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What condition does vector parallel satisfy
X1/x2=y1/y2 vector a (x1, y1) vector b (x2, y2) a= b, i.e. x1= x2 y1= y2 or x1*y2-x2*y1=0
Let vector a=(2,1,2), b=(4,-1,10), c=b-λa, and a⊥b, then λ=
A⊥c bar.
C=b-λa=(4,-1,10)-λ(2,1,2)=(4-2λ,-1+λ,10-2λ)
From a⊥c,
I.e. a*c=2*(4-2λ)+1*(-1+λ)+2*(10-2λ)=0
I.e.λ=27/7
Vector product quantity product (Axb)·c=2 find ((a+b) x (a-b))·c Abc is a vector Note the distinction between vector product and quantity product Vector product quantity product (Axb)·c=2 Find ((a+b) x (a-b))·c Abc is a vector Note the distinction between vector product and quantity product Vector product quantity product (Axb)·c=2 Find ((a+b) x (a-b))·c Abc is a vector Pay attention to distinguish between vector product and quantity product
No problem, the vector product and the quantity product are obviously different:
(Axb)·c=2,(( a+b) x (a-b))·c=(a×(a-b)+b×(a-b))·c=(a×a-a×b+b×a-b×b)·c
=-2(A×b)·c=-4
What is the geometric meaning of the space vector quantity product
The quantity product a • b is the product of the length of a and the projection of b in the direction of a ||cos@.
The product product a•b is the product of the length of a and the projection of b in the direction of a |b|cos@.
How to Obtain the Definite Formula a·b=|a b cos < a, b > Is this documented or artificial? The math book only says |a b cos < a, b > is written as a `b, but why does this a default this a` b to mean a times b? (See the example of Math Compulsory 2 B in Human Education Edition. He tore down a multiplication b obtained in a peaceful way.)
The product of general vectors that we learned in high school refers to the product of quantities, if you want to be more detailed: see below
A·a=|a| squared.
A⊥b <=> a·b=0.
| A·b a b|.
Number product of vectors Vector a + vector b + vector c = zero vector And the module of vector a=4, the module of vector b=3, the module of vector c=5, Find the product of vector a and vector c Um... I feel like the answer is -16 That angle should be 127 degrees. Number product of a vector Vector a + vector b + vector c = zero vector And the module of vector a=4, the module of vector b=3, the module of vector c=5, Find the product of vector a and vector c Um... I feel like the answer is -16 That angle should be 127 degrees. Number product of vectors Vector a + vector b + vector c = zero vector And the module of vector a=4, the module of vector b=3, the module of vector c=5, Find the product of vector a and vector c Well... I feel like the answer is -16 That angle should be 127 degrees.
Of course it's 0. COS90=0.
3 Square +4 square =5 square
Pythagorean theorem
And the number product of the vector = the two vectors of Mo by COS their angle, Mo means absolute value is also the length.
I'm sorry I did n' t read the title.
The answer is 4*5*4/5=16
Of course it's 0. COS90=0.
3 Square +4 square =5 square
Pythagorean theorem
And the product of the number of vectors = the two vectors of Mo by COS their angle, Mo means the absolute value is the length.
I'm sorry I did n' t read the title.
The answer is 4*5*4/5=16
Of course it's 0. COS90=0.
3 Square +4 square =5 square
Pythagorean theorem
And the product of the number of vectors = two vectors of Mo by the angle of COS, Mo means the absolute value is the length.
I'm sorry I did n' t read the title.
The answer is 4*5*4/5=16
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