Vector problem

[Jade1886]

Hello! I am unsure how to deal with this given question: "Find the unit vector perpendicular to both (3k + j) and (2i-j-5k)".
I think I have the correct thinking, if I find the vector product of the two vectors (this is perpendicular to both vectors above) so if I find the modulus/length of the product, then I have the unit vector that is perpendicular to both. I laid out the numbers in this form:

i	j	k
0	1	3
2	-1	-5

from this I got the vector (-8,-6,2) and the modulus: 2 x (Root 26), however this is wrong. I think my error is that I just reversed the order of 3k + j, but I am really unsure how to deal with this when the order of the vector is reversed as such.

Thank you for your time, any pointers would be a great help! Oh and correct answer is: 1/root(11) (-1,3-1).

 

[Dr.Peterson]

You did fine handling the order: Just rewrite in the proper order, so that 3k + j is j + 3k, which is <0, 1, 3>, so that row is right.

But check all your signs. You're either using the wrong signs, or just mishandling them, in more than one place. If you still don't get it right, show us the details of your calculation for each component (what did you multiply, add, etc.).

 

[Jade1886]

Hello and thanks for replying so quickly! I will attach a picture of my working (which I hope is easy enough to read) ,I realised that I forgot to include a negative sign in my first set of brackets, so this should instead be (-5 - (-3) ) but this wouldn't result in the correct answer anyways, I cannot spot where else I may have mishandled the information given.

 

[pka]

Hello! I am unsure how to deal with this given question: "Find the unit vector perpendicular to both (3k + j) and (2i-j-5k)".

Here is a general solution: If \(\displaystyle \vec{v}_1~\&~\vec{v}_2\) are two non-colinear vectors then
the vector \(\displaystyle \frac{\vec{v}_1\times \vec{v}_2}{\|\vec{v}_1\times \vec{v}_2\|}\) is a unit vector that is perpendicular to both \(\displaystyle \vec{v}_1~\&~\vec{v}_2\).

 

[Dr.Peterson]

With the correction you identified, and some obvious sign errors you didn't mention, the cross product becomes <-2, 6, 2>:

(-5+3)i - (0-6)j + (0-2)k = -2i + 6j - 2k​

The length of this is sqrt(4 + 36 + 4) = 2sqrt(11), so the answer is exactly what you quoted.

 

[Jade1886]

Here is a general solution: If \(\displaystyle \vec{v}_1~\&~\vec{v}_2\) are two non-colinear vectors then
the vector \(\displaystyle \frac{\vec{v}_1\times \vec{v}_2}{\|\vec{v}_1\times \vec{v}_2\|}\) is a unit vector that is perpendicular to both \(\displaystyle \vec{v}_1~\&~\vec{v}_2\).

Thank you for the reply!

 

[Jade1886]

With the correction you identified, and some obvious sign errors you didn't mention, the cross product becomes <-2, 6, 2>:

(-5+3)i - (0-6)j + (0-2)k = -2i + 6j - 2k​

The length of this is sqrt(4 + 36 + 4) = 2sqrt(11), so the answer is exactly what you quoted.

Thank you for the helpful explanation, however the answer in my textbook states that the vector is (-1,3-1), this must be a mistake in the answers section? One more thing could you explain what is happening in the middle bracket, are you multiplying -6 by the negative sign to obtain a positive answer?

 

[lev888]

Thank you for the helpful explanation, however the answer in my textbook states that the vector is (-1,3-1), this must be a mistake in the answers section? One more thing could you explain what is happening in the middle bracket, are you multiplying -6 by the negative sign to obtain a positive answer?

No, (-1,3,-1) is not the answer. It's 1/root(11) (-1,3-1), per your original post. This is what we get when we normalize (-2, 6, -2).

 

[Dr.Peterson]

Thank you for the helpful explanation, however the answer in my textbook states that the vector is (-1,3-1), this must be a mistake in the answers section? One more thing could you explain what is happening in the middle bracket, are you multiplying -6 by the negative sign to obtain a positive answer?

That is not what you said yourself:

Oh and correct answer is: 1/root(11) (-1,3-1).

(Or are you saying they left out a comma, since you've done so twice now?)

Yes, of course.

 

[Jade1886]

No, (-1,3,-1) is not the answer. It's 1/root(11) (-1,3-1), per your original post. This is what we get when we normalize (-2, 6, -2).

Thanks for your clarification, however I did not state that the answer was (-1, 3, -1), the way the answer appeared in the textbook seemed to imply that (-1, 3, -1) was the vector that you multiply by (1/the length of the vector) to obtain the unit vector but I understand now that is not the case.

 

[Jade1886]

That is not what you said yourself:

(Or are you saying they left out a comma, since you've done so twice now?)

Yes, of course.

The answer seemed to imply (-1, 3, -1) is the vector obtained from the cross-product which you multiply by (1/the length of the vector) to find the unit vector, I did not mean (-1, 3, -1) was the full answer.