What is the unit vector that is normal to the plane containing (3i + 4j - k)(3i+4jk) and (2i+ j - 3k)(2i+j3k)?

2 Answers

=\pm(\frac{-11i+7j-5k}{\sqrt{195}})=±(11i+7j5k195)

Explanation:

Taking the cross product of given vectors (3i+4j-k)(3i+4jk) & (2i+j-3k)(2i+j3k) as follows

(3i+4j-k)\times (2i+j-3k)(3i+4jk)×(2i+j3k)

=-11i+7j-5k=11i+7j5k

Above vector will be normal to the given vectors hence the unit normal vector

\hat n=\frac{-11i+7j-5k}{|-11i+7j-5k|}ˆn=11i+7j5k|11i+7j5k|

=\frac{-11i+7j-5k}{\sqrt{(-11)^2+7^2+(-5)^2}}=11i+7j5k(11)2+72+(5)2

=\frac{-11i+7j-5k}{\sqrt{195}}=11i+7j5k195

hence the unit vector \hat nˆn normal to the plane containing given vectors will be

=\pm(\frac{-11i+7j-5k}{\sqrt{195}})=±(11i+7j5k195)

Jul 28, 2018

The unit vector is = <〈-11/sqrt195,7/sqrt195,-5/sqrt195〉>=<11195,7195,5195>

Explanation:

The vector normal to a plane containing 22 vectors is given by the cross product of 2 vectors, This is calculated with the determinant

| (veci,vecj,veck), (d,e,f), (g,h,i) |

where veca=〈d,e,f〉 and vecb=〈g,h,i〉 are the 2 vectors

Here, we have veca=〈3,4,-1〉 and vecb=〈2,1,-3〉

Therefore,

| (veci,vecj,veck), (3,4,-1), (2,1,-3) |

=veci| (4,-1), (1,-3) | -vecj| (3,-1), (2,-3) | +veck| (3,4), (2,1) |

=veci((4)*(-3)-(1)*(-1))-vecj((3)*(-3)-(2)*(-1))+veck((3)*(1)-(4)*(2))

=〈-11,7,-5〉=vecc

Verification by doing 2 dot products

〈-11,7,-5〉.〈3,4,-1〉=(-11)*(3)+(7)*(4)+(-5)*(-1)=0

〈-11,7,-5〉.〈2,1,-3〉=(-11)*(2)+(7)*(1)+(-5)*(-3)=0

So,

vecc is perpendicular to veca and vecb

||vecc||= ||〈-11,7,-5〉|| = sqrt((-11)^2+(7)^2+(-5)^2)

=sqrt(121+49+25)

=sqrt(195)

The unit vector is

hatc=vecc/(||vecc||)=1/sqrt(195)〈-11,7,-5〉