![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
1. Given that a and b are two nonzero vectors, and a+3b is perpendicular to 7a-5b, a-4b is perpendicular to 7a-2b, find the angle between a and b?
(A+3b)(7a-b)=0
(A-4b)(7a-2b)=0
At the same time, eliminate a to get 11b^2=50ab
Remove b at the same time to obtain -77a^2=70ab
So:-5a^2=b^2
There's a problem with this.
(A+3b)(7a-b)=0
(A-4b)(7a-2b)=0
Remove a at the same time to get 11b^2=50ab
Remove b at the same time to obtain -77a^2=70ab
So:-5a^2=b^2
There's a problem with this.
To know that ab is a non-zero vector, a+3b is perpendicular to 7-5b, and a-4b is perpendicular to 7a-2b, calculate the cosine value of the included angle between a and b To know that ab is a non-zero vector, a+3b is perpendicular to 7-5b and a-4b is perpendicular to 7a-2b, find the cosine value of the included angle between a and b
First I'll show you a theorem:
Let a=(x1, y1), b=(x2, y2)
If a is perpendicular to b, then x1x2+y1y2=0
In the same way that your a+3b is represented by the coordinates of the vector (a,3b),7-5b is represented by (7,-5b), a-4b is represented by (a,-4b)7a-2b is represented by (7a,-2b)
Because ab is a non-zero vector, there are a*7+3b*(-5b)=0 and a*7a+(-4b)*(-2b)=0 according to the vertical condition above
Find the values of a and b, and the cosine of the angle between a and b is naturally found.
Given A (2,3), B (4,5), then The unit vector of AB collinear is ___.
AB=(2,2),
And
AB collinear unit vector =±
AB
|
AB |=±(2,2)
8=±(
2
2,
2
2).
Therefore, the answer is:±(
2
2,
2
2).
———————————————————————————————— Given two points A (2,3), B (-4,5), find the coordinates of the unit vector e collinear with the vector AB. ———————————————————————————————— ——————————[ The specific process, the details can be scored ]————————— ———————————————————————————————— Given two points A (2,3), B (-4,5), find the coordinate of unit vector e collinear with vector AB. ———————————————————————————————— ——————————[ The specific process, the details can be scored ]—————————
Vector AB=(-6,2)
Let e = x * vector AB (because e and AB are collinear)
=X*(-6,2)
=(-6X,2x)
So (-6x)^2+(2x)^2=1(unit vector length)
Jiede
X=(positive and negative)1/under 40
So e =(-3/ under 10,1/ under 10)
Or
E=(3 per 10,-1 per 10)
Given that for any plane vector ab=(x, y), the vector ab is rotated by an angle a about its starting point in the counterclockwise direction to obtain the vector AP=(xcosa-a+ycosa), which is called the point B is rotated by an angle a about the point A in the counterclockwise direction to obtain the ysina,xsina P. Let each point on the in-plane curve C be the curve x^2-y^2=3, and then the equation of the original curve C is obtained. Is the vector AB, not ab-
Let the point of the original curve C be (x, y)
X'= xcos45°-ysin 45°
Y'= xsin 45°+ ycos 45°
The trajectory of the point after 45° rotation is the curve x^2-y^2=3
Then, x'^2-y'^2=3
Substitute:(xcos45°-ysin45°)^2-(xsin45°+ycos45°)^2=3
To simplify:
Xy=-3/2
The direction vector e =(-4/5,3/5) of the straight line l on the known plane, If the projectiles of points O (0,0) and A (1,-2) on l are O' and A', respectively, then the vector O' A'=λe, where λ is equal to A.11/5 B.-11/5 C.2 D.-2
The direction vector e of the straight line L=(-4/5,3/5),
Slope of line L k =(3/5)/(-4/5)=-3/4
The projectiles of points O (0,0) and A (1,-2) on L are O' and A', respectively.
OO'⊥L, AA'⊥L
Koo'= Kaa'=-1/k =4/3.1
Let O'(a, b), A'(m, n)
B/a =4/3, i.e. b =4a/3
(N+2)/(m-1)=4/3, i.e. n =(4m-10)/3
O'A'=(m-a, n-b)=λe=(-4λ/5,3λ/5)
I.e. m-a=-4λ/5.2
N-b=[4(m-a)-10]/3=3λ/5.3
According to 3,4(m-a)=9λ/5+10.4
Substitute 1 into 4 to obtain:λ=-2.
----------------------------------------------------
Vector OA=(1,-2)
Vector OA=(1,-2)*e=(-4/5,3/5)=-4/5-6/5=-2
The projection of vector OA on vector e * the module of vector e=2
The projection of vector OA on vector e =2
O'A' vector is opposite to vector e,λ=-2
RELATED INFORMATIONS
- 1. Who can explain why the direction of vector a×b is the plane determined by the vertical vector ab It's a big question, because it's hard to understand what the teacher said in normal high school classes, but I just want to know why.
- 2. Given A (2,5,-1), B (5,1,11), find the direction cosine and direction angle of the AB vector,
- 3. In △ABC, it is known that BD and CF are high, M is the midpoint of BC and N is the midpoint of DF. Yes, In △ABC, it is known that BD and CF are high, M is the midpoint of BC and N is the midpoint of DF. The answer is yes,
- 4. M (3,-2), N (-5,-1) is known, and MP=1 2 MN, the coordinates of point P are ______.
- 5. Given m >0, n >0, vector a=(1,1), vector b=(m, n-3), and a⊥(a+b), then the minimum value of 1/m+4/n is
- 6. Given that a and b are non-collinear non-zero vectors, if there is a vector c such that c is parallel to a and c is parallel to b, then c equals?
- 7. The side length of equilateral triangle ABC is 1, vector AB=a, vector BC=b, vector CA=c, then a*b+b*c+c*a is equal to
- 8. In the equilateral triangle ABC with 1 side length, let BC vector be a vector, CA vector be b vector, AB vector be c vector, then a.b+b.c+c.a=? Why the included angle is 120
- 9. How to prove the judgment theorem of face perpendicularity by vector method?
- 10. What condition does vector parallel satisfy
- 11. Ab is a non-zero vector and |a+b|=|a||b|, Then A: a is parallel to b, and a and b have the same direction B.a=b C.a=-b D. None of the above is correct Ab is a nonzero vector and |a+b|=|a||b|, Then A: a is parallel to b, and a and b have the same direction B.a=b C.a =-b D. none of the above Ab is a nonzero vector and |a+b|=|a||b|, Then A: a is parallel to b, and a and b have the same direction B.a=b C.a=-b D. None of the above is correct
- 12. The non-zero vectors a, b are known to satisfy A Given the non-zero vector a, b satisfies the absolute value of a+b = the absolute value of a-b, prove a perpendicular b Solution by analytical method
- 13. Given the vector set M={ a|a=(1,2)+b (3,4), b∈R}, N={ a|a=(-2,-2)+P (4,5), P∈R}, then the intersection of M and N is? Given the set of vectors M={ a|a=(1,2)+b (3,4), b∈R}, N={ a|a=(-2,-2)+P (4,5), P∈R}, then the intersection of M and N is?
- 14. 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?
- 15. AB vertical AC, vector BD=5/3 vector BC, AC=2, find vector AC* vector AD
- 16. Known A =(2,-1), B=(λ,3), if A and If the angle of b is obtuse, the value range of λ is ___.
- 17. Known vector A =(1,2), B=(-2,3), C=(4,1), if used A and B indicates C, then C =___.
- 18. Is the zero vector perpendicular to the zero vector According to the textbook of Tongji University, zero vector is perpendicular to all vectors.
- 19. Why is the direction of the zero vector specified as arbitrary? What's the point of this?
- 20. Vector a and vector b are known to be two non-zero vectors if │ Vector a+t vector b│(t∈R) with minimum value (1) Find t (2) Prove that vector b is vertical (vector a+t vector b)