Question about determining the angles of triangle given two vectors

In summary, the problem is that you were not taking into account the length of the hypotenuse when working with the cross products.
  • #1
RoboNerd
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<<Mentor note: Missing template due to originally being posted elsewhere>>

Hello everyone.

I have the following problem:

Determine the angles of a triangle where two sides of a triangle are formed by the vectors

A = 3i -4j -k and B=4i -j + 3k

I thought that I would find the third side being represented by vector R which would be equal to A-B, that is
R = -i -3j -4k.

I would then take the cross products of each combination of the two vectors and find the angle between them and these would be the angles of the triangle.

Dot producting A and B vectors, I get the angle between them to be 60 degrees.

Dot producting the A and R vectors, I also get 60 degrees.

Thus with the final angle = 180 - 60 - 60, the last angle should be 60 degrees also.

This is blatantly wrong as my solutions tell me that the answer is:
arcos 7 /sqrt(75), arcos sqrt(26)/sqrt(75), 90 degrees, or 36degrees4', 53degrees56', 90 degrees

Could anyone please direct me as to what I did wrong and what mistakes need to be fixed? Thanks in advance.
 
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  • #2
Your method and results are fine.
 
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  • #3
RoboNerd said:
<<Mentor note: Missing template due to originally being posted elsewhere>>

Hello everyone.

I have the following problem:

Determine the angles of a triangle where two sides of a triangle are formed by the vectors

A = 3i -4j -k and B=4i -j + 3k

I thought that I would find the third side being represented by vector R which would be equal to A-B, that is
R = -i -3j -4k.

I would then take the cross products of each combination of the two vectors and find the angle between them and these would be the angles of the triangle.

Dot producting A and B vectors, I get the angle between them to be 60 degrees.

Dot producting the A and R vectors, I also get 60 degrees.

Thus with the final angle = 180 - 60 - 60, the last angle should be 60 degrees also.

This is blatantly wrong as my solutions tell me that the answer is:
arcos 7 /sqrt(75), arcos sqrt(26)/sqrt(75), 90 degrees, or 36degrees4', 53degrees56', 90 degrees

Could anyone please direct me as to what I did wrong and what mistakes need to be fixed? Thanks in advance.

Your solution is correct.

However, a slightly easier way might have been to compute the lengths |A|, |B| and |R| = |A-B|; you would find these to all be equal, so your triangle is equilateral.
 
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  • #4
Thank you everyone for your inputs. It is a relief to know that I did not mess something up.
 

1. How do you determine the angles of a triangle given two vectors?

The angles of a triangle can be determined using the dot product formula: cos⁡θ = (u · v) / (‖u‖‖v‖), where u and v are the two vectors and θ is the angle between them.

2. What is the dot product formula?

The dot product formula is a mathematical formula used to calculate the angle between two vectors. It is represented by the symbol · and is defined as: u · v = u1v1 + u2v2 + u3v3, where u and v are two vectors with three components each.

3. Can the dot product formula be used for any type of triangle?

Yes, the dot product formula can be used to calculate the angles of any type of triangle, whether it is acute, right, or obtuse.

4. Do the two vectors have to be perpendicular to each other for the dot product formula to work?

No, the two vectors do not have to be perpendicular to each other for the dot product formula to work. The formula can be applied to any two given vectors.

5. Is there an alternative method for determining the angles of a triangle given two vectors?

Yes, there is another method called the law of cosines which can also be used to calculate the angles of a triangle given two vectors. This formula is: c^2 = a^2 + b^2 - 2ab cos⁡C, where c is the side opposite to angle C and a and b are the other two sides of the triangle.

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