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

1 Answer
David G.
Jan 27, 2016

There are two steps in finding this solution: 1. Find the cross product of the two vectors to find a vector orthogonal to the plane containing them and 2. normalise that vector so that it has unit length.

Explanation:

The first step in solving this problem is to find the cross product of the two vectors. The cross product by definition finds a vector orthogonal to the plane in which the two vectors being multiplied lie.

#(i−2j+3k) xx (i−j+k)#

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

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

= #(i+2j+k)#

This is a vector orthogonal to the plane, but it is not yet a unit vector. To make it one we need to 'normalise' the vector: divide each of its components by its length. The length of a vector #(ai+bj+ck)# is given by:

#l = sqrt(a^2+b^2+c^2)#

In this case:

#l = sqrt(1^2+2^2+1^2) = sqrt6#

Dividing each component of #(i+2j+k)# by #sqrt6# yields our answer, which is that the unit vector orthogonal to the plane in which #(i−2j+3k) and (i−j+k)# lie is:

#(i/sqrt6+2/sqrt6j+k/sqrt6)#

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