B***y
发帖数: 83 | 1
scalar.
这题目在3-维下的罢？
Without loss let |a| = 1, a unit vector. then
let y = y1 a + b, where a and b is orthogonal.
then y \dot a = d = y1,
y \corss a = b \corss a = c
thus c \cross a = -b ,
so y = d a - c \cross a.
now if |a| \ne 1, then the result is
y = \frac{d a }{|a|^2} - c \cross \frac{a}{|a|^2}, i.e, differs by a
factor of |a|^2. | s***e
发帖数: 911 | 2
y{\dot}a=d is one equation:
a1*y1+a2*y2+a3*y3=d;
y{\cross}a are three equations:
a3*y2-a2*y3=c1
-a3*y1+a1*y3=c2
a2*y1-a1*y2=c3.
The matrix A={{0,a3,-a2},{-a3,0,a1},{a2,-a1,0}}is singhular(the rank is 2.
So you have to solve this equation accampany with a1*y1+a2*y2+a3*y3=d.
Express y2,y3 by y1:
y2=(a2/a1)*y1-c3/a1;
y3=(a3/a1)*y1+c2/a1.
Then
a1*y1+a2*y2+a3*y3={y1*a1+[(a2^2/a1)*y1-c3*a2/a1]+[(a3^2/a1)*y1+c2*a3/a1]}
=d
so,
y1=[a1*d-(c \cross a)_1]/|a|^2;
Then u can get y2,y3 by sim | s***e
发帖数: 911 | 3
Actually by the symmetrcity u can directly write out the solution:
y1=[a1*d-(c \cross a)_1]/|a|^2
y2=[a2*d-(c \cross a)_2]/|a|^2
y3=[a3*d-(c \cross a)_3]/|a|^2
U can check it urself.
Finally,
y=(d*a-c \cross a)/|a|^2
【在 s***e 的大作中提到】
:
: y{\dot}a=d is one equation:
: a1*y1+a2*y2+a3*y3=d;
: y{\cross}a are three equations:
: a3*y2-a2*y3=c1
: -a3*y1+a1*y3=c2
: a2*y1-a1*y2=c3.
: The matrix A={{0,a3,-a2},{-a3,0,a1},{a2,-a1,0}}is singhular(the rank is 2.
: So you have to solve this equation accampany with a1*y1+a2*y2+a3*y3=d.
: Express y2,y3 by y1:
| r*f
发帖数: 731 | 4 I got another method:
from (y*a)*a=(a.y)a-(a.a)y
and subsititute y*a=c and a.y=d
we can easily get c*a=da-a^2*y
so, y=(da-c*a)/a^2
thank all you guys.
【在 s***e 的大作中提到】
:
: Actually by the symmetrcity u can directly write out the solution:
: y1=[a1*d-(c \cross a)_1]/|a|^2
: y2=[a2*d-(c \cross a)_2]/|a|^2
: y3=[a3*d-(c \cross a)_3]/|a|^2
: U can check it urself.
: Finally,
: y=(d*a-c \cross a)/|a|^2
|
|
