What is the unit vector that is orthogonal to the plane containing `(3i – 4j + 4k)` and `(2i+j+2k)`?

A vector which is orthogonal (perpendicular, norma) to a plane containing two vectors is also orthogonal to the given vectors. We can find a vector which is orthogonal to both of the given vectors by taking their cross product. We can then find a unit vector in the same direction as that vector.

First, write each vector in vector form:

`veca=<3,-4,4>` and `vecb=<2,1,2>`

The cross product, `vecaxxvecb` is found by:

`vecaxxvecb=abs((veci,vecj,veck),(3,-4,4),(2,1,2))`

For the i component, we have:

`(-4*2)-(4*1)=(-8)-(4)=-12`

For the j component, we have:

`-[(3*2)-(4*2)]=-[6-8]=2`

For the k component, we have:

`(3*1)-(-4*2)=3-(-8)=11`

Therefore, `vecn=<-12,2,11>`

Now, to make this a unit vector, we divide the vector by its magnitude. The magnitude is given by:

`|vecn|=sqrt((n_x)^2+(n_y)^2+(n_z)^2)`

`|vecn|=sqrt((-12)^2+(2)^2+(11)^2)`

`|vecn|=sqrt(144+4+121)=sqrt(269)`

The unit vector is then given by:

`vecu=(vecaxxvecb)/(|vecaxxvecb|)=(vecn)/(|vecn|)`

`vecu=(< -12,2,11 >)/(sqrt(269))`

`vecu=< -12/(sqrt(269)),2/(sqrt(269)),11/(sqrt(269)) >`

We can rationalize the denominators to get:

`vecu=< (-12sqrt(269))/(269),(2sqrt(269))/(269),(11sqrt(269))/(269) >`