What is the unit vector that is orthogonal to the plane containing (2i + 3j – 7k) and (3i + 2j - 6k) ?
1 Answer
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
For the
(3*-6)-(2*-7)=-18+ 14=-4
For the
-[(2*-6)-(3*-7)]=-[-12+21]=-9
For the
(2*2)-(3*3)=4-9=-5
Our vector is
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.
