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O is any point in the plane, if A, B and C are collinear, it is proved that there exists a set of ordered number pairs (X, Y) such that vector OA=x vector OB+y vector OC, and x+y=1.
There are obviously: vector AB=vector OB-vector OA, vector BC=vector OC-vector OB.
A, B, C are collinear, vector AB=k vector BC, where k is a nonzero real number.
Vector OB-vector OA=k (vector OC-vector OB),
Vector OA=vector OB-k vector OC+k vector OB=(1+k) vector OB-k vector OC.
Let y=-k, x=1+k, then x+y=1.
Vector OA=x vector OB+y vector OC, and x+y=1.
Let O, B, and C be four points on the plane, vector OA=a, vector OB=b, vector OC=c, and a+b+c=0, a, b, and c are all -1. Find the absolute value of a + the absolute value of b + the absolute value of c Let O, B, C be the four points on the plane, vector OA=a, vector OB=b, vector OC=c, and a+b+c=0, a, b, c are all -1. Find the absolute value of a + the absolute value of b + the absolute value of c
Because a+b+c=0, therefore:
A*(a+b+c)=a*0(note: zero vector)
I.e.|a|2+a*b+a*c=0(Note: Exponent 0)
Also a*b=a*c=-1
So:|a|2=-a*b-a*c=2
Solution:|a|=√2
Similarly, b*(a+b+c)=b*0 and c*(a+b+c)=c*0 can also be solved as follows:
|B|=|c|=√2
So:|a b c|=3√2
Known vector OA=(3,-4), OB=(6,-3), OC=(5-m,-3-m). (1) If points A, B and C are collinear, calculate the value of m; (2) If △ABC is a right triangle, and ∠C=90°, take the value of m. Known vector OA=(3,-4), OB=(6,-3), OC=(5-m,-3-m). (1) If points A, B and C are collinear, calculate the value of m; (2) If △ABC is a right triangle and ∠C=90°, take the value of m.
(1) If points A, B, and C are collinear, then
AB =λ
AC,λ is a non-zero real number, so (3,1)=λ(2-m,1-m),
2λ-Mλ=3,λ-mλ=1,λ=2, m=1
2.
(2) ABC is a right triangle, and ∠C=90°,
CA•
CB=(m-2, m-1)•(m+1, m)=0,
M=1±
3.
Given the vector OA=(3,4) OB=(6.-3), OC=(5-m,-3-m). If ABC can form a right triangle, angle A=90, then m should satisfy? Given the vector OA=(3,4) OB=(6.-3), OC=(5-m,-3-m). If ABC can form a right triangle, the angle A=90, then m should satisfy?
Because angle A is a right angle, vector AB*vector AC=0
Vector AB=(3,-7)
Vector AC=(2-m,-7-m)
So 6-3m+49+7m=0
So m=-55/4
Given vector OA=(3,-4) vector OB=(6-3) vector OC=(5-m,-3-m) if point A, B, C can form right triangle with ∠A as right angle , Find the value of m Given vector OA=(3,-4) vector OB=(6-3) vector OC=(5-m,-3-m) if points A, B, C can form a right triangle with ∠A as a right angle , Find the value of m
If points A, B, C form a right triangle with ∠A as a right angle
Then AB*AC =0
I.e.(6-3,-3+4)*(5-m-6,-3-m+3)=0
3(-M-1)+1*(-m)=0
Solution m=-3/4
Let O be inside the triangle ABC and have 4 vectors OA+Vector OB+Vector OC=0, then the ratio of the area of triangle ABC to the area of triangle OBC is
Let BC midpoint be D,
Vector OB + Vector OC =2 Vector OD
4 Vector OA+vector OB+vector OC=vector 0
4 Vector OA+2 vector OD=vector 0
Vector OD=-2 Vector OA
So | A, O, D are collinear
|AD|=3/2|OD|
The ratio of the area of the triangle ABC to the area of the triangle OBC is 3/2
Let BC midpoint be D,
Vector OB + Vector OC =2 Vector OD
4 Vector OA + Vector OB + Vector OC = Vector 0
4 Vector OA+2 vector OD=vector 0
Vector OD=-2 Vector OA
So | A, O, D are collinear
| AD |=3/2| OD |
The ratio of the area of the triangle ABC to the area of the triangle OBC is 3/2
Let BC midpoint be D,
Vector OB + Vector OC =2 Vector OD
4 Vector OA+vector OB+vector OC=vector 0
4 Vector OA+2 vector OD=vector 0
Vector OD=-2 Vector OA
So | A, O, D are collinear
| AD |=3/2| OD |
The ratio of the area of the triangle ABC to the area of the triangle OBC is 3/2
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