Find the equation for the plane through the points Po(3,-2,5), Qo(-3,-1,-5) and Ro(0,-4,4). The equation of the plane is:

21x-24y-15z=36

Step-by-step explanation:

The equation of a plane is given a s

ax+by+cz=d

where a,b,c and d are  gotten from the vector product of the vector define by subtracting  one of the given points from the other two

Hence we define the vectors as follow  

<3,-2,5> - <-3,-1,-5> = <6,-1,10>

also <0,-4,4> - <-3,-1,-5>=<3,-3,9>

Next we need to carry out the cross product of the newly formed vector

<6,-1,10> X <3,-3,9> =<21,-24,-15>

The newly formed vector is in orthogonal to both vector and in  direction to the normal vector to the plane.

Since ax+by+cz=d  is the normal vector, we can conclude that

a=21, b=-24 c=-15

Hence we have

21x-24y-15z=d

if we plug in the point <-3,-1,-5> to solve for "d" we arrive at

21(-3)-24(-1)-15(-5)=36

Hence the final equation is

21x-24y-15z=36

let m = monomial

let t = trinomial

so, m times t = t.

monomial times a trinomial = another trinomial.

30005

yes ok