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

1. 3.4/4

2. 0.8/7

3. 0.5 x 3

4. 0.9 x 6

5. 10.8

6. 2.35

7. 0.567

8. 9.35

9. 0.02

10. 0.06

11. 0.16

12. 0.42

13. 0.63

14. 5.6

15. 8.1

16. 9.9

17. 0.28

18. 0.48

19. 0.27

20. 0.49

21. 0.45

22. 0.72

23. 0.9

24. 0.88

25. 0.75

26. 0.16

27. 0.328

28. 0.495

29. 0.72

30. 1.068