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

1 Answer
Narad T.
Nov 4, 2016

The unit vector is ((11veci)/sqrt486-(14vecj)/sqrt486-(13veck)/sqrt486)⎛⎝11→i√486−14→j√486−13→k√486⎞⎠

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〉→u=⟨1,−2,3⟩ and vecv=〈-4,-5,2〉→v=⟨−4,−5,2⟩
The cross product vecu→uxvecv→v ==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)

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