Question

Find the intercepts on the coordinate axis by the plane 2x-3y+5z=4

Answer 1

The plane 2x+3y-4z=5 intersects the x-axis at (a,0,0), the y-axis at (0,b,0), and the Z-axis at (0,0,c).

Therefore, it is required for us to get the values of a, b and c.

Substitute with each co-ordinate in the equation of the plane:

1. (a,0,0):

$2a+0+0=5\\a=5/2$

2. (0,b,0):

$0+3b+0 = 5\\b = 5/3$

3. (0,0,c):

$0+0=4c = 5\\c = -5/4$

Answer 2

Answer:

$a = 2 \\ \\ b = \frac{ - 4}{3} \\ \\ c = \frac{4}{5}$

Solution:

To find the intercepts on the coordinates axis,convert the equation into intercept form.

we know that standard intercept equation of plane is

$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$
here a,b,c are intercept of x,y and z axis respectively

So

$2x - 3y + 5z = 4 \\ \\ \frac{2x}{4} + \frac{3y}{ - 4} + \frac{5z}{4} = 1 \\ \\ \frac{x}{ \frac{4}{2} } + \frac{y}{ \frac{ - 4}{3} } + \frac{z}{ \frac{4}{5} } = 1$
So, x -axis intercept (2,0,0)

Y-axis intercept (0,-4/3,0)

z-axis intercept(0,0,4/5)

Hope it helps you.