In a tetrahedral o-abc with edge length 2, the square of (OA vector + ob vector + OC vector) is The answer is 24

24. Make the vertical line of the bottom triangle from the vertex o, and the vertical line indicates that the bottom triangle is in H, then the value of the correlation vector can be written as Oh vector + HA vector. The other two are the same, then the problem is transformed into the square of (3OH vector), that is 9OH ^ 2, and the oh ^ 2 of the regular tetrahedron is 8 / 3, so the result is 24

It is known that P and Q are two points on the line AB, and AP: Pb = 3:5, AQ: QB = 3:4. If PQ = 6cm, find the length of ab

（3/7)x -（3/8)x=6
x=9/28

Given that the angle between vectors OA and ob is Pai / 3, | OA | = 4, | ob | = 1, if the point m is on the straight line ob, then the minimum value of | OA minus om |, is

|Oa-om | is the line segment of am, point a is fixed, and M is on the line ob,
This is the shortest distance from the point to the straight line, which is equal to the height of ob in direction a, and the included angle is π / 3, so we know that the minimum is 2 times root 3

Given that the angle between vectors OA and ob is 30 degrees, | OA | = 2, if the point m is on the straight line ob, then the minimum value of | vector OA - vector om |, is
There should be a more detailed analysis

Let two vectors have the same starting point, go through a to make the vertical line of ob, and the vertical foot is CAC, which is the minimum value. The answer is 1

In a tetrahedral o-abc with edge length 2, the square of (OA vector + ob vector + OC vector) is The answer is 24