Answer to Question #335172 in Analytic Geometry for math

Question

By considering the angles between the vectors, show that a + b and a – b are

perpendicular when (a) = (b)

Answer 1

We remind that for any two vectors $f_1$ and $f_2$we have: $\cos\,(\alpha)=\frac{(f_1,f_2)}{|f_1||f_2|},$ where $\alpha$ is the angle between $f_1$ and $f_2$, $|.|$ denotes a norm of a vector and $(.,.)$ denotes a dot (scalar) product. Consider the scalar product: $(a+b,a-b).$ Using the properties of the scalar product, we get: $(a+b,a-b)=(a,a-b)+(b,a-b)=(a,a)-(a,b)+(b,a)-(b,b)=|a|^2-|b|^2$.

Thus, in case $a=b$ we receive: $(a+b,a-b)=0$. It means that $cos(\beta)=0$, where $\beta$ is the angle between $a+b$ and $a-b$. Thus, $\beta=90^0$ and it means that vectors are perpendicular.