Question

given u=(0,4,3)^T a unit vector that is perpendicular to u is...

Answer 1

@satellite73 plz help

Answer 2

@thomaster i really need help

Answer 3

what is ^T ?

Answer 4

transpose

Answer 5

he means a column vector , since it is a row right now

Answer 6

there are infinitely many perpindicular vectors

Answer 7

hehe jus testing u :)
how would u find that?

Answer 8

a vector A is perpindicular to vector B when A dot B = 0.
so we can first find a perpindicular vector
(0,4,3) . (a,b,c) = 0
0a + 4b + 3c = 0
4b + 3c= 0. set a = 1 , b =-3 . then we have
4(-3) + 3c = 0
3c = 12
c = 4.
so a perpindicular vector is <1,-3,4> . and you can normalise that by dividing by length of the vector.
sqrt( 1^2 + (-3)^2 + 4^2 ) = sqrt( 1 + 9 + 16 )

More generally we have
a = a ,b=b , c = -4b/3
so a perpindicular vector is (a, b, -4b/3) = a(1 , 0 , 0 ) + b ( 0, 1, -4/3) . now normalise that

Answer 9

bit different explanation:
(0,4,3) . (a,b,c) = 0
0a + 4b + 3c = 0
for the last expretion to be 0, a can be anything and: 4b=-3c so
c=-4/3b
What this means? It says, take a R^3 vector in the form (a,b,c) where a is anything, b is anything and c=-4/3b and it will be perpendicular to u

Answer 10

but you need a unit vector, so you have to normalize that

Answer 11

ya, that's the last step

Answer 12

to write it all:
\(v=\frac{(a,b,-4/3b)}{\sqrt{a^2+b^2+(16/9)b^2}}\)

Answer 13

thank guys