Show that the vector a+b is perpendicular to the line $AB$.

Question

The position vectors of points $A$ and $B$ relative to the origin $O$ are a and b respectively and |a|=|b| .

How does one show that the vector a+b is perpendicular to the line $AB$?

I know that a$\cdot$b $=0$ when vector a and b are perpendicular, but how do I use this information to solve the above question?

Answer 1

The direction of the line $AB$ can be represented by the vector $\mathbf b - \mathbf a$. (Thanks @amd!)

Now, $(\mathbf b - \mathbf a)\cdot(\mathbf a+\mathbf b) = -|\mathbf a|^2 + |\mathbf b|^2 = 0$ ( as $|\mathbf a| = |\mathbf b|$)

Answer 2

Assume the points $A$ and $B$ are distinct.We'll use parentheses for for points and brackets for vectors. Write $$A=(x_*,y_*),B(x_{**},y_{**}).$$ Then $$a=[x_*,y_*],b=[x_{**},y_{**}]$$ $$|a|=\sqrt(x_*^2+y_*^2),|b|=\sqrt(x_{**}^2+y_{**}^2)$$ $$\text {Since }|a|=|b|,\sqrt(x_*^2+y_*^2)=\sqrt(x_{**}^2+y_{**}^2) $$ $$\text{and thus }x_*^2+y_*^2=x_{**}^2+y_{**}^2$$ $$\text{and so }x_{**}^2-x_*^2+y_{**}^2-y_*^2=0.$$ The vector $a+b$ is $[x_*+x_{**},y_*+y_{**}]$. A vector in the direction of the line $AB$ is $\mathbf v=[x_{**}-x_*,y_{**}-y_*]$. Showing that a vector is perpendicular to the line $AB$ is the same as showing that that vector is perpendicular to the vector $\mathbf v.$ Showing that two vectors are perpendicular is equivalent to showing that their dot product is 0. $$\mathbf v \bullet (a+b)=[x_{**}-x_*,y_{**}-y_*]\bullet [x_*+x_{**},y_*+y_{**}]$$ $$=x_{**}^2-x_*^2+y_{**}^2-y_*^2=0,$$ $$Q.E.D..$$