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

1 Answer
Dec 30, 2016

Answer:

#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

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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 >#