Writing in component notation,
<1,0,0>=i
<0,1,0>=j
<0,0,1>=k
and using rules
ixj=k, jxk=i, kxi=j
jxi=-k, kxj=-i, ixk=-j
and the magnitude of all unit vectors is always one.
Above reduces to
<0,0,1>.<0,1,-3> + <1,0,-4>.<1,0,-4> - 8|ixk|
you can complete the computations and post your answer for a check if you wish.
if i,j,k are the standard unit basic vectors, in 3 space, determine the value of
k dot(j-3k)+(i-4k) dot (i-4k)-8 |i cross -k|
(Simplify without using components)
5 answers
thanks! This is my answer, is it correct?
(0,0,1).((0,1,-3) +(1,0,-4).(1,0,-4)-8|(1,0,0)x(0,0,-1)|
=(0)(0)+(0)(1)+(1)(-3)+(1)(1)+(0)(0)+(-4)(-4)-8|(0,1,0)|
=-3+1+16-8
=6
(0,0,1).((0,1,-3) +(1,0,-4).(1,0,-4)-8|(1,0,0)x(0,0,-1)|
=(0)(0)+(0)(1)+(1)(-3)+(1)(1)+(0)(0)+(-4)(-4)-8|(0,1,0)|
=-3+1+16-8
=6
agree
was I supposed to multiple -8 with |ixk| as I did above or subtract eight from the dot products?
oh nevermind I get the same answer either way, thanks everyone