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Jabatheblob1

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Joined: 2015-03-01

Posts: 856

Math Help

This seems like a stupid question, but in my calc class we are learning about vectors and there are some intelligent people on this website. Can anyone help me with this problem: if the angle between vectors c=ai+2j and d=3i+j is 45 degrees, what are the two possible answers for a?
I haven't done vectors in a while so i need some help. (I know the answer from the back of the book but i wanna know how to get it)

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Kaslai

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Joined: 2015-02-17

Posts: 787

Re: Math Help

At the core of the problem, you need to realize that you can treat vectors like right triangles.

If you know your basic trig relationships, tangent is opposite over adjacent, which is what you'll be wanting to use.

You can use arctangent (often abbreviated to atan or tan^-1) to perform an inverse tangent and get an angle. Since you have d=3i+1j, you can get the angle for that using atan(1/3). We'll call this A.

Using A, you can then add or subtract 45deg to get your new desired angles and take the tangent to get your side ratio using tan(A+/-45deg). We'll call this B.

Last, you just need to apply the tangent relationship (opposite over adjacent) to extract the value of a from this result. Since you have the values for opposite (2) and opposite/adjacent (B), you can find adjacent with the expression opposite/(opposite/adjacent), or rather, 2/B.

Using this method I got 1 and -4 for the value of a.

Ratburntro44

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Joined: 1970-01-01

Posts: 1,383

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Re: Math Help

magnitude of c is sqrt(a^2+4)
magnitude of d is sqrt(9+1)=sqrt(10)
c dot d = sqrt(10)(sqrt(a^2+4)cos(45 degrees) = sqrt(10a^2+40)/sqrt(2) = sqrt(5a^2+20) // by c . d = |c||d|cos(theta)
By other definition of dot product,
c dot d = 3a+2

SO solve

3a+2=sqrt(5a^2+20)
9a^2+12a+4=5a^2+20 // Squared both sides
4a^2+12a-16=0 // Moved to left
a^2+3a-4=0 // Divided by 4
(a-1)(a+4)=0 // Factored
a=1 or -4

You can plug these values in to see if they work in order to check.

edit: Kaslai's method is valid too btw, just adding another way to do it (and the way that I would first think to do it) [:)

Kaslai

Official Caroler

From: SEAͩT̓͑TLͯͥͧͪ̽ͧE͑̚

Joined: 2015-02-17

Posts: 787

Re: Math Help

magnitude of c is sqrt(a^2+4)
magnitude of d is sqrt(9+1)=sqrt(10)
c dot d = sqrt(10)(sqrt(a^2+4)cos(45 degrees) = sqrt(10a^2+40)/sqrt(2) = sqrt(5a^2+20) // by c . d = |c||d|cos(theta)
By other definition of dot product,
c dot d = 3a+2

SO solve

3a+2=sqrt(5a^2+20)
9a^2+12a+4=5a^2+20 // Squared both sides
4a^2+12a-16=0 // Moved to left
a^2+3a-4=0 // Divided by 4
(a-1)(a+4)=0 // Factored
a=1 or -4

You can plug these values in to see if they work in order to check.

edit: Kaslai's method is valid too btw, just adding another way to do it (and the way that I would first think to do it) [:)

Your method is the calculus 3 method, mine is the pre-calc method