# Vector problem

• Jul 2nd 2009, 06:09 PM
chengbin
Vector problem
Given $\displaystyle \triangle$ABC, where points D, E, F are the midpoints of the sides BC, CA, and AB, respectively, and given an arbitrary point P, show that $\displaystyle \overrightarrow {PD}+\overrightarrow {PE}+\overrightarrow {PF}=\overrightarrow {PA}+\overrightarrow {PB}+\overrightarrow {PC}$.

(solution in the book)

Letting $\displaystyle \overrightarrow {a}, \overrightarrow {b}$, and $\displaystyle \overrightarrow {c}$ be the position vectors of the vertices A, B, and C, respectively (originating from point P)

$\displaystyle \overrightarrow {PD} = \frac {1}{2}(\overrightarrow {b}+\overrightarrow {c})$

$\displaystyle \overrightarrow {PE} = \frac {1}{2}(\overrightarrow {a}+\overrightarrow {c})$

$\displaystyle \overrightarrow {PF} = \frac {1}{2}(\overrightarrow {a}+\overrightarrow {b})$

Why? How do they get that?
• Jul 2nd 2009, 06:44 PM
malaygoel
Quote:

Originally Posted by chengbin

$\displaystyle \overrightarrow {PD} = \frac {1}{2}(\overrightarrow {b}+\overrightarrow {c})$

Why? How do they get that?

$\displaystyle \overrightarrow {PD} = (\overrightarrow {b}+\overrightarrow {BD})$

$\displaystyle \overrightarrow {PD} = (\overrightarrow {c}+\overrightarrow {CD})$

$\displaystyle 2\overrightarrow {PD} = (\overrightarrow {b}+\overrightarrow {BD}+\overrightarrow {c}+\overrightarrow {CD})$

now,
$\displaystyle \overrightarrow {CD}+\overrightarrow {BD}=\overrightarrow {0}$
since D is midpoint....hence the equation follows
• Jul 3rd 2009, 10:43 AM
chengbin

It makes sense, but I don't get how you got your explanation. I can't imagine an arbitrary point in my mind, which is the main reason I have so much trouble learning vectors.

Do you mind explaining this?

$\displaystyle \overrightarrow {PD} = (\overrightarrow {b}+\overrightarrow {BD})$
• Jul 3rd 2009, 11:09 AM
running-gag
Quote:

Originally Posted by chengbin

It makes sense, but I don't get how you got your explanation. I can't imagine an arbitrary point in my mind, which is the main reason I have so much trouble learning vectors.

Do you mind explaining this?

$\displaystyle \overrightarrow {PD} = (\overrightarrow {b}+\overrightarrow {BD})$

Well since $\displaystyle \overrightarrow {b}$ is the position vector of B from point P it comes $\displaystyle \overrightarrow {PB} = \overrightarrow {b}$

And then $\displaystyle \overrightarrow {PD} = \overrightarrow {PB}+\overrightarrow {BD} = \overrightarrow {b}+\overrightarrow {BD}$
• Jul 3rd 2009, 12:35 PM
chengbin
Is a picture representation possible? I really don't understand this.
• Jul 3rd 2009, 03:49 PM
yeongil
Here's a picture with triangle ABC, the midpoints D, E & F, showing the last thing running-gag posted:
$\displaystyle \overrightarrow {PD} = \overrightarrow {PB}+\overrightarrow {BD} = \overrightarrow {b}+\overrightarrow {BD}$

Vector $\displaystyle \overrightarrow {PB}$ is in red, and $\displaystyle \overrightarrow {PD}$ is in blue.

I didn't want to show all of the vectors because it would get cluttered.

01
• Jul 3rd 2009, 06:30 PM
chengbin
Thanks so much. I get it now.