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

##### 1 Answer

#### Explanation:

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=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,

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) >#