Find an equation of the plane with x-intercept a, y-intercept b, and z-intercept c. What is the distance between the origin and the plane?

The equation is:

(1/a)x + (1/b)y + (1/c)z = 1

Step-by-step explanation:

The direction vector between the points (a, 0, 0) and (0, b, 0) is given as:

<0 - a, b - 0, 0 - 0>

<-a, b, 0> (1)

The direction vector between (0, 0, c) and (0, b, 0) is given as:

<0 - 0, b - 0, 0 - c>

= <0, b, -c> (2)

To obtain the direction vector that is normal to the surface of the plane, we take the cross product of (1) and (2).

Doing this, we have:

<-a, b, 0> × <0, b, -c> = <-bc, -ac, -ab>

To find the scalar equation of the plane we can use any of the points that we know. Using (0, b, 0), we have:

(-bc)x + (-ac)y + (-ab)z = (-bc)0 + (-ac)b + (-ab)0

(bc)x + (ac)y + (ab)z = (ac)b

Dividing both sides by abc, we have:

(1/a)x + (1/b)y + (1/c)z = 1

this "question" looks more like a project is there any other info you can give me about this. it could be too complex for someone else to do on this website