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O is the inner triangle if and only if aOA vector + bOB vector + cOC vector =0 vector.
Sufficiency: Given aOA vector + bOB vector + cOC vector =0 vector, extend CO to AB to D, add according to the vector to get: OA=OD+DA, OB=OD+DB, substitute into the known: a (OD+DA)+b (OD+DB)+ cOC =0, because OD is collinear with OC, so OD=kOC can be set, the formula can be transformed into (ka+kb+c) OC+(aDA+bDB)=0 vector, direction...
Three vectors can form a triangle if and only if? Such as title
0
Using Vector Method to Prove that the Centerline of a Triangle Intersects at a Point
The following provides you with two kinds of proof methods, please feel free to use them.(The vector sign can not be obtained, which may cause inconvenience to you in reading, etc.) The proof method 1 is to draw first, and make two middle lines of B and C, which are respectively to AC at M and AB at N, so M, N are the midpoint of AC and AB. Connect MN and set vector BP=λ vector PM, vector CP=μ direction...
The following provides you with two kinds of proof methods, please feel free to use them.(The vector sign can not be obtained, which may cause inconvenience to you in reading, etc.) The proof method 1 is to draw first, and make two middle lines of B and C, and respectively cross AC at M and AB at N, so M and N are the midpoint of AC and AB. Connect MN and set vector BP=λ vector PM, vector CP=μ direction...
The following provides you with two methods of proof, please feel free to use them.(The vector sign can not be obtained, which may cause inconvenience to you in reading, etc.) The method of proof 1 is to draw first, and make two center lines of B and C, and then to AC at M and AB at N respectively, so M and N are the midpoint of AC and AB. Connect MN and set vector BP=λ vector PM, vector CP=μ direction...
Why does the necessary and sufficient condition of vector collinearity emphasize that a is non-zero? So what if a is zero in b=ka?
The direction of the zero vector is arbitrary...
Vector a (2,-1), b (4,3) Find the Value of a·b
Where vector a x b =2 x 4+(-1) x 3=5. Vector multiplication is abscissa multiplication by abscissa plus ordinate multiplication by ordinate.
Where vector a x b =2 x 4+(-1) x 3=5. Vector multiplication is abscissa multiplied by abscissa plus ordinate multiplied by ordinate.
Given vector A =(4,3), vector B =(-1,2) the cosine of the angle D between vector A and vector B
Let vector A=a, vector B=b, from a*b=|a||b|cosD
CosD=[ a*b ]/[|a||b|]=[4x (-1)+3x2]/[5×√5]=2√5/25
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