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

1 Answer
Dec 19, 2016

Answer:

The answer is #=〈0,-3/sqrt13,-2/sqrt13〉#

Explanation:

We do a cross product to find the vector orthogonal to the plane

The vector is given by the determinant

# | (hati,hatj,hatk), (3,2,-3), (1,-2,3) |#

#=hati(6-6)-hatj(9--3)+hatk(-6-2)#

#=〈0,-12,-8〉#

Verification by doing the dot product

#〈0,-12,-8〉.〈3,2,-3〉=0-24+24=0#

#〈0,-12,-8〉.〈1,-2,3〉=0+24-24=0#

The vector is orthgonal to the other 2 vectors

The unit vector is obtained by dividing by the modulus

#∥〈0,-12,-8〉∥=sqrt(0+144+64)=sqrt208=4sqrt13#

Thre unit vector is #=1/(4sqrt13)〈0,-12,-8〉#

#=〈0,-3/sqrt13,-2/sqrt13〉#