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

1 Answer
Jul 1, 2016

Answer:

#= (-2 hat i + hat j + 7 hat k)/(3 sqrt(6))#

Explanation:

you'll do this by calculating the vector cross product of these 2 vectors to get the normal vector

so #vec n = (- 3 i + j -k) times (2i - 3 j +k )#

# = det [(hat i, hat j , hat k), (-3,1,-1),(2,-3,1)]#

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

#= -2 hat i + hat j + 7 hat k#

the unit normal is #hat n = (-2 hat i + hat j + 7 hat k)/(sqrt((-2)^2 + 1^2 +7^2))#

#= (-2 hat i + hat j + 7 hat k)/(3 sqrt(6))#

you could check this by doing a scalar dot product between the normal and each of the original vectors, should get zero as they are orthogonal.

so for example

#vec v_1 * vec n #

#= (- 3 i + j -k) * (-2i + j + 7k )#

#= 6 + 1 - 7 = 0#