If the vectors a and b are not collinear, a·b=0, and c=a-(a·a)/(a·b) b, what is the included angle between vectors a and c This equation can be directly calculated c=0 ah, why there is the assumption of c=0? Doesn't a.a.b/a.b equal a? A-a is equal to zero. Ask for answers! Thank you! If vectors a and b are not collinear, a·b=0, and c=a-(a·a)/(a·b) b, then what is the angle between vectors a and c This equation can be directly calculated c=0 ah, why there is the assumption of c=0? Doesn't a.a.b/a.b equal a? A-a is equal to zero. Ask for answers! Thank you!

If the vectors a and b are not collinear, a·b=0, and c=a-(a·a)/(a·b) b, what is the included angle between vectors a and c This equation can be directly calculated c=0 ah, why there is the assumption of c=0? Doesn't a.a.b/a.b equal a? A-a is equal to zero. Ask for answers! Thank you! If vectors a and b are not collinear, a·b=0, and c=a-(a·a)/(a·b) b, then what is the angle between vectors a and c This equation can be directly calculated c=0 ah, why there is the assumption of c=0? Doesn't a.a.b/a.b equal a? A-a is equal to zero. Ask for answers! Thank you!

No. a.a=|^2a.b=|a||b|cosθ These are both constants. c=a-[(a.a)/(a.b)] b is the vector a minus (a.a)/(a.b) times the vector b. A and b are not in common lines. Do you think they are direct approximations? The vector can not be approximated by a constant. This problem is calculated as follows. a*c=a*[ a-[(a*a)/(a...

No. a.a=|^2a.b=|a||b|cosθ These are both constants. c=a-[(a.a)/(a.b)] b is the vector a minus (a.a)/(a.b) times the vector b. A and b are not in common lines. Do you think they are direct approximations? The vector can not be approximated by a constant. This is the case.

If ab is a non-zero non-collinear vector, then (a×b) a is collinear with b

It should be a dot, not a cross.
A.b is a quantity
(1) A.b=0
(A.b) a and b are not collinear vectors.
(2) A.b=0
Here (a.b) a and b are collinear vectors.

It should be a dot, not a cross.
A.b is a quantity
(1) A.b=0
(A.b) a and b are not collinear vectors.
(2) A.b=0
(A.b) a and b are collinear vectors.

A (2,3), b (-1,-1) c (6, k) is known, where K is a constant, and the modus operandi of vector AB is equal to the modus operandi of vector AC. Given A (2,3), b (-1,-1) c (6, k), where K is a constant, the modus operandi of vector AB is equal to the modus operandi of vector AC.

According to the meaning of the title, vector AB=(-1-2,-1-3)=(-3,-4) vector AC=(6-2, k-3)=(4, k-3) has:|vector AB|=|vector AC|, then 3*3+4*4=4*4+(k-3)*(k-3) can be obtained: k=0 or 6 When k=0, vector AC=(4,-3)*(-3,-4)=0, that is, vector AC is perpendicular to vector AB. When k=6, vector AC=(4,3), then, The angle a between the vector AC and the vector AB following relation: cosa=AC*AB/(|AC AB|)=(4,3)*(-3,-4)/5*5=-24/25, i.e. the angle between them is a=arc cos (-24/25)

The vector a (3,4) and the vector b (6,8) are equal to the included angle of the vector c (2, k) to obtain the value of K The vector a (3,4) and the vector b (6,8) are equal to the included angle of the vector c (2, k), obtain the value of K

A=b/2
Show that vectors a and b are collinear,
So k can take any value

Let α1,α2 and α3 be linearly independent of each other, then if the constants λ and k satisfy Under what conditions, the vector groups 2-α1, kα3-α2,α1-α3 are linearly independent.

α1,α2, And α3 are linearly independent of the formula A1α1+A2α2+A3α3=0 if A1=A2=A3=0
2-α1, Kα3-α2,α1-α3 are linearly independent of the formula k1(2-α1)+k2(kα3-α2)+K3(α1-α3)=0
After finishing,(K3-K1)α1+(K1λ-k2)α2+(K2k-K3)α3=0
K3-K1=0 because α1,α2,α3 are linearly independent
K1λ-k2=0
K2k-K3=0
K2(λk-1)=0 is obtained because 2-α1, kα3-α2,α1-α3 are linearly independent
So K1 K2 K3 must have a solution of 0 so that λk-1 is not equal to 0λk is not equal to 1

Given three points A [2,3] B [-1,-1] C [6, k], where k is a constant, if |AB|=|AC|.

|AB |=5,
|AC |^2=16+(k-3)^2=25,
(K-3)^2=9, k1=0, k2=6.
AB=(-3,-4)
AC=(4,-3) when k1=0,
Vector AB*AC=0,∠BAC=90°;
AC=(4,3) at k2=6, AB*AC=-24,
Cos BAC=-24/25,
∠BAC=180°-arc cos (24/25).