Find the intercepts cut off by the plane 2x + y

Home » Find the intercepts cut off by the plane 2x + y – z = 5.

Find the intercepts cut off by the plane 2x + y – z = 5.

Find the intercepts cut off by the plane 2x + y – z = 5.

Solution:

It is given that
The plane $2 x+y-z=5$
Let us express the equation of the plane in intercept form
$x / a+y / b+z / c=1$
Where $a, b, c$ are the intercepts cut-off by the plane at $x, y$ and $z$ axes respectively.
$2 x+y-z=5 \ldots \text { (1) }$
Dividing both sides of eq(1) by 5 , we obtain
$\begin{array}{l} 2 x / 5+y / 5-z / 5=5 / 5 \\ 2 x / 5+y / 5-z / 5=1 \\ x /(5 / 2)+y / 5+z /(-5)=1 \end{array}$
Here, $a=5 / 2, b=5$ and $c=-5$
As a result, the intercepts cut-off by the plane are $5 / 2,5$ and $-5$ .