What is the unit vector that is orthogonal to the plane containing # ( - 5 i + 4 j - 5 k) # and # (4 i + 4 j + 2 k) #?

1 Answer
Jan 23, 2016

Answer:

There are two steps: (1) find the cross product of the vectors, (2) normalise the resultant vector. In this case, the answer is:
#((28)/(46.7)i-(10)/(46.7)j-(36)/(46.7)k)#

Explanation:

The cross product of two vectors yields a vector that is orthogonal (at right angles) to both.

The cross product of two vectors #(a#i#+b#j#+c#k#)# and #(p#i#+q#j#+r#k#)# is given by #(b*r-c*q)i+(c*p-a*r)j+(a*q-b*p)k#

First step is to find the cross product:

#(−5i+4j−5k) xx (4i+4j+2k) = ((4*2)-(4*-5)i + ((-5*4)-(-5*2))j + ((-5*4)-(4*4))k =((8-(-20))i+(-20-(-10)j+((-20)-16)k)=(28i-10j-36k)#

This vector is orthogonal to both the original vectors, but it is not a unit vector. To make it a unit vector we need to normalise it: divide each of its components by the length of the vector.

#l=sqrt(28^2+(-10)^2+(-36)^2)=46.7# units

The unit vector orthogonal to the original vectors is:

#((28)/(46.7)i-(10)/(46.7)j-(36)/(46.7)k)#

This is one unit vector that is orthogonal to both the original vectors, but there is another - the one in the exact opposite direction. Simply changing the sign of each of the components yields a second vector orthogonal to the original vectors.

#(-(28)/(46.7)i+(10)/(46.7)j+(36)/(46.7)k)#

(but it's the first vector that you should offer as the answer on a test or assignment!)