Finding Prependicular Unit Vectors in 3D Space

brad sue
Messages
270
Reaction score
0
Hi please, someone help me with those problems?

It's about three dimension space and it is about vectors properties:

Find the unit vectors u that are prependicular to both i+2j+k and 3i-4j+2k.

the second is:

Find two mutually perpendicular unit vectors that are perpendicular to 2i+3j

Thank you for your help

Brad
 
Mathematics news on Phys.org
Express the vector(s) you seek in terms of unknown coefficients (e.g. ai+bj+ck) and use the definition of 'perpendicular' (i.e. the dotproduct is zero) to find equations you can solve for these unknowns.
 
You can also use the fact that the outer product (vectorial) of two vectors gives a new vector which is perpendicular to the first two. Divide by the norm to make it a unit vector.
 
thanks

OK I going to try your suggestion


da_willem said:
Express the vector(s) you seek in terms of unknown coefficients (e.g. ai+bj+ck) and use the definition of 'perpendicular' (i.e. the dotproduct is zero) to find equations you can solve for these unknowns.
 
a vector perpendicular to two separate vectors is created by using the cross product
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top