Question

Determine the distance between the x-intercept and the z-intercept of the plane whose equation is 2x+9y-3z=18

Answer 1

Given Equation is 2x + 9y - 3z = 18.

(1) On x - intercept y = 0 and z = 0.

So, x = 18/2 = 9.

(2) On y - intercept x = 0 and z = 0.

So, y = 18/9 = 2.

(3) On z - intercept x = 0 and y = 0.

So, z = 18/-3 = -6.

Now,

Distance between x - intercept and z - intercept is given by:

$= > \sqrt{x^2 + z^2}$

$= > \sqrt{9^2 + (-6)^2}$

$= > \sqrt{81 + 36}$

$= > \sqrt{117}$

$= > \boxed{10.81}$

Hope this helps!

Answer 2

Given:

Plane: 2 x + 9 y - 3 z = 18

To find:

• X intercept
• Z intercept
• Distance

Solution:

To find x intercept,

Put,

• y = 0
• z = 0

Substituting,

We get,

x = 18 / 2

x = 9

To find z intercept,

Put x = 0 and y = 0.  

Substituting,

We get,

z = 18 / -3

z = -6

Distance = √ ( ( x )^2 + ( z )^2 )

√ ( 9 )^2 + ( -6 )^2

√ 81 + 36

√ 117

Hence, Distance = 10.8 units