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

answer: are you

step-by-step explanation:

a

step-by-step explanation:

when a number is between two straight lines it means the absolute value.

to find absolute value you need to know how many numbers is that number away from 0.

when finding absolute value the number will always come out positive

so absolute value of -3 is 3

absolute value of 3/2 is 3/2

absolute value of -2 is 2

absolute value of 3.5 is 3.5

absolute value of -1 is 1

now put these numbers in order:

a)   |-1|, |3/2|, |-2|, |-3|, |3.5|