Question

Find the point on the plane 4x+3y+z=10 that is nearest to​ (2,0,1). What are the values of​ x, y, and z for the​ point? x= 28 / 13 y = 3 / 26 z= 27 / 26 ​(Type integers or simplified​ fractions.)

Answer 1

Answer:

$\frac{1}{\sqrt{26}}$.

Step-by-step explanation:

The minimun distance between a point and a plane is the perpendicular distance. The formula is

d = $\frac{|Ax_0+By_0+Cz_0+D|}{\sqrt{A^{2}+B^{2}+C^{2}}}$

where $(x_0,y_0,z_0)=(2,0,1)$, A=4, B=3, C=1 and D=-10. So, the distance is

d = $\frac{|8+1-10|}{\sqrt{16+9+1}}$

d =  $\frac{|-1|}{\sqrt{26}}$

d = $\frac{1}{\sqrt{26}}$.