# In each of the following cases determine the direction cosines of the normal to the plane and the distance from origin(c) $2x+3y-z=5$

This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com

Toolbox:
• The equation of normal to a plane is $lx+my+nz=d$,where $l,m,n$ are the direction cosines and d is the distance of the perpendicular from the origin.
Step 1:
The given equation is $2x+3y-z=5$---(1)
The direction ratios of the normal are $(2,3,-1)$
Therefore $\sqrt{2^2+3^2+(-1)^2}=\sqrt{14}$
Step 2:
Divide equ(1) on both sides by $\sqrt{14}$
$\large\frac{2}{\sqrt{14}}$$x+\large\frac{3}{\sqrt{14}}$$y-\large\frac{z}{\sqrt{14}}=\frac{5}{\sqrt{14}}$
This equation of the form $lx+my+nz=d$,where $l,m$ and $n$ are the direction cosines of normal to the plane and $d$ is the distance of normal from the origin.
Therefore the direction cosines are $\big(\large\frac{2}{\sqrt {14}},\large\frac{3}{\sqrt {14}},\large\frac{-1}{\sqrt {14}}\big)$ and distance from the origin is $\large\frac{5}{\sqrt {14}}$units.