Question

What is the unit vector that is orthogonal to the plane containing `(8i + 12j + 14k)` and `(2i + 3j – 7k)`?

Answer 1

`vecu=< (-3sqrt(13))/13, (2sqrt(13))/13, 0 >`

Explanation:

A vector which is orthogonal (perpendicular, norma) 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=< 8,12,14 >` and `vecb=< 2,3,-7 >`, `vecaxxvecb`is found by

For the `i` component, we have

`(12*-7)-(14*3)=-84-42=-126`

For the `j` component, we have

`-[(8*-7)-(2*14)]=-[-56-28]=84`

For the `k` component, we have

`(8*3)-(12*2)=24-24=0`

Our normal vector is `vecn=< -126,84,0 >`

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((-126)^2+(84)^2+(0)^2)`

`|vecn|=sqrt(15878+7056+0)=sqrt(22932)=42sqrt(13)`

The unit vector is then given by:

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

`vecu=(< -126,84,0 >)/(42sqrt(13))`

`vecu=1/(42sqrt(13))< -126,84,0>`

or equivalently,

`vecu=< -3/(sqrt(13)),2/(sqrt(13)),0>`

You may also choose to rationalize the denominator:

`vecu=< (-3sqrt(13))/13, (2sqrt(13))/13, 0 >`