)}^{2}+{\left(-4-2\right)}^{2}}=9$