What is the cross product of -i x i? Is it negative 1 or is still just 0?
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May 7, 2008 #2 Defennder Homework Helper 2,591 5 Note that for any 2 vectors a,b: -a x b = -(a x b). This reduces the problem to -(i x i). Now what is the vector product of a vector with itself?
Note that for any 2 vectors a,b: -a x b = -(a x b). This reduces the problem to -(i x i). Now what is the vector product of a vector with itself?
May 7, 2008 #3 Ry122 565 2 well can you tell me how the projection of u on to v where u=-i+2j and v=i+2j is v=(3/5)i+(6/5)j ? The answer i got was (4/5)i + (8/5)j i used the equation w=v.((u.v)/(modulusv^2))
well can you tell me how the projection of u on to v where u=-i+2j and v=i+2j is v=(3/5)i+(6/5)j ? The answer i got was (4/5)i + (8/5)j i used the equation w=v.((u.v)/(modulusv^2))
May 7, 2008 #4 D H Staff Emeritus Science Advisor Insights Author 15,393 682 Ry122 said: i used the equation w=v.((u.v)/(modulusv^2)) This equation for the projection is [tex]\mathbf w = \mathbf v \frac {\mathbf u \cdot \mathbf v}{v^2}[/tex] Note well: The cross product is not involved when you compute the projection this way.
Ry122 said: i used the equation w=v.((u.v)/(modulusv^2)) This equation for the projection is [tex]\mathbf w = \mathbf v \frac {\mathbf u \cdot \mathbf v}{v^2}[/tex] Note well: The cross product is not involved when you compute the projection this way.
May 7, 2008 #5 Ry122 565 2 That's the same equation that I gave. yeah i realized my mistake after posting, i should have said dot product, not cross product.
That's the same equation that I gave. yeah i realized my mistake after posting, i should have said dot product, not cross product.