Vector proof

**TheJacksonater** · July 30th 2008, 02:50 PM

Hey guys,

I need help proving that if |a+b| = |a-b|, then a and b are perpendicular vectors.

I also need to prove that prove that the angle formed by joining any point on a circle to the end points of a diameter is 90 degrees.

Thanks!

**Plato** · July 30th 2008, 03:02 PM

$\left\| {A - B} \right\|^2 = \left( {A - B} \right) \cdot \left( {A - B} \right) = A \cdot A - 2A \cdot B + B \cdot B$
And $\left\| {A + B} \right\|^2 = \left( {A + B} \right) \cdot \left( {A + B} \right) = A \cdot A + 2A \cdot B + B \cdot B$
From the given we know that $\left\| {A + B} \right\|^2 = \left\| {A - B} \right\|^2 \Rightarrow \quad 4A \cdot B = 0 \Rightarrow \quad A \cdot B = 0$
QED

**TheJacksonater** · July 30th 2008, 06:49 PM

Originally Posted by Plato

$\left\| {A - B} \right\|^2 = \left( {A - B} \right) \cdot \left( {A - B} \right) = A \cdot A - 2A \cdot B + B \cdot B$
And $\left\| {A + B} \right\|^2 = \left( {A + B} \right) \cdot \left( {A + B} \right) = A \cdot A + 2A \cdot B + B \cdot B$
From the given we know that $\left\| {A + B} \right\|^2 = \left\| {A - B} \right\|^2 \Rightarrow \quad 4A \cdot B = 0 \Rightarrow \quad A \cdot B = 0$
QED

Thanks, but I'm still not following...

First off I still don't see how they're equal, (one had +2A, the other had -2A) then where did the 4A dot B = 0 come from??

**Soroban** · July 30th 2008, 06:54 PM

Hello, TheJacksonater!

Prove: .if $|\vec a+\vec b| \:= \:|\vec a-\vec b|$ , then: . $\vec a \perp \vec b$

Let: . $\begin{array}{ccc}\vec a &=& \langle x_1,y_1\rangle \\ \vec b &=& \langle x_2,y_2\rangle \end{array}$

Then: . $\vec a + \vec b \:=\:\langle x_1+x_2,\:y_1+y_2\rangle \quad\Rightarrow\quad |\vec a + \vec b| \:=\:\sqrt{(x_1+x_2)^2 + (y_1+y_2)^2}$

. And: . $\vec a - \vec b \:=\:\langle x_1-x_2,\:y_1-y_2\rangle \quad\Rightarrow\quad |\vec a - \vec b| \:=\:\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}$

Since $|a + b| \:=\:|a - b|$ , we have: . $(x_1+x_2)^2 + (y_1+y_2)^2 \:=\:(x_1-x_2)^2 + (y_1-y_2)^2$

. . $x_1^2 + 2x_1x_2 + x_2^2 + y_1^2 + 2y_1y_2 + y_2^2 \;=\;x_1^2 - 2x_1x_2 + x_2^2 + y_1^2 - 2y_1y_2 + y_2^2$

. . $4x_1x_2 + 4y_1y_2 \:=\:0 \quad\Rightarrow\quad x_1x_2 + y_1y_2\:=\:0 \quad\Rightarrow\quad \langle x_1,y_1\rangle \cdot \langle x_2,y_2\rangle \:=\:0$

Therefore: . $\vec a\cdot\vec b \:=\:0 \quad\Rightarrow\quad \vec a \perp \vec b$

Prove that the angle formed by joining any point
on a circle to the endpoints of a diameter is 90°.

Let the circle be: . $\begin{array}{ccc}x &=& r\cos\theta \\ y &=& r\sin\theta \end{array}$

Let $A(r,0) \text{ and }B(\text{-}r,0)$ be the endpoints of the diameter.

Let $C(r\cos\theta, r\sin\theta)$ be any point on the circle.

Draw segments $AC$ and $BC.$

Code:

                |
              * * *
          *     |     *   C
        *       |       o(r·cosθ,r·sinθ)
       *        |     /  *
                |   /
      *         | / θ     *
  - B o - - - - * - - - - o A - -
(-r,0)*         |         *(r,0)
                |
       *        |        *
        *       |       *
          *     |     *
              * * * 
                |

$\overrightarrow{AC} \;=\;\bigg\langle r\cos\theta - r,\:r\sin\theta\bigg\rangle \;=\;\bigg\langle r(\cos\theta - 1),\:r\sin\theta\bigg\rangle$

$\overrightarrow{BC} \;=\;\bigg\langle r\cos\theta + r,\:r\sin\theta\bigg\rangle \;=\;\bigg\langle r(\cos\theta + 1),\:r\sin\theta\bigg\rangle$

Then: . $\overrightarrow{AC}\cdot\overrightarrow{BC} \;=\;[r(\cos\theta - 1)][r(\cos\theta+1)] + (r\sin\theta)(r\sin\theta)$

. . . . . . . . . . . $= \;r^2(\cos^2\!\theta-1) + r^2\sin^2\!\theta$

. . . . . . . . . . . $=\;-r^2(1-\cos^2\!\theta) + r^2\sin^2\!\theta$

. . . . . . . . . . . $= \;-r^2\sin^2\!\theta + r^2\sin^2\!\theta$

. . . . . . . . . . . $= \;0$

Therefore: . $\overrightarrow{AC} \perp \overrightarrow{BC} \quad\Rightarrow\quad \angle C \:=\:90^o$

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