1. ## Geometric Vectors

In the diagram below, $\bigtriangleup ABC$ is equilateral and $D, E, F$ are the midpoints of its sides. Express each sum as a single vector.

If some of my answers are wrong, can you please thoroughly explain to tell me how to fix it, like where do you move the vectors in a diagram and why? (I'm struggling a bit with my homework right now. . .and I have no one to help me. . .)

a) ${\overrightarrow {AF} } + {\overrightarrow {DB} } = {\overrightarrow {DF} }$

b) ${\overrightarrow {DE} } + {\overrightarrow {DB} } = {\overrightarrow {AE} }$

c) ${\overrightarrow {FA} } + {\overrightarrow {EB} } = {\overrightarrow {DB} }$

d) ${\overrightarrow {DA} } + {\overrightarrow {EC} } = {\overrightarrow {FC} }$

e) ${\overrightarrow {AF} } + {\overrightarrow {DE} } = {\overrightarrow {AC} }$

f) ${\overrightarrow {EC} } + {\overrightarrow {FD} } = {\overrightarrow {0} }$

$\overrightarrow {FE} = \overrightarrow {DB} \quad \Rightarrow \quad \overrightarrow {AF} + \overrightarrow {DB} = \overrightarrow {AE}$

3. Originally Posted by Plato
$\overrightarrow {FE} = \overrightarrow {DB} \quad \Rightarrow \quad \overrightarrow {AF} + \overrightarrow {DB} = \overrightarrow {AE}$

That's not really a review. . .I have no idea what you did except that you moved one of the vectors, which I did the same thing for that question, but just in a different spot. . .

4. Originally Posted by Macleef
In the diagram below, $\bigtriangleup ABC$ is equilateral and $D, E, F$ are the midpoints of its sides. Express each sum as a single vector.

If some of my answers are wrong, can you please thoroughly explain to tell me how to fix it, like where do you move the vectors in a diagram and why? (I'm struggling a bit with my homework right now. . .and I have no one to help me. . .)

a) ${\overrightarrow {AF} } + {\overrightarrow {DB} } = {\overrightarrow {DF} }$

b) ${\overrightarrow {DE} } + {\overrightarrow {DB} } = {\overrightarrow {AE} }$

c) ${\overrightarrow {FA} } + {\overrightarrow {EB} } = {\overrightarrow {DB} }$

d) ${\overrightarrow {DA} } + {\overrightarrow {EC} } = {\overrightarrow {FC} }$

e) ${\overrightarrow {AF} } + {\overrightarrow {DE} } = {\overrightarrow {AC} }$

f) ${\overrightarrow {EC} } + {\overrightarrow {FD} } = {\overrightarrow {0} }$
The important point to realize is this: vectors have both direction and magnitude. In your addition of vectors, you neglected the first of these facts (and one important fact about the addition of vectors that results as a consequence of the direction of vectors). Namely, you should always add from "head to tail".

If you call a vector ${\overrightarrow {AF}}$, you must realize that you are saying that the vector goes in the direction from point A to point F. Conversely, vector ${\overrightarrow {FA}}$ goes in the exact opposite direction of ${\overrightarrow {AF}}$.

When you add vectors, the "head" of the first vector (the end point of that vector, which in the vector ${\overrightarrow {AF}}$ would be F), must be connected to the "tail" of the second vector (the beginning point of that vector, which in the vector ${\overrightarrow {DB}}$ would be D).

At this point, the resulting vector of the addition of these two goes from the tail of the first vector to the head of the second. In other words, in ${\overrightarrow {AF} } + {\overrightarrow {DB} }$, where ${\overrightarrow {DB}}$ is congruent to, and therefore equal to, ${\overrightarrow {FE}}$, the tail of the first is A, the head of the second is E, and so:

${\overrightarrow {AF} } + {\overrightarrow {FE} } = {\overrightarrow {AE} }$

5. Originally Posted by ecMathGeek
The important point to realize is this: vectors have both direction and magnitude. In your addition of vectors, you neglected the first of these facts (and one important fact about the addition of vectors that results as a consequence of the direction of vectors). Namely, you should always add from "head to tail".

If you call a vector ${\overrightarrow {AF}}$, you must realize that you are saying that the vector goes in the direction from point A to point F. Conversely, vector ${\overrightarrow {FA}}$ goes in the exact opposite direction of ${\overrightarrow {AF}}$.

When you add vectors, the "head" of the first vector (the end point of that vector, which in the vector ${\overrightarrow {AF}}$ would be F), must be connected to the "tail" of the second vector (the beginning point of that vector, which in the vector ${\overrightarrow {DB}}$ would be D).

At this point, the resulting vector of the addition of these two goes from the tail of the first vector to the head of the second. In other words, in ${\overrightarrow {AF} } + {\overrightarrow {DB} }$, where ${\overrightarrow {DB}}$ is congruent to, and therefore equal to, ${\overrightarrow {FE}}$, the tail of the first is A, the head of the second is E, and so:

${\overrightarrow {AF} } + {\overrightarrow {FE} } = {\overrightarrow {AE} }$
Okay, I get it, but can you please check my answers to see if I'm in the right track?

b) ${\overrightarrow {AE} }$

c) ${\overrightarrow {DA} }$

d) ${\overrightarrow {BF} }$

e) ${\overrightarrow {AC} }$

f) ${\overrightarrow {0} }$

By the way, the tail is the arrow and the head is the initial point? Also, is it possible to flip the vectors (opposites) in addition or does this only apply to subtraction?

6. Originally Posted by Macleef
a) ${\overrightarrow {AF} } + {\overrightarrow {DB} } = {\overrightarrow {DF} } ~No~ \rightarrow {\overrightarrow {AE} }~Yes$

b) ${\overrightarrow {DE} } + {\overrightarrow {DB} } = {\overrightarrow {AE} }~Yes$

c) ${\overrightarrow {FA} } + {\overrightarrow {EB} } = {\overrightarrow {DB} }~No~ \rightarrow {\overrightarrow {DA} }~No$

d) ${\overrightarrow {DA} } + {\overrightarrow {EC} } = {\overrightarrow {FC} }~No~ \rightarrow {\overrightarrow {FA} }~No~ \rightarrow {\overrightarrow {BF} }~Yes$

e) ${\overrightarrow {AF} } + {\overrightarrow {DE} } = {\overrightarrow {AC} }~Yes$

f) ${\overrightarrow {EC} } + {\overrightarrow {FD} } = {\overrightarrow {0} }~Yes$

7. Originally Posted by ecMathGeek
Could you please explain c and are there more than one answer for d because I get vector BF

8. Originally Posted by Macleef
By the way, the tail is the arrow and the head is the initial point? Also, is it possible to flip the vectors (opposites) in addition or does this only apply to subtraction?
The tail is the initial point, the head is the arrow (think "head of an arrow").

If you flip a vector that is being added to another, you have effectively subtracted that vector (or added its negative).

9. Originally Posted by Macleef
Could you please explain c and are there more than one answer for d because I get vector BF
BF is the only possible solution to problem d.

When doing these problems, if the two vectors you are adding are not already "head to tail" then you must find two congruent vectors that have the same directions as the two vectors being added but are also situated "head to tail." In the case of problem c, you must find the one and only set of vectors congruent to the two being added that appear "head to tail." I won't tell you which vectors those are. Since you have done all the other problems, I'm certain you can do this one on your own.

10. Originally Posted by ecMathGeek
BF is the only possible solution to problem d.

When doing these problems, if the two vectors you are adding are not already "head to tail" then you must find two congruent vectors that have the same directions as the two vectors being added but are also situated "head to tail." In the case of problem c, you must find the one and only set of vectors congruent to the two being added that appear "head to tail." I won't tell you which vectors those are. Since you have done all the other problems, I'm certain you can do this one on your own.
Thank you soooo much for your help!

Can you also check my answers for the following questions?

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The diagram below shows 3 congruent equilateral triangles. Express each difference as a single vector.

a) ${\overrightarrow {BA} } - {\overrightarrow {BC} } = {\overrightarrow {BA} } + {\overrightarrow {CB} } = {\overrightarrow {CA} }$

b) ${\overrightarrow {BA} } - {\overrightarrow {BD} } = {\overrightarrow {BA} } + {\overrightarrow {DB} } = {\overrightarrow {DA} }$

c) ${\overrightarrow {CE} } - {\overrightarrow {AE} } = {\overrightarrow {CE} } + {\overrightarrow {EA} } = {\overrightarrow {CA} }$

d) ${\overrightarrow {AE} } - {\overrightarrow {ED} } = {\overrightarrow {AE} } + {\overrightarrow {DE} } = {\overrightarrow {BE} }$

AND. . . . . . .

The diagram below contains two squares. Express each difference as a single vector.

a) ${\overrightarrow {SQ} } - {\overrightarrow {ST} } = {\overrightarrow {SQ} } + {\overrightarrow {TS} } = {\overrightarrow {SP} }$

b) ${\overrightarrow {QT} } - {\overrightarrow {QP} } = {\overrightarrow {QT} } + {\overrightarrow {PQ} } = {\overrightarrow {PT} }$

c) ${\overrightarrow {PR} } - {\overrightarrow {QS} } = {\overrightarrow {PR} } + {\overrightarrow {SQ} } = {\overrightarrow {UQ} }$

d) ${\overrightarrow {PT} } - {\overrightarrow {TS} } = {\overrightarrow {PT} } + {\overrightarrow {ST} } = {\overrightarrow {PU} }$

11. Originally Posted by Macleef
Thank you soooo much for your help!

Can you also check my answers for the following questions?

----

The diagram below shows 3 congruent equilateral triangles. Express each difference as a single vector.

a) ${\overrightarrow {BA} } - {\overrightarrow {BC} } = {\overrightarrow {BA} } + {\overrightarrow {CB} } = {\overrightarrow {CA} }$

b) ${\overrightarrow {BA} } - {\overrightarrow {BD} } = {\overrightarrow {BA} } + {\overrightarrow {DB} } = {\overrightarrow {DA} }$

c) ${\overrightarrow {CE} } - {\overrightarrow {AE} } = {\overrightarrow {CE} } + {\overrightarrow {EA} } = {\overrightarrow {CA} }$

d) ${\overrightarrow {AE} } - {\overrightarrow {ED} } = {\overrightarrow {AE} } + {\overrightarrow {DE} } = {\overrightarrow {BE} }$

AND. . . . . . .

The diagram below contains two squares. Express each difference as a single vector.

a) ${\overrightarrow {SQ} } - {\overrightarrow {ST} } = {\overrightarrow {SQ} } + {\overrightarrow {TS} } = {\overrightarrow {SP} }$

b) ${\overrightarrow {QT} } - {\overrightarrow {QP} } = {\overrightarrow {QT} } + {\overrightarrow {PQ} } = {\overrightarrow {PT} }$

c) ${\overrightarrow {PR} } - {\overrightarrow {QS} } = {\overrightarrow {PR} } + {\overrightarrow {SQ} } = {\overrightarrow {UQ} }$

d) ${\overrightarrow {PT} } - {\overrightarrow {TS} } = {\overrightarrow {PT} } + {\overrightarrow {ST} } = {\overrightarrow {PU} }$
Your answers to the first a,b,c and the second b,c,d are correct.