Recent content by Sandro Romualdez

Thanks for all the responses, I'll mark the problem as solved since I will just go with the solution using linear combinations and the properties of dot and cross products to prove the equation, since it seems like the simplest solution in terms of what we have studied recently. :smile:

Thanks. The numbers are quite big anyways so I assumed that they weren't too significant, and maybe served as a deterrent from actually calculating

Thanks for your help, and I understand how the determinant can help with this problem. The only limitation from me using this solution though is the fact that I can't use calculations, as Mark44 stated.

Using the distributive law, I can manipulate the equation to be: a • [(b x a) + (b x 2b)] I can further use the distributive law on this as well? Making it [a•(b x a)] + [a•(b x 2b)] Using my knowledge of the dot and cross product, the cross product of b and a will be perpendicular to both...

With this knowledge, would I be able to state that since c can be expressed as a linear combination of vectors a and b, all three vectors lie on the same plane? Following that, since the cross product is perpendicular to all three vectors, the dot product of a and (b x c) would be zero...

Thank you, I've been a member of PF for awhile, but for some reason my old account got deleted or something. What we've discussed in class is that the distributive property of the dot product is that a•(b+c) = a•b + c•a. I'm assuming that can also be applied with the cross product instead of...

How many optometrists does it take to change a light bulb? 1 or 2? 1... or 2...?

What did the pirate say when he turned 80? I'm eighty (Aye matey) It works better when you say it out loud ahhaha

Homework Statement Given that vector a = (1, 2, -5), b = (-12, 41, 75) and c = a + 2b, explain why (without doing any calculations whatsoever) the value of a•(b x c) = 0 Homework Equations No specific equations, as the question asks for the value without making any calculations. This problem...