What is the unit vector that is orthogonal to the plane containing # (2i + 3j – 7k) # and # (3i - j - 2k) #?

1 Answer
Nov 30, 2016

Answer:

The answer is #=1/sqrt579*〈-13,-17,-11〉#

Explanation:

To calculate a vector perpendicular to two other vectors, you have to calculate the cross product

Let #vecu=〈2,3,-7〉# and #vecv=〈3,-1,-2〉#

The cross product is given by the determinant

# | (i,j,k), (u_1,u_2,u_3), (v_1,v_2,v_3) | #

#vecw=| (i,j,k), (2,3,-7), (3,-1,-2) |#

#=i(-6-7)-j(-4+21)+k(-2-9)#

#=i(-13)+j(-17)+k(-11)#

#=〈-13,-17,-11〉#

To verify that #vecw# is perpendicular to #vecu# and #vecv#

We do a dot product.

#vecw.vecu=〈-13,-17,-11〉.〈2,3,-7〉=-26--51+77=0#

#vecw.vecv=〈-13,-17,-11〉.〈3,-1,-2〉=-39+17+22=0#

As the dot products #=0#, #vecw# is perpendicular to #vecu# and #vecv#

To calculate the unit vector, we divide by the modulus

#hatw=vecw/(∥vecw∥)=1/sqrt579*〈-13,-17,-11〉#