Vectorial Mechanics

[Brainwave]

Pls I don't know how to go about this problem. Find the unit vector perpendicular to both: a = 3i + 2j - k and b = 2i - j + 2k.

 

[pka]

Pls I don't know how to go about this problem. Find the unit vector perpendicular to both: a = 3i + 2j - k and b = 2i - j + 2k.

The unit vector \(\displaystyle \dfrac{{a \times b}}{{\left\| {a \times b} \right\|}}\) perpendicular to both.

 

[Brainwave]

Pls Pka, I don't get u

 

[pka]

The unit vector \(\displaystyle \dfrac{{a \times b}}{{\left\| {a \times b} \right\|}}\) perpendicular to both.

If you cannot do a vector product, a cross product, then you cannot work this problem.

By definition \(\displaystyle {\vec{a} \times \vec{b}}\) is a vector which is perpendicular to both \(\displaystyle {\vec{a} ~\& ~\vec{b}}\)

Divide by its length to form a unit vector.

 

[Brainwave]

Ofcourse I can do vector product. What am saying is that I can't get what u posted.

 

[Brainwave]

I mean type it in the way I can get.

 

[pka]

I mean type it in the way I can get.

You were given \(\displaystyle \vec{a}=3\vec{i}+2\vec{j}-\vec{k}~~\&~~\vec{b}=2\vec{i}-\vec{j}+2\vec{k}, \).

Find \(\displaystyle \vec{a}\times\vec{b}\). Find its length. \(\displaystyle \|\vec{a}\times\vec{b}\|\).

Now divide.

 

[Brainwave]

Thanks Pka, I got my answer to be ±(3i - 8j - 7k).

 

[Brainwave]

Then divide by what?

 

[Deleted member 4993]

Divide by ||a X b||

That will give you unit vector.

 

[Brainwave]

Thanks, am very grateful.