If IA+BI = IA-BI, find the angle between the vector A and B and show that the two vectos are perpendicular to each other. How to answer this question?

3 Answers
Mar 2, 2017

drawn

Let the angle between two vectors #A and B# be #theta#

So

#abs(A+B)^2=abs(A)^2+abs(B)^2+2abs(A)abs(B)costheta#

Again

#abs(A-B)^2=abs(A)^2+abs(B)^2+2abs(A)abs(B)cos(180-theta)#

By the given condition

#abs(A+B)=abs(A-B)#

So

#costheta=cos(180-theta)#

#=>theta=180-theta#

#=>theta=90^@#

Mar 2, 2017

Answer:

See below.

Explanation:

Considering that #A ne 0# and #B ne 0#

If #norm(A+B)# = #norm(A-B)# then

#norm(A+B)^2# = #norm(A-B)^2# or

#normA^2+2<< A, B >> + normB^2= normA^2-2<< A,B >> +normB^2#

or simplifying

#<< A,B >> = 0#

This means that the scalar product of #A# and #B# is null so the two vectors are orthogonal, and the angle between then is obtained knowing that

#<< A,B >> = cos(hat(AB))normA normB#. Now supposing that #normA ne 0# and #norm B ne 0# we have

# cos(hat(AB)) = << A, B >> /(normA normB) = 0# so #hat(AB)=pi/2#

Mar 25, 2017

We can use some properties of the Vector Norm.

# ||A|| = sqrt(A * A) => ||A||^2 = A * A #

We are given that:

# || A+B || = || A - B|| #

If we square both sides we get:

# || A+B ||^2 = || A - B||^2 #

Using the above property this becomes:

# (A+B) * (A+B) = (A - B) * (A-B) #

And the Vector Dot Product is distributive over vector addition so we can expand the above expression;

# (A+B) * A + (A+B) * B = (A - B) * A - (A - B) * B #
# :. A*A+B*A + A*B+B* B = A*A - B* A - A*B + B * B #
# :. B*A + A*B = - B* A - A*B #

And the vector product is Commutative so #A*B=B*A#;

# :. A*B + A*B = - A* B - A*B #
# :. 2A*B = -2 A* B #
# :. 4A*B = 0 #
# :. A*B = 0 #

And If #A*B=0# then either:

#A=0# or #B=0#; or
#A# and #B# are perpendicular. QED