Given the vector a = i + 2j - 2k, find a possible set of values for y and z such that b = yj + zk is a unit vector that is perpendicular to a. Should I use dot product or cross product? How do i identify which method to use? Please help.

1 Answer
Somebody N.
Apr 6, 2018

#color(blue)(1/sqrt(2)(bbj+bbk) )#

#color(blue)(1/sqrt(2)(-bbj-bbk) )#

Explanation:

I will use the scaler product( dot product) for this:

We know if two vectors are perpendicular, the their dot product is zero.

For vectors a and b

Dot Product

#bba*bb(b)=||bba||*||bb(b)||cos(theta)#

#bba*bb(b)=0+2y-2z=0#

For a non-zero solution:

#y=z#

Our vector will just be in the form:

#0bbi+ybbj+zbbk#

For arbitrary #y#

Check:

#bba*bb(b)#

For #y=3=z#

#(bbi+2bbj-2bbk)*(0bbi+3bbj+3bbk)=0+6-6=0#

The unit vector is:

#hat(bb(b))=(bb(b))/(||bb(b)||#

#||bb(b)||=sqrt(y^2+z^2)#

#bb(b)=(ybbj+zbbk)#

#(ybbj+zbbk)/(sqrt(y^2+z^2))#

For arbitrary #y#

#y= 1#

#(bbj+bbk)/(sqrt((1)^2+(1)^2))=1/sqrt(2)(bbj+bbk)#

#y=-1#

#(-bbj-bbk)/(sqrt((-1)^2+(-1)^2))=1/sqrt(2)(-bbj-bbk)#

The set of values will be:

#y in RR#

Notice since we have no #bbi# component in the unit vectors both lie in the #yz# plane.

PLOT:

enter image source here

You could use the cross product here. Since the cross product is non-commutative you would have to do both:

#bba xx bb(b)# and #bb(b) xx bba#

I would use the dot product where possible.

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