Given that the vector AB= 2i+3j and the vector CB= 5i + j

a. Show that AB is perpendicular to AC

how to do this????.... have no clue on how to start please help

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- Mar 20th 2012, 04:28 PMmathkid12Vector past paper question
Given that the vector AB= 2i+3j and the vector CB= 5i + j

a. Show that AB is perpendicular to AC

how to do this????.... have no clue on how to start please help - Mar 20th 2012, 07:44 PMProve ItRe: Vector past paper question
You should know that $\displaystyle \displaystyle \begin{align*} \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos{\theta} \end{align*}$, where $\displaystyle \displaystyle \begin{align*} \theta \end{align*}$ is the angle between $\displaystyle \displaystyle \begin{align*} \mathbf{a} \end{align*}$ and $\displaystyle \displaystyle \begin{align*} \mathbf{b} \end{align*}$.

When $\displaystyle \displaystyle \begin{align*} \theta = 90^{\circ} \end{align*}$ (i.e. when $\displaystyle \displaystyle \begin{align*} \mathbf{a} \end{align*}$ and $\displaystyle \displaystyle \begin{align*} \mathbf{b} \end{align*}$ are perpendicular), we have $\displaystyle \displaystyle \begin{align*} \cos{\theta} = \cos{90^{\circ}} = 0 \end{align*}$, which means that the dot product of the two vectors will also be 0.

Therefore, to show that two vectors are perpendicular, take the dot product and see if the dot product is 0. - Mar 25th 2012, 05:53 PMHallsofIvyRe: Vector past paper question
Of course, part of the problem is that you are

**not**given vector "AC". You have to use the fact that AC= AB+ BC= AB- CB and then take the dot product of AC with AB.