Vectors

[gwiyomi17]

Given |a|=3, |b|=5, and |a+b|=7, determine |a-b|.

(a and b are vectors)

 

how to start this?

[John]

I'd start by considering the geometric interpretations of vector addition and subtraction. There are a couple of ways to go about getting your answer, depending on what exactly you notice about the resulting shapes.

Edited December 3, 2013 by John

[studiot]

Hello gwiomi

 

I suggest you let the vectors be A= (x₁,y₁) and B=(x₂y₂) and consider the definitions of modulus and sum for vectors, and then form equations from these definitions.

[Olinguito]

Basically you are looking at a triangle with side lengths 3, 5, and 7. The angle between the two shorter sides (call it [latex]\theta[/latex]) can be found by the cosine rule:

[latex]c^2\ =\ a^2+b^2-2ab\cos\theta[/latex]

Now you want to look at the triangle two of whose sides have lengths 3 and 5, the angle between them being [latex]\pi-\theta[/latex]. The length of the opposite side can again be found by the cosine rule.

Edited December 3, 2013 by Olinguito

[Country Boy]

You can also think of this as a parallelogram with sides of length 3 and 5. |a+ b| is the length of one diagonal, |a- b| is the length of the other.

 

In particular, since |a+ b|= 7, one half of that parallelogram, the one with the larger angle between the sides, is a triangle with sides of length 3, 5, and 7. Taking [tex]\theta[/tex] to be the angle between the two sides, by the cosine law, [tex]7^2= 3^2+ 5^2- 2(3)(5)cos(\theta)[/tex]. You can use that to solve for [itex]\theta[\itex]. then the other angle, [tex]\phi[/tex], is the supplement of [tex]\theta[/tex]. And, letting x= |a- b|, [tex]x^2= 3^2+ 5^2- 2(3)(5)cos(\phi)[/tex].

[John]

You can also think of this as a parallelogram with sides of length 3 and 5. |a+ b| is the length of one diagonal, |a- b| is the length of the other.

 

In particular, since |a+ b|= 7, one half of that parallelogram, the one with the larger angle between the sides, is a triangle with sides of length 3, 5, and 7. Taking [tex]\theta[/tex] to be the angle between the two sides, by the cosine law, [tex]7^2= 3^2+ 5^2- 2(3)(5)cos(\theta)[/tex]. You can use that to solve for [itex]\theta[\itex]. then the other angle, [tex]\phi[/tex], is the supplement of [tex]\theta[/tex]. And, letting x= |a- b|, [tex]x^2= 3^2+ 5^2- 2(3)(5)cos(\phi)[/tex].

 

This is what I was alluding to earlier. There's actually an even easier formula to use.

[hypervalent_iodine]

!

Moderator Note

Hallsofivy,

 

The reason that John hadn't given the complete answer in his first post is because this is the homework help section. We ask that members not simply give out answers to such questions as it does not benefit the learning of the person asking the question. Please keep this in mind when posting in this section in the future.