What is the unit vector that is orthogonal to the plane containing # ( i - 2 j + 3 k) # and # ( - 4 i - 5 j + 2 k) #?

1 Answer
Nov 4, 2016

Answer:

The unit vector is #((11veci)/sqrt486-(14vecj)/sqrt486-(13veck)/sqrt486)#

Explanation:

Firstly, we need the vector perpendicular to other two vectros:
For this we do the cross product of the vectors:
Let #vecu=〈1,-2,3〉# and #vecv=〈-4,-5,2〉#
The cross product #vecu#x#vecv# #=#the determinant
#∣((veci,vecj,veck),(1,-2,3),(-4,-5,2))∣#
#=veci∣((-2,3),(-5,2))∣-vecj∣((1,3),(-4,2))∣+veck∣((1,-2),(-5,-5))∣#
#=11veci-14vecj-13veck#
So #vecw=〈11,-14,-13〉#
We can check that they are perpendicular by doing the dot prodct.
#vecu.vecw=11+28-39=0#
#vecv.vecw=-44+70-26=0#
The unit vector #hatw=vecw/(∥vecw∥)#
The modulus of #vecw=sqrt(121+196+169)=sqrt486#
So the unit vector is #((11veci)/sqrt486-(14vecj)/sqrt486-(13veck)/sqrt486)#