Question

find the perpendicular distance of the origin from plane x-3y+4z-6=0.

Answer 1

d=| 1×0+(-3)×0+4×0-(-6)|/√1^2+(-3)^2+4^2.
=6/√26 unit.

Answer 2

The perpendicular distance of the origin from plane x - 3y + 4z - 6 = 0 is 6/√26 unit

Given :

The equation of the plane x - 3y + 4z - 6 = 0

To find :

The perpendicular distance of the origin

Solution :

Step 1 of 2 :

Write down the given equation of the plane

The equation of the plane is

x - 3y + 4z - 6 = 0

Step 2 of 2 :

Find perpendicular distance of the origin

The perpendicular distance of the origin (0,0,0)

$\displaystyle \sf{ = \bigg| \frac{0 - (3 \times 0) + (4 \times 0) - 6}{ \sqrt{ {1}^{2} + {( - 3)}^{2} + {4}^{2} } } \bigg| \: \: unit }$

$\displaystyle \sf{ = \bigg| \frac{0 - 0+ 0 - 6}{ \sqrt{ 1 + 9 + 16}} \bigg| \: \: unit }$

$\displaystyle \sf{ = \bigg| \frac{- 6}{ \sqrt{ 26}} \bigg| \: \: unit }$

$\displaystyle \sf{ = \frac{6}{ \sqrt{26} } \: \: unit }$