# 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?

Mar 2, 2017

Let the angle between two vectors $A \mathmr{and} B$ be $\theta$

So

${\left\mid A + B \right\mid}^{2} = {\left\mid A \right\mid}^{2} + {\left\mid B \right\mid}^{2} + 2 \left\mid A \right\mid \left\mid B \right\mid \cos \theta$

Again

${\left\mid A - B \right\mid}^{2} = {\left\mid A \right\mid}^{2} + {\left\mid B \right\mid}^{2} + 2 \left\mid A \right\mid \left\mid B \right\mid \cos \left(180 - \theta\right)$

By the given condition

$\left\mid A + B \right\mid = \left\mid A - B \right\mid$

So

$\cos \theta = \cos \left(180 - \theta\right)$

$\implies \theta = 180 - \theta$

$\implies \theta = {90}^{\circ}$

Mar 2, 2017

See below.

#### Explanation:

Considering that $A \ne 0$ and $B \ne 0$

If $\left\lVert A + B \right\rVert$ = $\left\lVert A - B \right\rVert$ then

${\left\lVert A + B \right\rVert}^{2}$ = ${\left\lVert A - B \right\rVert}^{2}$ or

${\left\lVert A \right\rVert}^{2} + 2 \left\langleA , B\right\rangle + {\left\lVert B \right\rVert}^{2} = {\left\lVert A \right\rVert}^{2} - 2 \left\langleA , B\right\rangle + {\left\lVert B \right\rVert}^{2}$

or simplifying

$\left\langleA , B\right\rangle = 0$

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

$\left\langleA , B\right\rangle = \cos \left(\hat{A B}\right) \left\lVert A \right\rVert \left\lVert B \right\rVert$. Now supposing that $\left\lVert A \right\rVert \ne 0$ and $\left\lVert B \right\rVert \ne 0$ we have

$\cos \left(\hat{A B}\right) = \frac{\left\langleA , B\right\rangle}{\left\lVert A \right\rVert \left\lVert B \right\rVert} = 0$ so $\hat{A B} = \frac{\pi}{2}$

Mar 25, 2017

We can use some properties of the Vector Norm.

$| | A | | = \sqrt{A \cdot A} \implies | | A | {|}^{2} = A \cdot A$

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:

$\left(A + B\right) \cdot \left(A + B\right) = \left(A - B\right) \cdot \left(A - B\right)$

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

$\left(A + B\right) \cdot A + \left(A + B\right) \cdot B = \left(A - B\right) \cdot A - \left(A - B\right) \cdot B$
$\therefore A \cdot A + B \cdot A + A \cdot B + B \cdot B = A \cdot A - B \cdot A - A \cdot B + B \cdot B$
$\therefore B \cdot A + A \cdot B = - B \cdot A - A \cdot B$

And the vector product is Commutative so $A \cdot B = B \cdot A$;

$\therefore A \cdot B + A \cdot B = - A \cdot B - A \cdot B$
$\therefore 2 A \cdot B = - 2 A \cdot B$
$\therefore 4 A \cdot B = 0$
$\therefore A \cdot B = 0$

And If $A \cdot B = 0$ then either:

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