What is the unit vector that is orthogonal to the plane containing # (32i-38j-12k) # and # (41j+31k) #?

1 Answer
Jul 11, 2016

Answer:

#hat(n) = 1/(sqrt(794001))[-343vec(i) - 496vec(j) + 656vec(k)]#

Explanation:

The cross product of two vectors produces a vector orthogonal to the two original vectors. This will be normal to the plane.

#|(vec(i), vec(j), vec(k)),(32,-38,-12),(0,41,31)| = vec(i)|(-38,-12),(41,31)| - vec(j)|(32,-12),(0,31)| + vec(k)|(32,-38),(0,41)|#

#vec(n)=vec(i)[-38*31 - (-12)*41] - vec(j)[32*31 - 0] + vec(k)[32*41 - 0]#

#vec(n) =-686vec(i) - 992vec(j) + 1312vec(k)#

#|vec(n)| = sqrt((-686)^2+(-992)^2+1312^2) = 2sqrt(794001)#

#hat(n) = (vec(n))/(|vec(n)|)#

#hat(n) = 1/(sqrt(794001))[-343vec(i) - 496vec(j) + 656vec(k)]#