# Alternative Vector Method to Finding Angles of Vertices of Triangle?

• ltkach2015
The accepted solution find the three angles of the triangle with vertices (2 -1 1) (1 -3 -5) (3 -4 -4) using the dot product.In summary, the accepted solution is to use the dot product to find the three angles of the triangle.f

#### ltkach2015

NOTE: I know the solution, but I challenge it. Specifically, I feel that the generally accepted solution to this problem
is not intuitive. I have shown the accepted solution, and below that I have shown what I feel seems more
intuitive (from a geometric standpoint)

1. The problem statement

find the three angles of the triangle with vertices (2 -1 1) (1 -3 -5) (3 -4 -4).

## Homework Equations

Dot Product: AB = ||A|| ||B|| cosθ

## The Attempt at a Solution

Analysis:

define the points:
ptA = (2 -1 1)
ptB = (1 -3 -5)
ptC = (3 -4 -4)
create (position) vectors wrt origin (0,0,0):
A = <2-0,-1-0,1-0> = <2, -1, 1>
B = <1, -3, -5>
C = <3, -4, -4>
define the angles:
α (between vectors A&B) (alpha)
β (between vectors B&C) (beta)
γ (between vectors C&A) (gamma)
apply the dot product of each (position) vector wrt to one another:
A*B = norm(A)*norm(B)*cos(α) ∴ α = 90.00°
B*C = norm(B)*norm(C)*cos(β) ∴ β = 22.49°
C*A = norm(A)*norm(B)*cos(γ) ∴ γ = 67.50°

1. The problem statement
AS ABOVE SO BELOW

## Homework Equations

Dot Product: AB = ||A|| ||B|| cosθ
Vector Addition: B = A + (B - A)

## The Attempt at a Solution

Flowchart: define the points:
create (position) vectors wrt origin (0,0,0):
find displacement vectors:
-start point is point A, followed by B, C, and back to A.
-then employ vector addition (final minus initial)
-then employ vector addition of the entire "displacement-vectored" triangle
define the angles & Please view second image:
apply the dot product of each (displacement) vector wrt to one another:

Analysis:
*skipped to step 3 of Flowchart*

find displacement vectors:
"path": Start at ptA, the ptB, ptC and then return to ptA

displacement vectors: B = A + (B - A)
=> dBA = B - A
=> dCB = C - B
=> dAC = A - C

triangle described by vector sum: dBA + dCB = dAC

define the angles then solve: (Please see image: https://www.physicsforums.com/forums/calculus-and-beyond-homework.156/attachments/my-understanding-of-the-vectors-png.77059/?temp_hash=21a6e771909902657d7fa4fa3b0d6ae7 [Broken])

α = acosd(sum(dAC.*dCB)*inv(norm(dAC)*norm(dCB))) ∴ α = 90.00°
β = acosd(sum(dBA.*dAC)*inv(norm(dBA)*norm(dAC))) ∴ β = 157.5085°
γ = acosd(sum(dCB.*dBA)*inv(norm(dCB)*norm(dBA))) ∴ γ = 112.4915°

Adjustments: (I am off by 180°)

END

Intuitive Explanation anyone?
How is my geometry wrong? What intuitive (geometric) reasons justify subtracting off 180°. Also, I am ignoring the sign of the angle, hence the abs operator above..I don't feel this is an effective solution. Can you please help me?

HAPPY NEW YEAR!

#### Attachments

Last edited by a moderator:
I stopped reading already when you start talking about position vectors. The vectors to use in the dot product definition are supposed to be relative vectors, i.e. in your case vector A could for instance be the relative vector from vertex A to vertex B, and vector B could be from vertex A to vertex C. You seem to use position vectors instead, which I don't think makes much sense.

Thanks for the response. Please see the second part of my original post.

And you're totally right about using the relative vectors. In fact my displacement vectors should amount to the relative vectors you described. Again further described in my original post.

My own approach/analysis is not as straight forward as just immediately finding the relative vectors (vertex-vertex).

My process: 1)find position vectors 2)find three displacement vectors
3) apply dot product and find the angles.

However, I am 180deg off for each vector.

Whats wrong here? Thank you.

Hi there. I think your approach is correct. The reason you have to take 180 - the angle found is that you have to flip one of the difference vectors if you want the angles between two of them (you can see it in the second picture).

Now, why do the 'accepted solution' and your 'correct approach' come out the same ? That's because there is something peculiar with the three original vectors, namely C = A + B. That means all the vectors are coplanar. OACB is a parallellogram (*). And triangle OBA has the same angles as trangle ABC.

( O = origin, (0,0,0) )

And the 'accepted solution' way fails miserably for e.g. A = 1,0,0 B = 0,1,0 C = 0,0,1 Your way comes out fine. (Check it!)

This leads me to a critical remark on this 'accepted solution' : many students will now think that's the right way to do it !

(*) with angle BOA = 90 degrees it's even a rectangle! The figure obscures that by misrepresenting vector/point A [edit:] I'm not doing it justice. It's just hard to see.

Last edited:
NOTE: I know the solution, but I challenge it. Specifically, I feel that the generally accepted solution to this problem
is not intuitive. I have shown the accepted solution, and below that I have shown what I feel seems more
intuitive (from a geometric standpoint)

1. The problem statement

find the three angles of the triangle with vertices (2 -1 1) (1 -3 -5) (3 -4 -4).

## Homework Equations

Dot Product: AB = ||A|| ||B|| cosθ

## The Attempt at a Solution

Analysis:

define the points:
ptA = (2 -1 1)
ptB = (1 -3 -5)
ptC = (3 -4 -4)
create (position) vectors wrt origin (0,0,0):
A = <2-0,-1-0,1-0> = <2, -1, 1>
B = <1, -3, -5>
C = <3, -4, -4>
define the angles:
α (between vectors A&B) (alpha)
β (between vectors B&C) (beta)
γ (between vectors C&A) (gamma)
apply the dot product of each (position) vector wrt to one another:
A*B = norm(A)*norm(B)*cos(α) ∴ α = 90.00°
B*C = norm(B)*norm(C)*cos(β) ∴ β = 22.49°
C*A = norm(A)*norm(B)*cos(γ) ∴ γ = 67.50°

1. The problem statement
AS ABOVE SO BELOW

## Homework Equations

Dot Product: AB = ||A|| ||B|| cosθ
Vector Addition: B = A + (B - A)

## The Attempt at a Solution

Flowchart: define the points:
create (position) vectors wrt origin (0,0,0):
find displacement vectors:
-start point is point A, followed by B, C, and back to A.
-then employ vector addition (final minus initial)
-then employ vector addition of the entire "displacement-vectored" triangle
define the angles & Please view second image:
apply the dot product of each (displacement) vector wrt to one another:

Analysis:
*skipped to step 3 of Flowchart*

find displacement vectors:
"path": Start at ptA, the ptB, ptC and then return to ptA

displacement vectors: B = A + (B - A)
=> dBA = B - A
=> dCB = C - B
=> dAC = A - C

triangle described by vector sum: dBA + dCB = dAC

define the angles then solve: (Please see image: https://www.physicsforums.com/forums/calculus-and-beyond-homework.156/attachments/my-understanding-of-the-vectors-png.77059/?temp_hash=21a6e771909902657d7fa4fa3b0d6ae7 [Broken])

α = acosd(sum(dAC.*dCB)*inv(norm(dAC)*norm(dCB))) ∴ α = 90.00°
β = acosd(sum(dBA.*dAC)*inv(norm(dBA)*norm(dAC))) ∴ β = 157.5085°
γ = acosd(sum(dCB.*dBA)*inv(norm(dCB)*norm(dBA))) ∴ γ = 112.4915°

Adjustments: (I am off by 180°)

END

Intuitive Explanation anyone?
How is my geometry wrong? What intuitive (geometric) reasons justify subtracting off 180°. Also, I am ignoring the sign of the angle, hence the abs operator above..I don't feel this is an effective solution. Can you please help me?

HAPPY NEW YEAR!

$$\angle\, A = \arccos \left( \frac{(\vec{B-A}) \cdot (\vec{C-A})}{|B-A| \, |C-A|} \right) \\ \angle\, B = \arccos \left( \frac{(\vec{A-B}) \cdot (\vec{C-B})}{|A-B| \, |C-B|} \right) , \;\text{etc.}$$

Last edited by a moderator:
NOTE: I know the solution, but I challenge it. Specifically, I feel that the generally accepted solution to this problem
is not intuitive. I have shown the accepted solution, and below that I have shown what I feel seems more
intuitive (from a geometric standpoint)

1. The problem statement

find the three angles of the triangle with vertices (2 -1 1) (1 -3 -5) (3 -4 -4).

## Homework Equations

Dot Product: AB = ||A|| ||B|| cosθ

## The Attempt at a Solution

Analysis:

define the points:
ptA = (2 -1 1)
ptB = (1 -3 -5)
ptC = (3 -4 -4)
create (position) vectors wrt origin (0,0,0):
A = <2-0,-1-0,1-0> = <2, -1, 1>
B = <1, -3, -5>
C = <3, -4, -4>
define the angles:
α (between vectors A&B) (alpha)
β (between vectors B&C) (beta)
γ (between vectors C&A) (gamma)
apply the dot product of each (position) vector wrt to one another:
A*B = norm(A)*norm(B)*cos(α) ∴ α = 90.00°
B*C = norm(B)*norm(C)*cos(β) ∴ β = 22.49°
C*A = norm(A)*norm(B)*cos(γ) ∴ γ = 67.50°
Above is just the part you call the "accepted solution".

It has been pointed out (by Filip Larsen and by BvU) that this solution has serious flaws as it's presented here. It shouldn't be accepted by anyone as a solution to this problem.

So, you might ask, - - why does it give the correct answer for the angles of this triangle?

Part of the reason was given by BvU. The plane determined by the triangle's vertices passes through the origin. This results in the position vectors lying in the same plane and the sum of the two smaller angles formed by them is equal to the third. But that's not enough to give the correct angles. For this triangle, the larger of these three angles is a right angle, so that the sum of the three angles formed by the position vectors is 180° as required.

It seems likely that the coordinates you give for the three vertices are actually the components for the three displacement vectors for the three sides of some triangle. Notice that <2, -1, 1> + <1, -3, -5> = <3 -4, -4> .

The "accepted solution" is correct for such triangles.

(You also have some mis-labeled items in the firgure for the"accepted solution". )

Last edited by a moderator:
But that's not enough to give the correct angles.
Whereas I stated
OACB is a parallellogram. And triangle OBA has the same angles as trangle ABC.
-- the "And" to be interpreted as "therefore".

@SammyS: If am wrong, could you show me where ?

Whereas I stated -- the "And" to be interpreted as "therefore".

@SammyS: If am wrong, could you show me where ?
Interpreting the "And" as "Therefore" does clarify the statement. I'm not sure which way I read it initially, but I agree with your argument about the two triangles.

... triangle OBA has the same angles as triangle ABC.​

And that does follow directly from OACB being a parallellogram

The "accepted solution" given by OP doesn't directly find the angles of either of these triangles. It finds the three angles between the directions of the three position vectors taken in pairs. With the four points being co-planar, the sum of the two smallest angles is equal to the third. The sum of the three angles is 180° only because the largest of the three is 90°. I.e., this method only gave three angles of correct measure, because the parallelogram was indeed a rectangle.

So, I conclude that you were not wrong with anything you stated.

Thanks. Must admit that I missed that the 'accepted solution' does indeed calculate angles BOC and AOC and not ABO and BOA. So the 90 degrees is needed to let it end up with the correct angles. Thanks. Must admit that I missed that the 'accepted solution' does indeed calculate angles BOC and AOC and not ABO and BOA. So the 90 degrees is needed to let it end up with the correct angles. Yes. Not only that, the angle ∠BOC has the same measure as the angle at vertex A of triangle ABC, so that it's likely that the measures of the two acute angles would be switched.

why don't you use the law of cosines. After finding the distance between all three vertices, plug them in and find all three angles.

@BvU

Aww yes it totally makes sense to subtract 180 degrees. A 2D vector analysis scenario would more easily provide a geometric justification for such a subtraction.

So may I conclude that when seeking the angles between the vertices of any shape & using the vectors method, specifically Displacement Vectors, one must ensure that any pair of vectors being considered: only one vector is modified (made negative) so that the pair vectors may then have a relative initial point (tail) extending to the another vertex (head) of the shape.

Basically, (only when using displacement vectors) flip one of the vectors as you said. Or simply subtract 180 degrees which is the same thing.

Yes I think the accepted solution is not acceptable, and I understand that the given interior angles of the triangle happened to be the same angles of the parallelepiped.

• Ray Vickson
$$\angle\, A = \arccos \left( \frac{(\vec{B-A}) \cdot (\vec{C-A})}{|B-A| \, |C-A|} \right) \\ \angle\, B = \arccos \left( \frac{(\vec{A-B}) \cdot (\vec{C-B})}{|A-B| \, |C-B|} \right) , \;\text{etc.}$$

@Ray Vickson

Yes totally makes sense.

I did not notice the distinction between the latter analysis and your guidance, until now.

I can see now that this explicitly confirms my conclusion that an application of negative sign onto one of the vectors is necessary when using displacement vectors.

Thanks for you help.