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Given points A (6,1), B (1,3) C (3,1), the projection of vector AB on vector BC is _____?
AB=OB-OA=(1,3)-(6,1)=(-5,2), BC=(2,-2),
According to the formula, the projection of AB on BC is AB*BC/|BC|=(-10-4)/(2√2)=-7√2/2.
Given three points A (0,2,3), B (-2,1,6), C (1,-1,5). Let M (x, y, z) be any point of the plane ABC and find that x, y, z satisfies the relation
(1) Let a (x, y, z)
AB=(2,1,-3)
AC=(-1,3,2)
AB, a⊥AC
2X+y-3z=0
-X+3y-2z =0
| A |=√3
X^2+y^2+z^2=3
Solution of simultaneous equations: x=1, y=1, z=1
A=(1,1,1)
The Projection Formula of A on B Vector
The projection matrix.
A projection on B vector =(BB'/ B' B) A, where B' is the transposition of B
The formula applies not only to vectors but also to subspaces
The projection matrix.
A projection on B vector =(BB'/ B' B) A, where B' is the transposition of B
This formula applies not only to vectors but also to subspaces
Vector a, b, c, d satisfies:|a|=1,|b|=root2, the projection of b in a direction is 1/2, vector (a-c)(b-c)=0,|d-c|=1, find the maximum value of |d|
Make the triangle ABO so that the vector OA = a vector OB = b takes AB as the diameter of the circle M and takes any point C on the circle, then c = OC must have (a-c)(b-c)=AC*BC =0, then the radius of the circle M is R=|AB|/2=√2/...
Make the triangle ABO so that the vector OA = a vector OB = b takes AB as the diameter of the circle M and takes any point C on the circle, then c = OC must have (a-c)(b-c)=AC*BC =0, then the radius of the circle M is R =|AB|/2=√2/...
Given a+b=(1,2), c=(-3,-4), and b is perpendicular to c, then the projection of a in the c direction is (all above are vectors) Given a+b=(1,2), c=(-3,-4), and b is perpendicular to c, then the projection of a in the c-direction is (all above are vectors)
A+b=(1,2), c=(-3,-4), b⊥c
B ● c=0
(A+b)●c=a●c+b●c=-3-8=-11
A ●c=-11
A ●c=|a||c|cos,|c|=5
A The projection in the c direction is
|A|cos=a●c/|c|=-11/5
(A+b)·(c+d)=? Abcd is a vector. Why?
(A+b)·(c+d)=)= a · c + a · d + b · c + b · d [
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