Question

Given that A = B and AperpendicularB. What is the angle
between (A+B) and (A-B) ?​

Answer 1

Answer:

Squaring both sides,

|A+B|2=|A−B|2  

Since  A.A=|A|2  

|A+B|2=|A−B|2  

(A+B).(A+B)=(A−B).(A−B)  

A.A+A.B+B.A+B.B=A.A−A.B−B.A+B.B  ( Using distributive property)

|A|2+2A.B+|B|2=|A|2−2A.B+|B|2  

4A.B=0  

A.B=0  

|A||B|cos(θ)=0  

Explanation:

Squaring both sides,

|A+B|2=|A−B|2  

Since  A.A=|A|2  

|A+B|2=|A−B|2  

(A+B).(A+B)=(A−B).(A−B)  

A.A+A.B+B.A+B.B=A.A−A.B−B.A+B.B  ( Using distributive property)

|A|2+2A.B+|B|2=|A|2−2A.B+|B|2  

4A.B=0  

A.B=0  

|A||B|cos(θ)=0  

Since A and B are non-zero vectors,  cos(θ) must be zero. It implies that A and B are perpendicular as in the interval  [0,π]  cos vanishes only at  π/2