What is the unit vector that is orthogonal to the plane containing # ( i - 2 j + 3 k) # and # (4 i + 4 j + 2 k) #?

1 Answer
Feb 22, 2016

Answer:

There are two steps in solving this question: (1) taking the cross product of the vectors and then (2) normalizing the resultant. In this case, the final unit vector is #(-16/sqrt500i+10/sqrt500j+12/sqrt500k)# or #(-16/22.4i+10/22.4j+12/22.4k)#.

Explanation:

First step: cross product of the vectors.

#(i-2j+3k)xx(4i+4j+2k) = (((-2)*2-3*4))i+(3*4-1*2)j+(1*4-(-2)*4)k)=((-4-12)i+(12-2)j+(4-(-8))k)=(-16i+10j+12k)#

Second step: normalize the resultant vector.

To normalize a vector we divide each element by the length of the vector. To find the length:

#l=sqrt((-16)^2+10^2+12^2)=sqrt500~~22.4#

Putting it all together, the unit vector orthogonal to the given vectors can be represented as:

#(-16/sqrt500i+10/sqrt500j+12/sqrt500k)# or #(-16/22.4i+10/22.4j+12/22.4k)#