Question

Find a vector perpendicular to each of the vector `a=2i+j+3k` and `b=3i+5j-2k`, which has magnitude of `10` units?

Answer 1

`(10sqrt(3))/39(-17hat(i)+13hat(j)+7hat(k))`

Explanation:

The cross product of two non-parallel vectors will be perpendicular to both:

`(2hat(i)+hat(j)+hat(k)) xx (3hat(i)+5hat(j)-2hat(k)) = abs((hat(i), hat(j), hat(k)), (2, 1, 3), (3, 5, -2))`

`color(white)((2hat(i)+hat(j)+hat(k)) xx (3hat(i)+5hat(j)-2hat(k))) = -17hat(i)+13hat(j)+7hat(k)`

This vector has modulus:

`sqrt((-17)^2+13^2+7^2) = sqrt(289+169+49) = sqrt(507) = 13sqrt(3)`

So scaling it, a suitable vector is:

`10/(13sqrt(3))(-17hat(i)+13hat(j)+7hat(k))= (10sqrt(3))/39(-17hat(i)+13hat(j)+7hat(k))`

Answer 2

The vector is `=10/sqrt(507)<-17,13,7>`

Explanation:

A vector perpendicular to `2` other vectors is calculated with he cross product, that is 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=〈2,1,3〉` and `vecb=〈3,5,-2〉`

Therefore,

`| (veci,vecj,veck), (2,1,3), (3,5,-2) |`

`=veci| (1,3), (5,-2) | -vecj| (2,3), (3,-2) | +veck| (2,1), (3,5) |`

`=veci((1)*(-2)-(3)*(5))-vecj((-)*(-2)-(3)*(3))+veck((2)*(5)-(1)*(3))`

`=〈-17,13,7〉=vecc`

Verification by doing 2 dot products

`〈-17,13,7〉.〈2,1,3〉=(-17)*(2)+(13)*(1)+(7)*(3)=0`

`〈-17,13,7〉.〈3,5,-2〉=(-17)*(3)+(13)*(5)+(7)*(-2)=0`

So,

`vecc` is perpendicular to `veca` and `vecb`

The magnitude of the vector is

`||vecc||=||<-17,13,7>|| = sqrt((-17)^2+(13^2)+(7^2))`

`=sqrt507`

The unit vector is

`hatc=1/sqrt(507)<-17,13,7>`

Therefore,

`k*sqrt((-17/sqrt507)^2+(13/sqrt507)^2+(7/sqrt507)^2)=10`

`k=10`

The vector of magnitude `=10` is

`=10/sqrt(507)<-17,13,7>`