Find the line orthogonal to plane

Question

Find the line orthogonal to plane spanned by vector v= i-5j and vector 3k passing through (2,3,1)

Answer 1

This type of problem lends itself to analysis using the techniques of linear algebra.

 

A vector orthogonal to the plane can be constructed by forming the vector cross product, C, of  (i - 5 j) and  3 k,   that is  C =  ( i -5j )  X 3 k   .

The vector C works out to be    C =  - 15 i  - 3j

 

The vector, V, corresponding to each point on the desired line is a function of a single numerical parameter, β, according to:

  V(β)  = 2 i + 3 j + k  +  β  C  =   (2 - 15 β) i + (3 - 3 β ) j + k  

  Notice that   the point ( 2,3,1) is generated by  β =0.  

 

This equation can be written out in component form as:

   x = 2 -15 β

   y = 3 - 3β

   z = 1

 

The parameter β can be eliminated between the first two of these three equations resulting in

    y =  (x + 13) /5     

 

    This last equation, along with  z =1   ,  are the two equations needed to specify the desired line in the three dimensional space.    The alternate description of the line is the expression for  V(β)  given above.  

 

 

 

Answer 2

Let v = < 1, -5, 0 >   w = < 0, 0, 3 >, where < 1, -5, 0 > is the vector 1i - 5j + 0k and, <0, 0, 3 > is the vector 0i + 0j + 3k.    

 

Then, v x w is orthogonal to the plane.

 

         v x w = < -15, -3, 0 >

 

Let (x,y,z) be a point on the line.  Then, < x-2, y-3, z-1 > is parallel to the line and is parallel to < -15, -3, 0 >.

 

So, < x-2, y-3, z-1 > = c<-15, -3, 0 > (c = any real number)

 

Equating corresponding components of the vectors, we have the equations:  

 

     x = -15c+2

     y =  -3c +3

     z = 1         (here, c is an arbitrary constant)