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Vector a=(1,-3), b=(-2, m), and vector a is perpendicular (a-b). Fact the value of Vector a=(1,-3), b=(-2, m), and vector a is perpendicular (a-b). Fact the value of m
Vector a-b =(3,-3-m)
Because a is perpendicular to a-b, a (a-b)=0
So (1,-3)(3,-3-m)=0
3+3(3+ M)=0
M =-4
The vector a=(2,1) b=(minus 3,4) is known and the vector ma+b is perpendicular to a-b, for the value of m
Ma+b=(2m, m)+(-3,4)=(2m-3, m+4)
A-b =(5,-3)
Because they're vertical.
So (ma+b)(a-b)=0
So 5(2m-3)-3(m+4)=0
Solution m=27/7
The known vector a=(3, m), b=(1,-4), and the quantity product ab=5, try to find the value of real number m The known vector a=(3, m), b=(1,-4), and the quantity product ab=5, try to find the value of the real number m
3*1+(-4)*M=5
Solution m=(-1/2)
Product formula of quantity ab=x1y1+x2y2
Given vector a=(m+1), vector b (1, m-1), if (vector a+vector b)⊥(vector a-vector b). Factoring the value of m Vector Ah =(m+1,-3) Given vector a=(m+1), vector b (1, m-1), if (vector a+vector b)⊥(vector a-vector b). Fact value of number m Vector Ah =(m+1,-3)
Your vector a doesn't have all the coordinates,
Give you a basic idea.
(Vector a+vector b)⊥(vector a-vector b).
I.e.(vector a + vector b).(vector a - vector b)=0
I.e. vector a2- vector b2=0
That is,|vector a|=|vector b|
Then solve the equation.
Please check the question and ask.
Let vector a=(10,-4), vector b=(3,1) vector c=(-2,3)(1), vector b, c denote vector a (2) Verify that the vectors b, c can serve as a set of bases representing all vectors in the same plane
Let vector a=xb+yc a=x (3,1)+y (-2,3)=(3x-2y, x+3y), i.e.3x-2y=10x+3y=-4, then x=2, y=-2, then c=2b-2c, because 3X3-1X (-2)=11 is not equal to 0, then vector b and vector c are not collinear. Then vectors b, c can be used as a set of bases for all vectors in the same plane.
Let α1=(a,2,10),α2=(-2,1,5),α3=(-1,1,4),β3=(1, b, c). (1)β Can be expressed linearly by α1,α2,α3, and the representation is unique; (2)β Can not be expressed linearly by α1,α2,α3. Let α1=(a,2,10),α2=(-2,1,5),α3=(-1,1,4),β3=(1, b, c). (1)β Can be expressed by α1,α2,α3 linearly and uniquely; (2)β Can not be expressed linearly by α1,α2,α3.
(α3^T,α2^T,α1^T,β)=-1-2a11 1 2B4 5 10Cr1+r2, r3-4r20-1a+2 1+B1 1 2B0 1 2C-4br1+r30 0A +4 1-3B+c1 1 2B0 1 2C-4b so that when a =-4,β can be expressed linearly by α1,α2,α3 and the representation is unique when a =-4 and 3b-c=1,...
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