Projecting a vector to another vector
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I would like to project a vector to another vector. You can find more information here:
For example I would like to project vector A to vector B. I have used these tricks but it does not work: Any comment is appreciated.
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Solution 1)
A=[-10,10,0];
B=[0,0,1];
C=(dot(A,B)/norm(B)^2)*B
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Solution 2)
A=[-10,10,0];
B=[0,0,1];
CosTheta = dot(A,B)/(norm(A)*norm(B));
ThetaInDegrees = acos(CosTheta)*180/pi;
c=norm(A)*cos(ThetaInDegrees)
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1 件のコメント
採用された回答
Jan
2011 年 3 月 1 日
You have to tell DOT, that it must work on the 2nd dimension. Finally BSXFUN is needed to multiply the [88 x 1] and the [88 x 3] arrays. Unfortunalely NORM replies a matrix norm when operating on a matrix. Either use DNorm2: http://www.mathworks.com/matlabcentral/fileexchange/29035-dnorm2 or create the vector norm manually:
A = repmat([10,10,-10] ,[88,1]);
B = repmat([1,1,-1], [88,1]);
lenC = dot(A, B, 2) ./ sum(B .* B, 2); % EDITED, twice
C = bsxfun(@times, lenC, B)
4 件のコメント
その他の回答 (9 件)
Paulo Silva
2011 年 2 月 28 日
A=[-10,10,0];
B=[0,0,1];
%calculation of the projection of A into B
C=(sum(A.*B)/(norm(B)^2))*B;
%versors of each vector
An=A/norm(A);
Bn=B/norm(B);
Cn=C/norm(C);
%graphic representation
clf
line([0 A(1)],[0 A(2)],[0 A(3)],'LineWidth',10,'Color',[0 0 1])
line([0 B(1)],[0 B(2)],[0 B(3)],'LineWidth',8,'Color',[0 1 0])
line([0 C(1)],[0 C(2)],[0 C(3)],'LineWidth',5,'Color',[1 0 0])
legend('A','B','proj A into B')
xlabel('X')
ylabel('Y')
zlabel('Z')
view(80,10)
5 件のコメント
Jan
2011 年 3 月 1 日
What exactly does "but it does not work" mean?
Your solution 1:
A = [-10,10,0];
B = [0,0,1];
C = (dot(A,B)/norm(B)^2)*B
This looks ok. If you get C = [0,0,0], the method works. A and B are orthogonal, such that the projection is zero.
Your solution 2: wrong
CosTheta = dot(A,B)/(norm(A)*norm(B));
ThetaInDegrees = acos(CosTheta)*180/pi;
c=norm(A)*cos(ThetaInDegrees)
Now c is a scalar, but you wanted a vector. Converting Theta in degrees is not correct here: COS works win radians. Use COSD for degerees. Improved:
CosTheta = dot(A,B) / (norm(A)*norm(B));
C = norm(A) * CosTheta * B / norm(B);
And as expected: If you insert CosTheta in the 2nd line, you get your solution 1 again.
2 件のコメント
Paulo Silva
2011 年 3 月 1 日
I failed somehow to find the function dot and done sum(A.*B) instead :) but the results are the same
Jan
2011 年 3 月 1 日
sum(A.*B) and A*B' are faster then DOT. But for [1 x 3] vectors this does not matter.
Foday Samura
2020 年 5 月 1 日
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0 件のコメント
fatema hasan
2020 年 12 月 13 日
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fatema hasan
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fatema hasan
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0 件のコメント
fatema hasan
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0 件のコメント
Brandon O'Neill
2021 年 3 月 26 日
At a certain time of day the radiant energy from the sun reaches the roof along the direction given by the unit vector
The fraction of the sun’s energy which is falling perpendicularly on the roof is the projection of vector (A) onto the direction perpendicular to the roof – this is the dot product of (A) with the unit vector.
Q1) Use Matlab to calculate the fraction of the sun’s energy which is falling perpendicularly on the roof.
can anyone help with this?
the univ vector is [0.7627;0.5509;0.3390]
the vector A = 1/sqrt(21)[1;2;-4]
0 件のコメント
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