Question

Find the cross product and show it is orthogonal to both u and v.
u= <-3,2,3>
v= <0,1,0>
I dont understand! Can someone please help?

Answer 1

Do you know how to find the cross product?

Answer 2

\[
\langle-3,2,3\rangle
\times
\langle 0,1,0\rangle
=
\det\left(
\begin{bmatrix}
\mathbf i&\mathbf j&\mathbf k\\
-3&2&3\\
0&1&0
\end{bmatrix}
\right)
\]

Answer 3

\[u \times v=\]U x v=|■(i&j&k@-3&2&3@0&1&0)|={(2)(0)-(3)(1)}i-{(3)(0)-(-3)(0)}j+{(-3)(1)-(2)(0)}k
=-3i-3k

Answer 4

So the correct answer is
-3i - 3k ?

Answer 5

correct
now again find dot product of (-3,2,3) and (-3,0,3) ,it should be (0,0,0)
similarly find dot product of (0,1,0) with (-3,0,3),it should also be( 0,0,0)

Answer 6

Okay great! Thanks for your help! I appreciate it

Answer 7

Wait a minute, do you even know how to get the answer on your own?

Answer 8

Yes I know how to get it on my own. He just explained it to me. Thanks for your concern

Answer 9

correction it is <-3,0,-3> not (-3,0,3)

Answer 10

If so, then what is the cross product of \(\langle 1,1,0\rangle\) and \(\langle 0,1,1\rangle\)?

Answer 11

I can tell out right that isn't correct, because \(\langle 0,0,1\rangle \cdot \langle 0,1,1\rangle = 1\neq 0\)

Answer 12

He showed me how to do it. That doesn't mean I am an expert at it. Sorry for the disappointment.

Answer 13

Do you know the method? You don't need to be an expect.

Answer 14

I certainly wouldn't understand how to do it based on his post alone. If you did then you're pretty clever.

Answer 15

Find the cross product and show it is orthogonal to both u and v.

u= 6k
v= -i+3j+k

This is the last question I have. I have attached the work below. I just want to make sure that I have it right.

Answer 16

your cross product is correct but why are you finding unit normal to u cross v

Answer 17

That is where I have my issue. I know how to do the steps, but I never know if my answer is write or not.

Answer 18

again find dot product with u and v separately, if 0 then u cross v is orthogonal to both u and v.

Answer 19

\[u.(u \times v)=\left( 6k \right).\left( -18i-6j \right)=0+0+0=0\]
\[v.\left( u \times v \right)=\left( -i+3j+k \right).\left( -18i-6j \right)=\left( -1 \right)\left( -18 \right)+3\left( -6 \right)+1\left( 0 \right)=18-18=0\]
i think now you are clear.

Answer 20

When you find the dot product separately, you are looking for u and v to equal zero? Then, the answer would be 1?

Answer 21

So, the cross product is 0? I am totally confused now :0

Answer 22

I need to find the cross product, and show it is orthogonal to both u and v.

Answer 23

you have found cross product as correct.
Now you have to show this cross product is orthogonal to both i.e., u cross v is perpendicular to both u and v.
for this you have to find dot product of u and u cross v.
again find dot product of v with u cross v.

Answer 24

this dot product as i have shown above should be 0.