Question

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)

Answer 1

Writing in component notation,
<1,0,0>=<b>i</b>
<0,1,0>=<b>j</b>
<0,0,1>=<b>k</b>

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.

Answer 2

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

Answer 3

agree

Answer 4

was I supposed to multiple -8 with |ixk| as I did above or subtract eight from the dot products?

Answer 5

oh nevermind I get the same answer either way, thanks everyone