Find a vector x perpendicular to the vectors v=<-1,2,-3> and u = <1,-3,-2>
do you know how to compute a cross product of two vectors ?
For more info, see http://www.khanacademy.org/math/linear-algebra/vectors_and_spaces/dot_cross_products/v/linear-algebra--cross-product-introduction
so i only need to do the cross product ?
yes
okay thanks i wanted to find the angle between two vectors, <-2 , -4> <4 , -3> and i got 79.695, but it marked me wrong
you use the dot product \[ a \cdot b = |a| |b| \cos\theta\] you need to find |a| , |b|, and a dot b
i did that, and i got \[\cos \theta = \frac{ 4 }{ \sqrt{20}*5 } = 79.695\] but idk what i did wrong
maybe they want the answer in radians ?
it said find the angle but is that answer correct?
yes, 79.695 degrees is correct.
\[" {\color{red}{\alpha}}"->~~ does ~~this~~symbol ~~mean ~~radian ?\]
so would this be it's radian -->1.390942827002
\( \alpha\) (alpha) is the first letter of the Greek alphabet.(notice that alphabet is made up of alpha and beta jammed together... (though it got trimmed to alpha bet ). alpha and beta \( \beta \) are the first two letters of the Greek alphabet It is common to use Greek letters to denote angles. It does not mean they are measured in radians or degrees. To know what units are expected, you have to look at the question or examples. But if degrees do not work, I would try radians. Here, about 1.391 radians
Another thing to check is do they want you to round the answer to a particular decimal place. It may be they want the answer to the nearest degree or to the nearest tenth of a degree... in which case they may not accept your original 79.695
alright thanks
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