Vector problem

**chengbin** · July 2nd 2009, 06:09 PM

Given $\triangle$ 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 $\overrightarrow {PD}+\overrightarrow {PE}+\overrightarrow {PF}=\overrightarrow {PA}+\overrightarrow {PB}+\overrightarrow {PC}$ .

(solution in the book)

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

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

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

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

Why? How do they get that?

**malaygoel** · July 2nd 2009, 06:44 PM

Originally Posted by chengbin

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

Why? How do they get that?

from triangle addition of vectors,

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

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

adding above two equations,

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

now,
$\overrightarrow {CD}+\overrightarrow {BD}=\overrightarrow {0}$
since D is midpoint....hence the equation follows

**chengbin** · July 3rd 2009, 10:43 AM

maylaygoel, thanks for your explaination.

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?

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

**running-gag** · July 3rd 2009, 11:09 AM

Originally Posted by chengbin

maylaygoel, thanks for your explaination.

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?

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

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

And then $ \overrightarrow {PD} = \overrightarrow {PB}+\overrightarrow {BD} = \overrightarrow {b}+\overrightarrow {BD} $

**chengbin** · July 3rd 2009, 12:35 PM

Is a picture representation possible? I really don't understand this.

**yeongil** · July 3rd 2009, 03:49 PM

Here's a picture with triangle ABC, the midpoints D, E & F, showing the last thing running-gag posted:
$\overrightarrow {PD} = \overrightarrow {PB}+\overrightarrow {BD} = \overrightarrow {b}+\overrightarrow {BD} $

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

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

01

**chengbin** · July 3rd 2009, 06:30 PM

Thanks so much. I get it now.

Thread: Vector problem

LinkBack

Thread Tools

Display

Vector problem

Similar Math Help Forum Discussions

vector problem

3D vector problem

Vector problem

vector problem

vector problem sum

Search Tags