How do you find a unit vector perpendicular to a 3-D plane formed by points (1,01),(0,2,2) and (3,3,0)?

1 Answer
sente
Oct 17, 2016

#(-1/sqrt(3), 1/(5sqrt(3)), -7/(5sqrt(3)))#

Explanation:

Our strategy will be to find two vectors in the plane, take their cross product to find a vector perpendicular to both of them (and thus to the plane), and then divide that vector by its measure to make it a unit vector.

Step 1) Find two vectors in the plane.

We will do this by finding the vector from #(1,0,1)# to #(0,2,2)# and from #(1,0,1)# to #(3,3,0)#. As all three points are in the plane, so will each of those vectors.

#vec(v_1) = (0,2,2)-(1,0,1) = (-1,2,1)#

#vec(v_2) = (3,3,0)-(1,0,1)=(2,3,-1)#

Step 2) Find a vector perpendicular to the plane.

If a vector is perpendicular to two vectors in a plane, it must be perpendicular to the plane itself. As the cross product of two vectors produces a vector perpendicular to both, we will use the cross product of #vec(v_1)# and #vec(v_2)# to find a vector #vec(u)# perpendicular to the plane containing them.

#vec(u) = vec(v_1)xxvec(v_2)#

#= |(hat(i), hat(j), hat(k)), (-1, 2, 1), (2, 3, -1)|#

#=(2(-1)-1(3))hat(i)-((-1)(-1)-(1)(2))hat(j)+((-1)(3)-(2)(2))hat(k)#

#=-5hat(i)+hat(j)-7hat(k)#

#=(-5, 1, -7)#

Step 3) Turn #vec(u)# into a unit vector.

A unit vector is a vector whose measure is #1#. Using the fact that for any vector #vec(v)# and scalar #c#, we have #||cvec(v)|| = c||vec(v)||#, we will find #||vec(u)|| = u#, then divide by #u#.

#||vec(u)/u|| = ||vec(u)||/u = u/u = 1#

As multiplying by a scalar does not change the direction of a vector, this will be a unit vector perpendicular to the plane. Proceeding,

#||vec(u)|| = sqrt((-5)^2+1^2+(-7)^2) = sqrt(75)=5sqrt(3)#

Thus, our final result is

#vec(u)/u = ((-5","1","-7))/(5sqrt(3)) = (-1/sqrt(3), 1/(5sqrt(3)), -7/(5sqrt(3)))#

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