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In the triangle ABC, we know that D is a point on the edge of ab. if vector ad = 2 vector dB, vector CD = 1 / 3 vector Ca + X vector CB, then x is equal to A.2/3 B.1/3 C.-1/3 D.-2/3
According to the basic theorem of plane vector, any two non collinear vectors in the plane can be used as the base, any direction can be represented by base vector, and the expression is unique. Because vector ad = 2 vector dB, vector CD - vector CA = 2 (vector CB - vector CD) so 3 vector CD = vector Ca + 2 vector CB, so vector CD
In the triangle ABC, it is known that D is a point on the edge of ab. if the vector ad = 2dB, the vector CD = 1 / 3CA + λ CB, then λ is equal to () A.2/3 B.1/3 C.-1/3 D.-2/3
λ=2/3
Ad = 2dB, so D is ab trisection
If CE = 1 / 3CA, e is on Ca, then E is ca trisection
DE//CB
According to the law of vector addition,
There is CF = 2 / 3CB such that CEDF is a parallelogram,
So λ = 2 / 3
In the triangle ABC, we know that D is a point on the edge of ab. if the vector ad = 2dB, the vector CD = 1 / 3CA + λ CB, then λ is equal to?
Vector AB = cb-ca,
Vector 2ad = 2dB,
Then the vector ad = 2 / 3AB = 2 / 3 (cb-ca) = 2 / 3cb-2 / 3CA,
The vector CD = Ca + ad = 1 / 3CA + 2 / 3CB,
That is, λ = 2 / 3
In the triangle ABC, if AB = 2, AC = 3, D is the midpoint of the edge BC, then the vector ad * vector BC =?
We can extend the segment ad, do de = ad, and then connect be and CE
Vector ad = 1 / 2 vector AE
Vector ad · vector BC = 1 / 2 vector AE · vector BC
=1 / 2 (vector AC + vector AB) (vector AC - vector AB)
=1 / 2 (modulus of AC - modulus of AB)
=1/2（9-4）
=2.5
In the triangle ABC, we know that D is a point on the edge of ab. if vector ad = 2 vector dB, vector CD = 1 / 3 vector Ca + μ vector CB, then μ =?
Change the following λ into μ in your topic. From vector CD = 1 / 3 vector Ca + λ vector CB, vector Ca + vector ad = 1 / 3 vector Ca + λ (vector Ca + vector AB) (λ - 2 / 3) vector CA = λ vector ab - vector ad because vector ad = 2 vector dB, we get (λ - 2 / 3) vector CA = (3 λ / 2-1) vector AD due to vector Ca and vector ad
In the triangle ABC, a (7,8). B (3,5), C (4,3), m, N, D are AB, AC, BC. The intersection of Mn and adj at f is used to calculate the vector DF
In the triangle ABC, a (7,8), B (3,5), C (4,3), m, N, D are the midpoint of AB, AC, BC respectively. Mn and ad intersect F, and the vector DF is calculated
M (5,6.5), n (5.5,5.5), D (3.5,4), f (5.25,6), so the vector DF = (1.75,2)
It is known that the vertices a (7,8) B (3,5) C (4,3), m and N are the midpoint of AB and AC respectively, D is the midpoint of BC, and Mn intersects ad with F to find the coordinates of vector DF Please help with this. Thank you
Process omits vector 2 words:
AB=OB-OA=(3,5)-(7,8)=(-4,-3),AC=OC-OA=(4,3)-(7,8)=(-3,-5)
Ad = (AB + AC) / 2 = ((- 4, - 3) + (- 3, - 5)) / 2 = (- 7 / 2, - 4), while DF = - AD / 2 = (7 / 4,2)
(in the triangle ABC, D is the midpoint of BC and G is the midpoint of AD. any arbitrary line Mn crosses AB, AC and Mn respectively, if vector am = x, vector AB, vector an= Y vector AC, ask whether 1 / x + 1 / y is a fixed value, why?
Proof: because vector am and vector AB are co direction vectors, x = | vector am | / | vector ab | = am / AB (here am, AB is a line segment) similarly, y = an / NC (here an, NC is also a line segment). Therefore, the problem of 1 / x + 1 / y = (AB / AM) + (AC / an) is transformed into a plane geometry problem called BF ∥ Mn intersecting AC with F, and de ∥ Mn intersecting AC
It is known that in the triangle ABC, de parallels BC, DF parallels AC, AF and de intersect at point m, be and DF intersect at point n. verification: Mn is parallel to ab
DF parallel AC
So: BN / NE = BF / FC
De parallel BC
So: DM / BF = ad / AB = AE / AC = me / FC
So: BF / FC = DM / me
So: BN / NE = DM / me
So: Mn is parallel to ab
It is known that m, n are two trisection points on one side BC of the triangle ABC. If vector AB = a, vector AC = B, then vector Mn =?
Vector AC vector AB = vector BC = B-A
Vector Mn = 1 / 3 (B-A)
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