Vector problem

chengbin · #1 $Old$ July 2nd, 2009, 08:09 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 · #2 $Old$ July 2nd, 2009, 08:44 PM

Quote:

Originally Posted by chengbin $View Post$

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

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 · #3 $Old$ July 3rd, 2009, 12:43 PM

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 · #4 $Old$ July 3rd, 2009, 01:09 PM

Quote:

Originally Posted by chengbin $View Post$

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 · #5 $Old$ July 3rd, 2009, 02:35 PM

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

yeongil · #6 $Old$ July 3rd, 2009, 05:49 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 · #7 $Old$ July 3rd, 2009, 08:30 PM

Thanks so much. I get it now.