Finding the Angle Between Vectors a and b

[Jane K]

1. Homework Statement
The vector a=2 and vector b=1. The vectors a+5b and 2a-3b are perpendicular. Determine the angle between a and b .

Homework Equations

The dot product a•b=lallblcosθ

The Attempt at a Solution

I've tried a few things but none of it really makes sense. I'm worried that maybe this question doesn't call for the dot product method but I've become fixed on it.

 

[LCKurtz]

1. Homework Statement
The vector a=2 and vector b=1. The vectors a+5b and 2a-3b are perpendicular. Determine the angle between a and b .

Homework Equations

The dot product a•b=lallblcosθ

The Attempt at a Solution

I've tried a few things but none of it really makes sense. I'm worried that maybe this question doesn't call for the dot product method but I've become fixed on it.

You don't have a vector a = 2 and b = 1, those are scalars. Perhaps you mean their magnitudes are 2 and 1? What do you get if you dot a+5b and 2a-3b together?

 

[Jane K]

"the vectors a and b have lengths 2 and 1, respectively." I'm trying to find the angle between the two. The dot product for the perpendicular vectors a+5b and 2a-3b would be zero... I am stuck:P.

 

[LCKurtz]

"the vectors a and b have lengths 2 and 1, respectively." I'm trying to find the angle between the two. The dot product for the perpendicular vectors a+5b and 2a-3b would be zero... I am stuck:P.

Show us what you get in terms of a and b when you dot those two vectors together and set it equal to 0.

 

[SammyS]

"the vectors a and b have lengths 2 and 1, respectively." I'm trying to find the angle between the two. The dot product for the perpendicular vectors a+5b and 2a-3b would be zero... I'm stuck:P.

"FOIL" works for vectors .

$\left(\vec{a}+\vec{b}\right)\cdot\left(\vec{c}+ \vec{d}\right)=\vec{a}\cdot\vec{c}+\vec{a}\cdot \vec{d}+\vec{b}\cdot\vec{c}+\vec{b}\cdot\vec{d}$​

 

[Deveno]

what SammyS means is that the dot product is bilinear, it is linear in each variable:

if a,b,c are vectors, and r is a scalar:

a.(b+c) = a.b + a.c
(a+b).c = a.c + b.c

a.(rb) = r(a.b)
(ra).b = r(a.b)

also, a.b = b.a (the dot product is symmetric).

thus (a+5b).(2a-3b) = 2(a.a) + 5(b.a) - 3(a.b) - 15(b.b)

a.a = |a|², for any vector a.

you are given |a| and |b|, and you are given that the dot product (a+5b).(2a-3b) = 0.

if you can deduce what a.b is, then you can figure out the angle:

θ = arccos[(a.b)/(|a||b|)]

 

[Jane K]

Thank-you everyone!
This really helped:)