Checking result on review exercise

Question

We have four points M1(1, 1, −1), M2(−1, 3, 2), M3(2, −1, 0) и M4(0, 1, −2)

1.Middle point A which is between M3M4

Solution:

A(-1, 1, -1)

2. Dot product M1M2 of M1M3

Solution:

M1M2(0, 2, 3), M1M3(1, -2, 1)

(M1M2, M1M3) = 0*1 +  2*-2 +  1* 3 = 0 

3. Length of M1M2;

Solution:

|M1M2| = sqrt(0 + 4 + 9) = sqrt(13)

4. Angle between M1M2 and M1M3

Solution:

cosa = M1M2*M1M3/ |M1M2||M1M3|

M1M2 = 0 => a = PI/2

5.Cross product of M1M2 × M1M4

Solution:

M1M2 × M1M4 = |1, 1, 1  | = 6 
              |0, 2, 3  |
              |2, -2, -1|

6.Area of M1M2M4

Solution: S = 3

7. Combined Product (M1M2 ,M1M3, M1M4):

Solution:

HOW DO I GET The Dot product of these two
|1, 1, 1|
|0, 2, 3|      * M1M4
|1, -2, 1|

Volume of M1M2M3M4:

Can someone check if these are correct and how can I solve the last two ?

Answer 1

"We have four points M1(1, 1, −1), M2(−1, 3, 2), M3(2, −1, 0) и M4(0, 1, −2)

1.Middle point A which is between M3M4

Solution: A(-1, 1, -1)"

No, this is not correct. The midpoint of a segment between two points is the point whose coordinates are the "average" of the coordinates of the given two points. Here, that would be ((2+ 0)/2, (-1+ 1)/2, (0- 2)/2)= (1, 0, -1).

"2.Dot product M1M2 of M1M3 Solution: M1M2(0, 2, 3), M1M3(1, -2, 1)

(M1M2, M1M3) = 0*1 + 2*-2 + 1* 3 = 0" No. To begin with M1M2 is not (0, 2, 3). It is (-1- 1, 3- 1, 2- (-1))= (-2, 2, 3). M1M3 is (2- 1, -1- 1, 0- (-1))= (1, -2, 1) so the dot product is (-2)(1)+ 2(-2)+ 3(1)= -2- 4+ 3= -3.

"3. Length of M1M2; Solution: |M1M2| = sqrt(0 + 4 + 9) = sqrt(13)" No, since M1M2 is NOT (0, 3, 2).

"4.Angle between M1M2 and M1M3 Solution: cosa = M1M2*M1M3/ |M1M2||M1M3| M1M2 = 0 => a = PI/2" No, again because you have 2 wrong.

"5.Cross product of M1M2 × M1M4 Solution: M1M2 × M1M4 = |1, 1, 1 | = 6 |0, 2, 3 | |2, -2, -1|

Not only because you have M1M2 and M1M4 wrong but you are completely mistaken as to what a "cross product" is! The cross product of two vectors is a vector, not a number.
With M1M3= (-2, 2, 3) and M1M4= (-1, 0, -1), the cross product is $\left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ -2 & 2 & 3 \\ -1 & 0 & -1\end{array}\right|= \vec{i}\left|\begin{array}{cc}2 & 3 \\ 0 & -1\end{array}\right|- \vec{j}\left|\begin{array}{cc}-2 & 2 \\ -1 & -1\end{array}\right|+ \vec{k}\left|\begin{array}{cc}-2 & 2 \\ -1 & 0 \end{array}\right|= (-2- 0)\vec{i}- (2+ 2)\vec{j}+ (0+ 2)\vec{k}= (-2, 4, 2)$

I don't know what is meant by "M1M2M4", the "combined product" or "M1M2M3M4".