$\begin{array}{1 1}(A) \large\frac{-5}{2},5,5 \\ (B) \large\frac{-5}{2},-5,5 \\(C) \large\frac{5}{2},5,-5 \\ (D) \large\frac{2}{5},5,-5\end{array} $

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

- Equation of a plane in intercept form is $\large\frac{x}{a}$$+\large\frac{y}{b}$$+\large\frac{z}{c}$$=1$
- Where $a,b,c$ are the intercepts cut off by the plane at $x,y$ and $z$ respectively.

Step 1:

The given equation is $2x+y-z=5$

Divide throughout by 5,we get

$\large\frac{x}{\Large\frac{2}{5}}$$+\large\frac{y}{5}$$+\large\frac{z}{-5}$$=1$

Step 2:

Equation of a plane in intercept form is $\large\frac{x}{a}$$+\large\frac{y}{b}$$+\large\frac{z}{c}$$=1$

Where $a,b,c$ are the intercepts cut off by the plane at $x,y$ and $z$ axes respectively.

Here for the given equation

$a=\large\frac{2}{5}$$,b=5$ and $c=-5$

Thus the intercepts cutoff by the plane are $\large\frac{5}{2}$$,5$ and $-5$

Ask Question

Tag:MathPhyChemBioOther

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...