Question

fasrgsdghd

Answer 1

the is the answer i got was <-3/sqrt109,-10/sqrt109> but apparently thats wrong

Answer 2

@surjithayer

Answer 3

find vector AB
\[find~ \left| vector~ AB \right|\]
\[then~ unit~vector~\in~the~direction~of~AB=\frac{ vector~AB }{ \left| vector ~AB \right| }\]

Answer 4

ok so it would be <-3,-10>/sqrt109

Answer 5

Vector AB=<0-3,-5-(-5)>=<-3,0>
\[\left| vector ~AB \right|=\sqrt{\left( -3 \right)^2+0^2}=3\]
now you can write.

Answer 6

<-1,0>?

Answer 7

correct.

Answer 8

yup thats it thanks. i do have another question though.
The unit vector oppositely directed to i is? The problem never mentions what vector i is so what should i do?

Answer 9

never mind i got it. vector i is the unit vector. i totally forgot

Answer 10

if you want unit vector in opposite direction to this vector is <1,0>

Answer 11

no it was asking for the opposite of <1,0> which is <-1,0>

Answer 12

i do have another question if you can help me?

Answer 13

correct.

Answer 14

find (1/magnitude of vector w)(vector w). vector w=−4i+4j=<-4,4>

Answer 15

what i got is (1/4sqrt2)(<-4,4>)=<-sqrt2,sqrt2> but this is wrong

Answer 16

\[\frac{ 1 }{ \left| vector~w \right| }vector~w=\frac{ -4i+4j }{ 4\sqrt{2} }=-\frac{ 1 }{ \sqrt{2} }i+\frac{ 1 }{\sqrt{2}}j\]

Answer 17

you got it.

Answer 18

thanks

Answer 19

yw.