Find the point on the plane 2x - y + 2z = 20 nearest the origin.

Solution:

Given, the equation of the plane is 2x - y + 2z = 20

We have to find the point on the plane nearest to the origin.

If the equation of the plane is given by ax + by + cz = d, then

Distance = \(\frac{ap_{x}+bp_{y}+cp_{z}}{\sqrt{a^{2}+b^{2}+c^{2}}}\)

The point on the plane ax + by + cz = d nearest the origin is

\((\frac{da}{\sqrt{a^{2}+b^{2}+c^{2}}},\frac{db}{\sqrt{a^{2}+b^{2}+c^{2}}},\frac{dc}{\sqrt{a^{2}+b^{2}+c^{2}}})\)

The point on the plane 2x - y + 2z = 20 nearest the origin is

\((\frac{1\times 2}{\sqrt{(2)^{2}+(-1)^{2}+(2)^{2}}},\frac{1\times -1}{\sqrt{(2)^{2}+(-1)^{2}+(2)^{2}}},\frac{1\times 2}{\sqrt{(2)^{2}+(-1)^{2}+(2)^{2}}})\)

= \((\frac{2}{\sqrt{4+1+4}},\frac{-1}{\sqrt{4+1+4}},\frac{2}{\sqrt{4+1+4}})\)

= \((\frac{2}{\sqrt{9}},\frac{-1}{\sqrt{9}},\frac{2}{\sqrt{9}})\)

= \((\frac{2}{3},\frac{-1}{3},\frac{2}{3})\)

Therefore, the point nearest the origin is \((\frac{2}{3},\frac{-1}{3},\frac{2}{3})\).

Find the point on the plane 2x - y + 2z = 20 nearest the origin.

Summary:

The point on the plane 2x - y + 2z = 20 nearest the origin is \((\frac{2}{3},\frac{-1}{3},\frac{2}{3})\).