Question

Find two unit vectors orthogonal to $$\vec{a}=<−5,3,2>$$ and $$\vec{b}=<3,−4,−4>$$

Answer 1

say one of the vectors you are looking for is
\[\vec c = \langle c_1,c_2,c_3\rangle\]
For orthogonality
\[\vec a\cdot\vec c=0\]
i.e.
\[\langle -5,3,2\rangle\cdot\langle c_1,c_2,c_3\rangle=0\]

Answer 2

compute this dot product,

and do the same for the vector \(\vec b\)

Answer 3

you should get two equations in-terms of \(c_1\), \(c_2\), and \(c_3\)

Answer 4

doesn't c need to be orthogonal to both a and b?

Answer 5

yes, the two equations you get should both be true

Answer 6

im attempting this now.. not sure i know what im doing though

Answer 7

i mean I'm attempting what you said, im very slow though

Answer 8

*\[\langle -5,3,2\rangle\cdot\langle c_1,c_2,c_3\rangle=0\\-5c_1+3c_2+2c_3=0\]

Answer 9

ok so the other one is $$3c_{1} -4c_{2} - 4c_{2}$$

Answer 10

= 0

Answer 11

yep

Answer 12

So the components of \(\vec c\) , must satisfy both
\[-5c_1+3c_2+2c_3=0\]and\[3c_1-4c_2-4c_3\]

Answer 13

set them equal to each other now since they both equal 0?

Answer 14

eliminate one of the components

and solve for another component in-terms of the other one

Answer 15

so i can multiply the top one by -2 right?

Answer 16

to get rid of c3?

Answer 17

yes! that is a good idea

Answer 18

awesome, I'm going to try this, ty so much :) I think i can get it now, i hope :) lol

Answer 19

@UnkleRhaukus, still not getting it, ty for the help though, is there another way to do this using the cross product?

Answer 20

I'm getting c1 = 2a/7
c2 = a
and c3 = -11a/14

Answer 21

Ah, I got it was not reading question correctly, I needed to find unit vectors, ty again :)