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

1 Answer
Feb 29, 2016

Answer:

#\hat{n}_{AB} = -1/\sqrt{62}(\hat{i}+6\hat{j}+5\hat{k})#

Explanation:

The unit vector perpendicular to the plane containing two vectors #\vec{A_{}}# and #\vec{B_{}}# is :

#\hat{n}_{AB} = \frac{\vec{A} \times \vec{B}} {|\vec{A} \times \vec{B}|}#

#\vec{A_{}} = 3\hat{i}+2\hat{j}-3\hat{k}; \qquad \vec{B_{}} = \hat{i}-\hat{j}+\hat{k};#

#\vec{A_{}}\times\vec{B_{}} = -(\hat{i}+6\hat{j}+5\hat{k});#
#|\vec{A_{}}\times\vec{B_{}}|=\sqrt{(-1)^2+(-6)^2+(-5)^2}=\sqrt{62}#

#\hat{n}_{AB} = -1/\sqrt{62}(\hat{i}+6\hat{j}+5\hat{k})#.