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Space vector minimum problem Given a=(1-t,1-t, t), b (2, t, t), then what is the minimum value of |b-a|? == Lx is wrong. And because it's (2t-1)^2 and how did you get that 1/25?
B-a=(t+1,2t+1,0)
|B-a|=Root [(t+1)^2+(2t-1)^2]=Root (5t^2-2t+2)>=Root 49/25=7/5
Don't you know the formula?
B-a=(t+1,2t+1,0)
|B-a|=Root [(t+1)^2+(2t-1)^2]=Root (5t^2-2t+2)>=Root 49/25=7/5
Don't you know the recipe?
Given A (1,2), B (-3,2), AM vector =3/5 vector A B, then the coordinates of point M are A.(-1,8/5) B.(-7/5,2) C.(5,2/5) D.5,-2/5)
B vector AB=vector B-vector A=(-3,2)-(1,2)=(-4,0)3/5 vector AB=(-12/5,0) AM vector=vector M-vector A vector M=(-12/5,0)-(1,2)=(-7/5,2)
Given points A (2,1), B (-1,-2), and vector AM =2/2 of vector A B, then the coordinates of point M are Given points A (2,1), B (-1,-2) and vector AM =2/2 of vector A B, then the coordinates of point M are Given points A (2,1), B (-1,-2) and vector AM=2/2 vector A B, then the coordinates of point M are
Since vector AM=2/3*AB,
So OM-OA=2/3*(OB-OA),
Therefore OM=1/3*OA+2/3*OB
=1/3*(2,1)+2/3*(-1,-2)
=(2/3,1/3)+(-2/3,-4/3)
=(0,-1) ,
That is, the M coordinate is (0,-1).
Let a1,a2...am be a standard orthogonal vector group of n dimensional Euclidean space V. It is proved that for any vector a in V there is ∑(a, a i)^2
The standard orthogonal basis a1,a2...am,...,an any vector a extending a1,a2...am to V can be expressed as a=k1a1+k2a2+...+kmam+...+knan (a, ai)=ki||a|^2=(a, a)=(a, k1a1+k2a2+...+kmam+...+knan)=∑(a, kiai)=∑ki (a, ai)=∑(a, ai)^2>=∑(a, a)|^2>
Given △ABC and point M, for any point O in in space, vector OM=1/3(vector OA+vector OB+vector OC), if vector AB+vector AC=m vector AM How much is m?
1.MA,MB,MC is coplanar
Just prove MA+MB+MC=0
MA=OA-OM
MB=OB-OM
MC=OC-OM
MA+MB+MC=OA+OB+OC-3OM=0
In fact, the first question can be judged that ABCM is coplanar, and the second question is to prove that M point is within ABC
If the M point is outside, the sum of any one or two vectors can not be reversed from the third vector
So it's the M point.
As shown in the figure, in triangle ABC, point O is the focus of BC, the lines passing through point O intersect line AB respectively, AC is at different two points M, N if vector AB=m vector AM Vector AC=n vector AN, then m+n=?
Point O is the midpoint of BC, so vector AO=1/2(AB+AC). Vector OM=AM-AO=1/mAB-1/2(AB+AC)=(1/m-1/2) A-1/2AC Vector ON=AN-AO=1/nAC-1/2(AB+AC)=-1/2 AB+(1/n-1/2) AC, as known, vector OM is collinear with vector ON, then (1/m-1/2)/(-1/2)=-1/2/(1/n-1/2)(...
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