It is known that the three side lengths of △ ABC are ab = 7, BC = 5 and Ca = 6 respectively, then AB• The value of BC is __

It is known that the three side lengths of △ ABC are ab = 7, BC = 5 and Ca = 6 respectively, then AB• The value of BC is __

From the cosine theorem, CoSb = AB2 + BC2 − ac2
AB × BC＝19
35，
AB•
BC＝|
AB||
BC|cos(π−B)＝−7 × five × nineteen
35＝−19
So the answer is: - 19

Mathematics elective course 2-1, known points a (2,0,0), B (0,5,0), C (0,0,3), find the unit normal vector of plane ABC____ The answer is to write the coordinates of vectors AB and AC, and let n (x, y, z) list two ternary primary equations, and change them to 0

A (2,0,0), B (0,5,0), C (0,0,3), ab = (- 2,5,0), AC = (- 2,0,3) let the normal vector of plane ABC n = (x, y, z) then n · AB = 0, n · AC = 0, so {- 2x + 5Y = 0 {- 2x + 3Z = 0, the solution is: y = 2 / 5 * x, z = 2 / 3 * x, then n = (x, 2 / 5 * x, 2 / 3 * x}, take x = 1, get: n = (1,2 / 5,2 / 3} do not set X = 0, at this time, n

High school mathematics known point a (3.0.0) B (0.4.0) C (0.0.5) to find the unit normal vector of plane ABC, there is a process

Letters all represent vectors AB = (- 3,4,0), AC = (- 3,0,5). Let: the normal vector is n = (x, y, z), then: n ⊥ AB, then: ab * n = 0, get: - 3x + 4Y = 0n ⊥ AC, then: AC * n = 0, get: - 3x + 5Z = 0, substitute x = 20, get: y = 15, z = 12, that is, a normal vector is: n = (20,15,12) the unit normal vector is: n '= (20

In the equilateral triangle ABC with side length 1, let vector AB = vector C, vector BC = vector a, and vector CA = vector B, then vector a * vector B + vector b * vector C+ Vector c * vector a =?

a.b+b.c + c.a
=BC.CA + CA.AB + AB.BC
=|BC||CA|cos120° + |CA||AB|cos120° + |AB||BC|cos120°
= -3/2

In the regular triangle ABC with side length of 1, let the vectors BC = a, CA = B, ab = C. Then AB + BC + Ca equals!

The vector obeys the triangle principle and gets 0

The three vertices of triangle ABC are a (4,1) B (2, - 1) C (0,5) point D. on AB, vector ad = 2 vector dB, find the coordinates of point E on AC so that de bisects the area of triangle ABC

A (4,1) B (2, - 1) point D on AB vector ad = 2 vector DBD is a near B trisection point of ab. it is easy to prove that D (8 / 3, - 1 / 3) easy to prove AC: x + Y-5 = 0, so the distance from D to AC = (4 root sign 2) / 2b to AC = 2 root sign 2|ac| = 4 root sign 2S triangle ABC = 8s triangle ade = 4|ae| = 2 root sign 22|ae| = |ac| e is the midpoint of a and c