What is the unit vector that is normal to the plane containing #( i +k)# and #(i+2j+2k) #?

1 Answer
Jun 10, 2017

Answer:

#vecn = 2/3i + 1/3j -2/3k#

Explanation:

The vector we're looking for is #vec n = aveci+bvecj+cveck# where #vecn * (i+k)=0# AND #vecn * (i+2j+2k)=0#, since #vecn# is perpendicular to both of those vectors.

Using this fact, we can make a system of equations:

#vecn * (i+0j+k) = 0#

#(ai+bj+ck)(i+0j+k)=0#

#a+c = 0#

#vecn * (i+2j+2k) = 0#

#(ai+bj+ck) * (i+2j+2k) = 0#

#a+2b+2c = 0#

Now we have #a+c = 0# and #a+2b+2c = 0#, so we can say that:

#a+c = a+2b+2c#

#0 = 2b+c#

#therefore a+c = 2b+c#

#a = 2b#

#a/2 = b#

Now we know that #b = a/2# and #c = -a#. Therefore, our vector is:

#ai + a/2j-ak#

Finally, we need to make this a unit vector, meaning we need to divide each coefficient of the vector by its magnitude. The magnitude is:

#|vecn|=sqrt(a^2+(a/2)^2+(-a)^2)#

#|vecn|=sqrt(9/4a^2)#

#|vecn| = 3/2a#

So our unit vector is:

#vecn = a/(3/2a)i + (a/2)/(3/2a)j + (-a)/(3/2a)k#

#vecn = 2/3i + 1/3j -2/3k#

Final Answer