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It is known that triangle ABC and point m satisfy Ma + MB + MC = 0. If there is a real number m such that ab + AC = mam holds, then M is equal to
A=OA B=OB C=OC M=OM
MA=OA-OM
MB=OB-OM
MC=OC-OM
MA+MB+MC=0
A+B+C=3M
3M-3A=B+C-2A
AB+AC=mAM
B-A+C-A=B+C-2A=m(M-A)
m=(B+C-2A)/(M-A)=3(B+C-2A)/(3M-3A)=3
If M is a point in the plane where △ ABC is located and satisfies (vector MB - vector MC) * (vector MB + vector MC) = 0, vector MB + vector MC + 2, vector Ma = 0 Then the shape of △ ABC is The answer is isosceles triangle. Please note that the first question is 0, and the second is 0 vector for complete analysis. Thank you
By (mb-mc) (MB + MC) = 0,
Get MB ²- MC ²= 0, i.e. | MB| ²-| MC| ²= 0
|MB|=|MC|,
So m is on the vertical bisector of edge BC
Thus, the diagonal of the diamond of the adjacent sides of the vector MB + MC,
That is, MB + MC is on the vertical bisector of segment BC,
And 2mA = - (MB + MC), collinear with MB + MC,
Thus, point a is on the vertical bisector of segment BC, so | ab | = | AC|
It is known that plane vectors a and B satisfy | a | = 3, | B | = 2, and the angle between vector a and vector B is 120 degrees. If (a + MB) ⊥ a, then the real number M=
|a|=3,|b|=2,=2π/3
Namely: a · B = |a| * |b| * cos (2 π / 3) = - 3
(a + MB) ⊥ a, that is: (a + MB) · a = 0
That is: | a | ^ 2 + Ma · B = 0
Namely: 9-3m = 0
Therefore: M = 3
It is known that a and B are two non-zero vectors. It is known that the included angle of vectors a and B is a, vector C = a + nomiga B, and the real number nomiga minimizes the absolute value of C ① It is known that a and B are two non-zero vectors. It is known that the included angle of vectors a and B is a, vector C = a + nomiga B, and the real number nomiga minimizes the absolute value of C ① find the value of nomiga ② when a = 45 degrees, it is proved that B is perpendicular to C
1. C ^ 2 = (a + XB) ^ 2 = (a + XB) ^ 2 + A ^ 2 + A ^ 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2. It is a quadratic function. It is easy to know that when x = - | a | a | cosa / b cosa, which is a second function, and when x = - | a | B (x)) ^ 2. 2. B * c * C = (a | a | B cos45 + XB ^ 2. At this time, x = - | a | cos45 / 124b 124; b * b * b * C = - | a | a | B | B | cos45 + (- | B | cos45 + (- | cos45 + (- | a a 124|b|cos45) = 0. Hope to adopt!
It is known that vectors a and B are non-zero vectors. When the modulus of a + TB takes the minimum value, find the value of T Thank you
u^2=a^2+t^2*b^2+2t*(ab)
Look at it as a univariate quadratic function about t, because t is a real number,
(1) When | u | gets the minimum value, the real number T = - (a • b) / b ^ 2,
For non-zero vectors a and B, a = (2,1) B = (1,2), find the value of the real number T when | a + TB |
|A + TB | minimum
Then (a + TB) ^ 2min = a ^ 2 + 2ta * B + T ^ 2B ^ 2 = 5 + 8t + 5T ^ 2
Finding the minimum value of quadratic function
Take at symmetry axis t = - 4 / 5
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