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

1 Answer
Mar 9, 2017

Answer:

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

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# 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) >#