Question

What is the unit vector that is orthogonal to the plane containing `(2i + 3j – 7k)` and `(3i + 2j - 6k)`?

Answer 1

`vecu=< -4/(sqrt(122)),-9/(sqrt(122)),-5/(sqrt(122))>`

Explanation:

A vector which is orthogonal to a plane containing two vectors is also orthogonal to the given vectors. We can find a vector which is orthogonal to both of the given vectors by taking their cross product. We can then find a unit vector in the same direction as that vector.

Given `veca=< 2,3,-7 >` and `vecb=< 3,2,-6 >`, `vecaxxvecb` is found by

For the `i` component, we have

`(3*-6)-(2*-7)=-18+ 14=-4`

For the `j` component, we have

`-[(2*-6)-(3*-7)]=-[-12+21]=-9`

For the `k` component, we have

`(2*2)-(3*3)=4-9=-5`

Our vector is `vecn=< -4,-9,-5 >`

Now, to make this a unit vector, we divide the vector by its magnitude. The magnitude is given by:

`|vecn|=sqrt((n_x)^2+(n_y)^2+(n_z)^2)`

`|vecn|=sqrt((-4)^2+(-9)^2+(-5)^2)`

`|vecn|=sqrt(16+81+25)=sqrt(122)`

The unit vector is then given by:

`vecu=(vecaxxvecb)/(|vecaxxvecb|)`

`vecu=(< -4,-9,-5 >)/(sqrt(122))`

`vecu=1/(sqrt(122))< -4,-9,-5 >`

or equivalently,

`vecu=< -4/(sqrt(122)),-9/(sqrt(122)),-5/(sqrt(122))>`

You may also choose to rationalize the denominator.