Question

Find the distance between the point and the plane. (round your answer to three decimal places.) (1, 2, 1) x − y + 2z = 2

Answer 1

Answer : Distance of plane from a given point =0.408

Explanation:

Since the equation of plane is x-y+2z-2=0

and the given point is (1,2,1)

And we know the formula of distance of plane from a point

i.e.

D= $\mid\frac{Ax_{1}+By_1+Cz_1-D}{\sqrt{A^2+B^2+C^2}}\mid$

where A,B,C are the coefficient of x, y, z respectively ,

and $(x_1,y_1,z_1)$ are the given points .

So,

$\mid\frac{1.1+(-1).2+1.2-2}{\sqrt{1+1+4}}\mid$

= $\mid\frac{1-2+2-2}{\sqrt6}\mid$

= $\mid\frac{-1}{\sqrt6}\mid$

= $\frac{1}{\sqrt6}$

Now by using calculator we can find the value of √6 i.e. 2.449 upto three decimal places.

∴ Distance = 0.408