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If vector a=(1,1), vector b=(1,-1), vector c=(-1,2), then a, b=?
Let c=xa+yb
Then -1=x+y
2=X-y
Solution x=1/2, y=-3/2
So c=1/2a-3/2b
Note: a, b and c above refer to vectors, x and y indices.
Let c=xa+yb
Then -1=x+y
2=X-y
Solution: x=1/2, y=-3/2
So c=1/2a-3/2b
Note: a, b and c above refer to vectors, x and y indices.
Given vector a=(1,-1), b=(1,2), vector c satisfies (c+b)⊥a,(c-a)∥b, then why is c equal to (2,1)
Let c (x, y)
According to (c+b)⊥a:
(X+1)·1+(y+2)·(-1)=0
X+1-y-2=0
I.e. x-y-1=0
X=y+1
C (y+1, y)
(C-a)∥b
Y+1=2(y+1-1)
Solution y=1
C (2,1)
A =(1,0,2) b =(0,2,1) Try to determine the normal vector of the plane
A, b for cross multiplication
Because a, b cross-multiplies the vector vertical a, b
Is the normal vector of the plane containing a, b
Find as follows
I, j, k
1,0,2
0,2,1
=(0*1-2*2) I-(1*1-0*2) j+(1*2-0*2) k
=-4I-j+2k
That is, the normal plane vector is (-4,-1,2)
A, b as cross multiplication
Because a, b cross-multiplies the vector perpendicular a, b
Is the normal vector of the plane containing a, b
Find as follows
I, j, k
1,0,2
0,2,1
=(0*1-2*2) I-(1*1-0*2) j+(1*2-0*2) k
=-4I-j+2k
That is, the normal plane vector is (-4,-1,2)
How to find the normal vector of plane?
Try to find any two lines (but not parallel) in the plane with vector (XYZ), and write their vector P1 P2.
When the product of normal vector and P1 P2 is 0, the ternary first-order equation (2) of XYZ is obtained. If any of the unknowns is known, for example, Z, then X and Y can be expressed by Z. At this time, the normal vector has only the unknowns of Z, and the value of Z can be set according to the situation.
Of course, it is best to set the value, and finally write out that the normal vector is the simplest, in other words, there is no common factor between their numbers.
Let A (2,0,0), B (1,2,1),2,1), C (−1,1,2) be the normal vector n =.
Let A (2,0,0), B (1,2,1),2,1), C (−1,1,2) be the normal vector n =.
Let {M, N, P} be the normal vector of the plane in question, where M, N, P are not simultaneously zero because the plane passes through B (1,2,1).
Therefore, the plane equation can be written as M (x-1)+N (y-2)+P (z-1)=0.(1)
Because A (2,0,0) and C (-1,1,2) are also in this plane, the following two conditions are obtained:
M (2-1)+ N (0-2)+ P (0-1)=0; M (-1-1)+ N (1-2)+ P (2-1)=0; reduced to:
M-2N-P=0.(2)
-2M-N+p=0.(3)
The condition for a nonzero solution to a homogeneous system of equations of M, N, P consisting of (1)(2)(3) is the following third-order determinant =0, i.e.
X-1.y-2.z-1
∣1.-2.-1..∣=0
∣-2.-1.1.∣
Expanded (x-1)(-2-1)-(y-2)(1-2)+(z-1)(-1-4)=-3(x-1)+(y-2)-5(z-1)=0
I.e.-3x+y-5z+6=0 is the equation of this plane, and its normal vector n={-3,1,-5}
Given that the module of vector a is equal to root number 3, the module of vector b is equal to 2, and the included angle of vector a vector b is pie/6, the included angle of vector a plus 2 vector b and minus 2 vector a plus 2 vector b is obtained
Given |a|=√3,|b|=2, a, b is π/6
Then a·b=|a||b|cosπ/6=√3*2 3/2=3
Then (a+2b)·(2a-b)
=2A·a+3a·b-2b·b
=2|A|2+3a·b-2|b|2
=2*3+3*3-2*2²
=7
Also |a+2b|2=(a+2b)·(a+2b)=|a|2+4a·b+4|b|2=3+4*3+4*22=31
|2A-b|2=(2a-b)·(2a-b)=4|a|2-4a·b+|b|2=4*3-4*3+22=4
Then |a+2b|=√31,|2a-b|=2
(A+2b)·(2a-b)=|a+2b||2a-b|cosC
Then cosC=7/(√31*2)=7√31/62
Then C=arc cos (7√31/62)
The angle between a+2b and 2a-b is arccos (7√31/62)
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