0

Guest

Courses New

find the equation of a plane which is at a distance of 3 root 3 units from the origin and the normal to which is equally inclined to the coordinate axes.

find the equation of a plane which is at a distance of 3 root 3 units from the origin and the normal to which is equally inclined to the coordinate axes.

Grade:12

3 Answers

Shobhit Varshney IIT Roorkee
askIITians Faculty 33 Points
9 years ago
Hi,

You know the direction cosine of normal to the plane and u know the point lying on that plane ( say, (3,3,3) in first quadrant) .So, you would get four such planes depending on the quadrant.

thanks................................................................................................
Ankit Jaiswal
165 Points
8 years ago
You know the direction cosine of normal to the plane and u know the point lying on that plane ( say, (3,3) in first quadrant) .So, you would get four such planes depending on the quadrant.
kush srivastava
13 Points
7 years ago
given tht direction ratios are (1,1,1) and perpendicular distance as given 3√3 so,, d/√(1²+1²+1²) =3√3=> d=9 so the required equation is x+y+z=9

Think You Can Provide A Better Answer ?

Other Related Questions on Vectors

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More