Linear Algebra: Vector Geometry Question

Question

Problem 1: Suppose ∥v∥ = 4 and ∥w∥ = 7. Use inner products to determine the largest and smallest possible values of ∥6 v − 5w∥. Hint. Start by expanding ∥6 v − 5 w∥ ^2.

Hi. I would really appreciate it if someone could tell me where I've gone wrong in this problem or how to correctly approach it. Thanks!

∥6 v − 5 w∥ ^2 = ∥6 v − 5w∥×∥6 v − 5w∥

= <6v, 6v> + <6v, -5w> + <-5w, 6v> + <-5w, -5w>

Sum of the following:

<6v, 6v> = 6 × 6 <v,v> = 36 ∥v∥ ^2 = (36)(16) = 576

<6v, -5w> = 6 × -5 <v,w> = -30<v,w>

<-5w, 6v> = -5 × 6 <w,v> = -30 <w,v>

<-5w, -5w> = -5 × -5 <w,w> = 25 ∥w∥ ^2 = (25)(49) = 1225

∥6 v − 5 w∥ ^2 = 576 -30 <v,w> -30<w,v> + 1225

= 1801-60<v,w>

Not quite sure what to do now...

Answer 1

We start by expanding ∥6v−5w∥2 using the properties of inner products. The square of the norm is:

∥6v−5w∥²=(6v−5w)⋅(6v−5w)

Using the distributive property of the dot product:

∥6v−5w∥²=6²∥v∥²−2(6)(5)(v⋅w)+5²∥w∥²

Simplifying:

∥6v−5w∥²=36∥v∥²−60(v⋅w)+25∥w∥²

Substitute the given norms. We are given ∥v∥=4 and ∥w∥=7 Substituting these values:

∥6v−5w∥²=36*4²−60(v⋅w)+25*7²

∥6v−5w∥²=1801−60(v⋅w)

We know that

v⋅w=∥v∥∥w∥cos⁡θ,

where θ\is the angle between v and w,

so:

v⋅w=4×7×cos⁡θ=28cos⁡θ = 4×7×cosθ =28cosθ

The cosine function cos⁡θvaries between -1 and 1, so:

−28≤v⋅w≤28-28

Now, the smallest value of v⋅w occurs when cos⁡θ=−1, which gives v⋅w=−28.

Substituting this into the equation for ∥6v−5w∥²

∥6v−5w∥²=1801−60(−28)

∥6v−5w∥²=1801+1680=3481

∥6v−5w∥= √3481​=59

Smallest value

The largest value of v⋅w occurs when cos⁡θ=1, so v⋅w=28. Substituting this into the equation for ∥6v−5w∥²

∥6v−5w∥²=1801−60(28)

∥6v−5w∥²=121

∥6v−5w∥=121=11

Final Answer

The largest possible value of ∥6v−5w∥ is 59, and the smallest possible value is 11.