- #1
The_Brain
- 42
- 3
I hate Vector Algebra!
I am having some trouble doing some vector algebra and any help or direction is greatly appreciated.
In our lecture we drew a vector C(s) = A + s(B-A) which described a line passing through points A and B and found that the vector of min. length (shortest dist. from origin to line) was given by so = [A (dot) (A-B)]/[|B-A|2. The actual problem is to plug so into C(s) to show that C(so) = (sqrt(|A|2|B|2 - (A dot B)2))/(|B-A|).
What I have so far consists of changing [A (dot) (A-B)]/[|B-A|2 to [-A (dot) (B-A)]/[|B-A|2 so I can square the (B-A) however, when I do that I do not know how to then multiply that by the dot product of A. I know that A dot (B-A) is A dot B minus A dot A (A squared) but I have no idea how to do A dot (B-A)2. Thanks for any help.
I am having some trouble doing some vector algebra and any help or direction is greatly appreciated.
In our lecture we drew a vector C(s) = A + s(B-A) which described a line passing through points A and B and found that the vector of min. length (shortest dist. from origin to line) was given by so = [A (dot) (A-B)]/[|B-A|2. The actual problem is to plug so into C(s) to show that C(so) = (sqrt(|A|2|B|2 - (A dot B)2))/(|B-A|).
What I have so far consists of changing [A (dot) (A-B)]/[|B-A|2 to [-A (dot) (B-A)]/[|B-A|2 so I can square the (B-A) however, when I do that I do not know how to then multiply that by the dot product of A. I know that A dot (B-A) is A dot B minus A dot A (A squared) but I have no idea how to do A dot (B-A)2. Thanks for any help.