What is the unit vector that is normal to the plane containing #(- 3 i + j -k)# and #( i +k )?

1 Answer
Nov 29, 2016

Answer:

The unit vector is:

#hatC = sqrt(6)/6hati + sqrt(6)/3hatj - sqrt(6)/6hatk#

Explanation:

Let #barA = -3hati + hatj - hat k#
Let #barB = hati + hatk#
Let #barC = # the vector perpendicular to both #barA and barB# such that the following is true:

#barC = barA xx barB#

#barC = | (hati, hatj, hatk, hati, hatj), (-3,1,-1,-3,1), (1,0,1,1,0) |#

#barC = {1(1) - (-1)(0)}hati + {-1(1) - (-3)(1)}hatj + {-3(0) - (1)(1)}hatk#

#barC = hati + 2hatj -hatk#

Check:

#barC*barA = (1)(-3) + (2)(1) + (-1)(-1) = 0#
#barC*barB = (1)(1) + (2)(0) + (-1)(1) = 0#

This checks but #barC# is not a unit vector.

To make it a unit vector we must divide #barC# by its magnitude.

#|barC| = sqrt(1^2 + 2^2 + (-1)^2)#

#|barC| = sqrt(6)#

Dividing by #sqrt(6)# is the same as multiplying by #sqrt(6)/6#

The unit vector is:

#hatC = sqrt(6)/6hati + sqrt(6)/3hatj - sqrt(6)/6hatk#