Vector addition (triangle law)

[Indranil]

Vector addition (triangle law):
If the resultant vector is c, then we can write c = √a² + b² + 2ab cosθ
1. But in the diagram no-1, there is only c = √a² + b² why?
2. In the diagram no-2, Why is there the formula c = √a² + b² - 2ab cosθ here? is the formula valid?
Please see the diagrams below:

 

[mmm4444bot]

… But in the diagram no-1, there is only c = √a² + b² why?

… In the diagram no-2, Why is there the formula c = √(a² + b² - 2ab cosθ) here? …

In diagram #1, they used the Pythagorean Formula to find c because that's a right triangle.

In diagram #2, they used the Law of Cosines to find c because that's not a right triangle.

This stumbling is preventable. You ought to know these things, before studying vector algebra. This is why we have encouraged you to adopt a structered form of study (eg: following textbooks, in order). You are putting yourself at a disadvantage, by jumping around from topic to topic (not a good way to learn undergraduate math). If I remember correctly, you're trying to jam a few year's worth of study into a single year, in order to take an entrance exam next year. I'm concerned that this plan will not succeed.

 

[Deleted member 4993]

Vector addition (triangle law):
If the resultant vector is c, then we can write c = √a² + b² + 2ab cosθ
1. But in the diagram no-1, there is only c = √a² + b² why?
2. In the diagram no-2, Why is there the formula c = √a² + b² - 2ab cosθ here? is the formula valid?
Please see the diagrams below:

Indranil - please read law of cosines (that you are using) in a triangle - very carefully. Identify each term (like θ), and locate those in the diagram.

Note that in (1) θ = 90 and cosθ = 0.

If you are still confused contact back with a new sketch with a triangle whose vertices are PQR and length of the corresponding sides are p, q and r. Identify each angle and side according to your new sketch and re-write law of cosine using these variables (p, q, r, etc.).

 

[MarkFL]

...This stumbling is preventable. You ought to know these things, before studying vector algebra. This is why we have encouraged you to adopt a structered form of study (eg: following textbooks, in order). You are putting yourself at a disadvantage, by jumping around from topic to topic (not a good way to learn undergraduate math). If I remember correctly, you're trying to jam a few year's worth of study into a single year, in order to take an entrance exam next year. I'm concerned that this plan will not succeed.

This is excellent advice. Mathematical knowledge at this level is generally cumulative, in that topics depend on material that should be learned prior to the given topic. There is a generally agreed upon progression by professionals in education of mathematics that should be followed.

 

[Indranil]

In diagram #1, they used the Pythagorean Formula to find c because that's a right triangle.

In diagram #2, they used the Law of Cosines to find c because that's not a right triangle.

This stumbling is preventable. You ought to know these things, before studying vector algebra. This is why we have encouraged you to adopt a structured form of study (eg: following textbooks, in order). You are putting yourself at a disadvantage, by jumping around from topic to topic (not a good way to learn undergraduate math). If I remember correctly, you're trying to jam a few year's worths of study into a single year, in order to take an entrance exam next year. I'm concerned that this plan will not succeed.

But according to the cosine rule, it should be c = √a² + b² + 2ab cosθ
But why is, in the diagram no-2, the formula 'c = √a² + b² - 2ab cosθ' applied?

 

[MarkFL]

But according to the cosine rule, it should be c = √a² + b² + 2ab cosθ
But why is, in the diagram no-2, the formula 'c = √a² + b² - 2ab cosθ' applied?

You have the Law of Cosines wrong, while the other formula, with the negative sign, is correct:

$\displaystyle c=\sqrt{a^2+b^2-2ab\cos(\theta)}$

 

[Indranil]

You have the Law of Cosines wrong, while the other formula, with the negative sign, is correct:

$\displaystyle c=\sqrt{a^2+b^2-2ab\cos(\theta)}$

When to use the cosine law 'c = √a² + b² - 2ab cosθ'? I mean is the law valid for both the addition and subtraction of vectors?

 

[MarkFL]

When to use the cosine law 'c = √a² + b² - 2ab cosθ'? I mean is the law valid for both the addition and subtraction of vectors?

It can be used any time we know two sides of a triangle ($\displaystyle a$ and $\displaystyle b$), and the angle subtended by the two known sides ($\displaystyle \theta$).

 

[Indranil]

It can be used any time we know two sides of a triangle ($\displaystyle a$ and $\displaystyle b$), and the angle subtended by the two known sides ($\displaystyle \theta$).

If you don't mind, could you please explain what 'the angle subtended by the two known sides' means?

 

[mmm4444bot]

… please explain what 'the angle subtended by the two known sides' means?

The verb "to subtend" has a technical meaning, but here Mark simply meant the angle between the two sides.

We use the Law of Cosines when we know two sides and the angle between them.

PS: You can google "subtended" to find a definition.