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Given A (2,5,-1), B (5,1,11), find the direction cosine and direction angle of the AB vector,
| A|= Root (2^2+5^2+(-1)^2)= Root (30)| B|= Root (5^2+1^2+11^2)= Root (147) The direction cosine of the AB vector is equal to the dot product of A and B divided by the product of their moduli, i.e. cos (A, B)= A·B/(|A B|)=[2*5+5*1+(-1)*11]/[ Root (30)*Root (147)]=2*Root (10)/105, i.e. find...
Given that the vector ab satisfies a+b=(2,8), a-b=(-8,16), find the cosine value of the angle between (1) a*b (2) a and b
A+b=(2,8), a-b=(-8,16)
So a =(-3,12)
B=(5,-4)
(1) A*b=-15-48=-63
(2) A*b=lallblcos (a, b) cos=-21/root 697
A+b=(2,8), a-b=(-8,16)
So a =(-3,12)
B=(5,-4)
(1) A*b=-15-48=-63
(2) A*b=lallblcos (a, b) cos=-21/Root 697
(In the triangle ABC, D is the midpoint of BC, G is the midpoint of AD, and the arbitrary passing point G makes the straight line MN intersect with AB respectively, AC is at MN, if vector AM=X vector AB, vector AN= Y vector AC, whether 1/X+1/Y is constant value, why? (In the triangle ABC, D is the midpoint of BC, G is the midpoint of AD, and the arbitrary passing point G is the straight line MN intersect with AB respectively, AC is at MN, if vector AM=X vector AB, vector AN= Y vector AC, ask if 1/X+1/Y is constant and why? (In the triangle ABC, D is the midpoint of BC, G is the midpoint of AD, and the arbitrary passing point G makes the straight line MN intersect with AB respectively, AC is at MN, if vector AM=X vector AB, vector AN= Y vector AC, if 1/X+1/Y is constant, why?
Prove: Because vector AM and vector AB are the same vector, so x=|vector AM|/|vector AB|=AM/AB (where AM, AB is a line segment) the same as y=AN/NC (where AN, NC is also a line segment), so 1/x+1/y=(AB/AM)+(AC/AN), the problem is transformed into plane geometry, i.e. BF//MN, AC to F, DE//MN, AC to...
Because vector AM and vector AB are the same vector, so x=|vector AM|/|vector AB|=AM/AB (where AM, AB is a line segment), the same as y=AN/NC (where AN, NC is also a line segment), so 1/x+1/y=(AB/AM)+(AC/AN), the problem is transformed into plane geometry.
Because vector AM and vector AB are the same vector, so x=|vector AM|/|vector AB|=AM/AB (where AM, AB is a line segment), y=AN/NC (where AN, NC is, NC is also a line segment), so 1/x+1/y=(AB/AM)+(AC/AN), the problem is transformed into plane geometry problem i.e. BF//MN, AC, F, DE//MN, AC.
In triangular ABC, DE is parallel to BC, DF is parallel to AC, AF and DE intersect at point M, BE and DF intersect at point N. Verification: MN is parallel to AB
DF Parallel AC
Therefore: BN/NE=BF/FC
DE parallel BC
Therefore: DM/BF=AD/AB=AE/AC=ME/FC
Therefore: BF/FC=DM/ME
Therefore: BN/NE=DM/ME
Therefore: MN parallel AB
Given M, N is two trisection points on one side BC of 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)
In △ABC, BD and CF are high M, respectively, which is the midpoint of BC and N, which is the midpoint of DF. Rt diagram, imagine yourself In △ABC, BD and CF are high M respectively, N is the midpoint of BC, N is the midpoint of DF, and MN⊥DF is verified. Rt diagram, imagine yourself In △ABC, BD and CF are high M respectively, which is the midpoint of BC and N is the midpoint of DF. Rt diagram, imagine yourself
It's a good proof.
The first thing to think about is proof: MD = MF.
Connect MD, MF.
In Rt△BCD, BM=MC,
So DM =1/2BC
(In a right triangle, the center line on the hypotenuse is equal to half the length of the hypotenuse)
Similarly, in Rt△BFC,
Available, FM =1/2 BC.
So DM = FM
Because N is the midpoint of the DF
So MN⊥DF
(Isosceles triangle three-in-one)
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