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

1 Answer
Dec 28, 2016

Answer:

#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

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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.