Scalar Multiples

Question

 The angle between vector u and vector v is 60 degrees. If magnitude of vector u is 4 times the magnitude of vector v, find a scalar multiple, m, such that:

vector u + m(vector)v is perpendicular to 2(vector)u+(vector)v

Answer 1

Hello Kam,

Solving this problem involves using a few properties of dot products u ⋅ v. We will make use of the following:

1. u ⋅ v = v ⋅ u
2. u ⋅ v = ||u|| ||v|| cos θ, where θ is the angle between u and v
3. u ⋅ u = ||u||²
4. u ⊥ v if and only if u ⋅ v = 0
5. distributive property

We are told θ = 60 and ||u|| = 4 ||v||. We want to find m so that (u + mv) ⊥ (2u + v). Starting with Property (4) above,

(u + mv) ⊥ (2u + v) = 0 if and only if (u + mv) ⋅ (2u + v) = 0

We now work through the algebra using the five properties above:

u ⋅ (2u) + u ⋅ v + (mv) ⋅ (2u) + (mv) ⋅ v = 0 by distributive property

2 ||u||² + (1 + 2m)u ⋅ v + m ||v||² = 0 by Properties (1), (3), & (5)

2 ||u||² + (1 + 2m) ||u|| ||v|| cos 60 + m ||v||² = 0 by Property (2)

32 ||v||² + (1 + 2m) (4) ||v||² (1/2) + m ||v||² = 0 since cos 60 = 1/2 and we are the given ||u|| = 4 ||v||

(34 + 5m) ||v||² = 0 by combining like terms

Last, because v is a non-zero vector, we must have 34 + 5m = 0 and so m = -34 / 5 is our desired scalar. This looks complicated, but it is just applying the above properties. I encourage you to work through this problem on your own and write back if you have any more questions.

Best,

Kevin

Answer 2

To show that u+mv is perpendicular to 2u+v, we can show that (u+mv)•(2u+v)=0.

Expanding this dot product, this condition becomes

2(u•u)+(u•v)+2m(u•v)+m(v•v) = 0

We know that for any vector a, (a•a)=|a|^2.

We also are given that |u| = 4|v| (and so |u|^2 = 16 |v|^2)

We know that (u•v)=|u||v|cos(ß), where ß is the angle between u and v. Therefore, (u•v)=4|v|^2 (cos(60))=4|v|^2 (1/2) = 2|v|^2.

Substituting this in,

2(16|v|^2) + 2|v|^2 +2m(2|v|^2)+m|v|^2 = 0,

so

|v|^2 ( 32 +2 +4m +m) = 0, hence 5m=-34, so m=-34/5.