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M (3,-2), N (-5,-1) is known, and MP=1 2 MN, the coordinates of point P are ______.
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Given M (3,-2), N (-5,-1), and vector MP=1/2 vector MN, then the coordinates of p are
Let P coordinate be (x, y)
Vector MP=(x, y)-(3,-2)=(x-3, y+2)
Vector MN=(3,-2)-(-5,-1)=(8,-1)
Because vector MP =1/2 vector MN
So (x-3, y+2)=1/2(8,-1)
X-3=4
Y+2=-1/2
So x=7, y=-5/2
So P's coordinates are (7,-5/2)
Given two points M (3,2), N (-5,5), vector MP=1/2 vector MN, then the coordinates of point P are () You knowll know what to choose.
Vector MN=(-8,3)
Vector MP=(-4,3/2)
Let P coordinate (x, y)
X-3=-4
Y-2=3/2
X=-1, y=3.5
I.e. the P coordinate is (-1,3.5)
Known vector A =(2,3), B=(-1,2) if m A+ B and A-2 B parallel, m equals ___.
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Let a, b be two non-collinear non-zero vectors, t be R (1) If a is the same as b, where t is, the midpoint of a, tb,1/3(a+b) is in a straight line? (2) If the absolute value a = the absolute value b and the included angle is 60°, then why is the value of the absolute value a-tb the smallest? Let a, b be two non-collinear non-zero vectors, t be R (1) If a is the same as b, where is t, the midpoint of a, tb,1/3(a+b) is in a straight line? (2) If the absolute value a = the absolute value b and the included angle is 60°, then why is the value of the absolute value a-tb the smallest?
(1) On a straight line, the point multiplication of the difference between the three vectors equals 0, i.e.(tb/2-a/2)*[1/6(a+b)-a/2]=0=> t=(ab-2a^2)/(b^2-2ab)(2)|a-tb|^2=(a-tb)*(a-tb)=a^2-2tab+b^2=2-2t|a||b||cos60 b|^2= a|^2(2-t) Because |a-tb|>=0,|a-tb|^2...
If the vectors a, b are two non-collinear non-zero vectors, t belongs to R. If the starting points of the vectors a, b are the same, and t is the value,1/3(a+b) of the three vector ends are straight... If the vectors a, b are two non-collinear non-zero vectors, t belongs to R. If the starting points of the vectors a, b are the same, what is the value of t?
Zero vector
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