$\begin{array}{1 1}(0,1,0);\frac{8}{5} units \\ (0,0,1);\frac{8}{5} units \\ (0,-1,0);\frac{8}{5} units \\ (D) (0,-1,0);\frac{8}{5} units. \end{array} $

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

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

$\begin{array}{1 1}(0,1,0);\frac{8}{5} units \\ (0,0,1);\frac{8}{5} units \\ (0,-1,0);\frac{8}{5} units \\ (D) (0,-1,0);\frac{8}{5} units. \end{array} $

0 votes

- 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 $5y+8=0$

This can be written as $0x+5y+0z=-8$

(i.e)$0x-5y+0z=8$-----(1)

Therefore $\sqrt{0^2+(-5)^2+0^2}=5$

Step 2:

Divide on both sides of eq(1) by 5

$\large\frac{0}{5}$$x-\large\frac{5}{5}$$y+\large\frac{0z}{5}=\frac{8}{5}$

On simplifying we get

$\Rightarrow -y=\large\frac{8}{5}$

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 of the normal to the plane are $(0,-1,0)$ and distance from the origin is $\large\frac{8}{5}$units.

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...