Vectors

Hi,

I'm struggling a bit with these two questions.

1) Let P and Q denote points in

\mathbb{R}^3

with position vectors

p

and

q

respectively. Find the position vector

a

of a point

A

on the line through

P

and

Q

which is

3

times as far from

Q

as from

P

: does this fix the point

A

uniquely?

2) Consider a triangle

ABC

. Let

M

be a point on

AB

such that

BM = 2MA

and

N

a point on

AC

such that

CN = 2NA

. Using vectors, prove that the line segment

MN

is parallel to

BC

and

MN = \dfrac{1}{3} BC

.

For the first one I got a result of

\vec{a} = \dfrac{3}{4}\vec{p} + \dfrac{1}{4}\vec{q}

but I have no clue about when it says "does this fix the point

A

uniquely".

For the second one I constructed a triangle and so far worked out:

\vec{BA} + \vec{AC} = \vec{BC}

\vec{BM} + 2\vec{MA} - 2\vec{NA} - \vec{CN} = \vec{BC}

\vec{BM} - \vec{CN} + 2(\vec{MA} - \vec{NA}) = \vec{BC}

\vec{BM} - \vec{CN} + 2\vec{MN} = \vec{BC}

Not sure if I'm on the right track with the latter. Any help is appreciated. Thanks!

(edited 8 years ago)

Reply 1

ghostwalker

17

Original post by r3l3ntl3ss

For the first one I got a result of

\vec{a} = \dfrac{3}{4}\vec{p} + \dfrac{1}{4}\vec{q}

but I have no clue about when it says "does this fix the point

A

uniquely".

OK, that's one point.

If you can recall back to your schooldays, you can divide a line internally, and externally, which may give you a clue.

For the second one I constructed a triangle and so far worked out:

\vec{BA} + \vec{AC} = \vec{BC}

\vec{BM} + 2\vec{MA} - 2\vec{NA} - \vec{CN} = \vec{BC}

\vec{BM} - \vec{CN} + 2(\vec{MA} - \vec{NA}) = \vec{BC}

\vec{BM} - \vec{CN} + 2\vec{MN} = \vec{BC}

Not sure if I'm on the right track with the latter. Any help is appreciated. Thanks!

Don't follow your second line there - check your diagram.

To show that two lines are parallel, i.e. the vectors representing those lines are paralllel, you need to show that one is a scalar multiple of the other.

Reply 2

r3l3ntl3ss

OP

10

Original post by ghostwalker

OK, that's one point.

If you can recall back to your schooldays, you can divide a line internally, and externally, which may give you a clue.

Don't follow your second line there - check your diagram.

To show that two lines are parallel, i.e. the vectors representing those lines are paralllel, you need to show that one is a scalar multiple of the other.

ah yeah for the second one I was being a bit stupid, so I've got:

\vec{BA} + \vec{AC} = \vec{BC}

\vec{BA} = \vec{BM} + \vec{MA}

\vec{BA} = 2\vec{MA} + \vec{MA}

\vec{BA} = 3\vec{MA}

\vec{AC} = -\vec{NA} -\vec{CN}

\vec{AC} = -\vec{NA} - 2\vec{NA} = -3\vec{NA}

\vec{BA} + \vec{AC} = 3\vec{MA} -3\vec{NA}

\vec{BC} = 3(\vec{MA} - \vec{NA}) = 3\vec{MN}

\vec{MN} = \dfrac{1}{3}\vec{BC}

I'm still a bit unsure about the first one though

Reply 3

ghostwalker

17

Original post by r3l3ntl3ss

I'm still a bit unsure about the first one though

Here's an image of the two possible positions of A.

You should be able to use a similar method for the second point to the one you used for the first - unless you just wrote it down. If you need more help let me know, but have a go first.

OP

Original post by ghostwalker

Here's an image of the two possible positions of A.

You should be able to use a similar method for the second point to the one you used for the first - unless you just wrote it down. If you need more help let me know, but have a go first.