How do you normalize (2 i -3j -k)(2i−3j−k)?

1 Answer
Morgan
Jan 9, 2018

hatu=<2/sqrt(14),-3/sqrt(14),-1/sqrt(14)>ˆu=<2√14,−3√14,−1√14>

Explanation:

To normalize a vector is to find unit vector (vector with magnitude/length of one) in the same direction as the given vector. This can be accomplished by dividing the given vector by its magnitude.

hatu=vecv/(|vecv|)ˆu=→v∣∣→v∣∣

Given vecv=<2,-3,-1>→v=<2,−3,−1>, we can calculate the magnitude of the vector:

abs(vecv)=sqrt((v_x)^2+(v_y)^2+(v_z)^2)∣∣→v∣∣=√(vx)2+(vy)2+(vz)2

=>=sqrt((2)^2+(-3)^2+(-1)^2)⇒=√(2)2+(−3)2+(−1)2

=>=sqrt(4+9+1)⇒=√4+9+1

=>=sqrt(14)⇒=√14

We now have:

hatu=(<2,-3,-1>)/sqrt(14)ˆu=<2,−3,−1>√14

=>hatu=<2/sqrt(14),-3/sqrt(14),-1/sqrt(14)>⇒ˆu=<2√14,−3√14,−1√14>

We can also rationalize the denominator for each of the components:

=>hatu=<(sqrt(14))/7,-(3sqrt(14))/14,-(sqrt(14))/14>⇒ˆu=<√147,−3√1414,−√1414>

Which may also be written ((sqrt(14))/7i-(3sqrt(14))/14j-(sqrt(14))/14k)(√147i−3√1414j−√1414k)

Hope that helps!

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