Question

let W be an inner product space and let w1 and w2 be vectors in W. suppose ll w1 ll =5^1/2 and ll w2 ll= 4 and the angle between w1 and w2 is pi/3. compute

• <w1,w1> and <w2,w2>
• <w1, w2>
• <w1+ 2w2, 3w1-w2>

Answer 1

Answer:

Given w₁ and w₂ are two vectors in W and

ll w₁ ll =5^1/2______(1)

 ll w₂ ll =4_________(2)

Angle between w₁ and w₂=$\frac {π} 3$.

To compute $\coloneq$

1. <w₁,w₁> and <w₂,w₂>
2. <w₁, w₂>
3. <w₁+ 2w2, 3w₁-w2>

We know that,

"If u₁ and u₂ are two vectors and $\theta$ be the angle between them, then

$\cos$ $\theta$ =$⟨u1, u2 ⟩⟩ ⟩ \over ∥u1, u2 ∥$ where $\theta$ is angle between w_{1 and} w₂

$\over$

where ∥u₁∥= $\sqrt {<u1, u1>}$

and ∥u₂∥= $\sqrt {<u2, u2>}$

1.

(a) <w₁,w₁> =? ; given  ll w₁ ll =$\sqrt 5$

using inner product property,

ll w₁ ll = $\sqrt {<w1, w1>}$

ll w₁ ll² = <w₁,w₁> (square both sides)

<w₁,w₁> =  ll w₁ ll²

= ($\sqrt 5$)²

=(5^1/2)² (multiply $1 \over 2$ by 2)

=5

Therefore, <w₁,w₁> =5

(b)  <w₂,w₂>=? ; given  ll w₂ ll =4

Using,

ll w₂ ll= $\sqrt {<u2, u2>}$

squaring both sides we get,

  ll w₂ ll² = <w₂,w₂>

Hence, <w₂,w₂>=  ll w₂ ll²

 <w₂,w₂>= (4)² = 16

Therefore,   <w₂,w₂> = 16

(2)To find <w₁, w₂>

we know,

 $\cos$ $\theta$ = $⟨w1, w2 ⟩ \over ∥w1, w2 ∥$ where $\theta$ is angle between w_{1 and} w₂ 

Given, ll w₁ ll = $\sqrt 5$

llw₂ ll = 4

and $\theta$ = $\frac {π} 3$

  $\cos$  $\frac {π} 3$= $<w1, w2> \over (\sqrt 5) (4)$  

$1 \over 2$ = $<w1, w2> \over (\sqrt 5) (4)$    (  $\cos$  $\frac {π} 3$=  $1 \over 2$)

<w₁, w₂> =  $1 \over 2$ ($\sqrt 5$) (4) (divide 4 by 2 to eliminate the half)

=2 $\sqrt 5$

Therefore, <w₁, w₂> = 2 $\sqrt 5$ ____(3)

(3)

To find <w₁+ 2w2, 3w₁-w2>.

We know, according to property of inner product.

<αa+βb, γc+δd> = <αa, γc+δd> + <βb, γc+δd>

where α, β, γ and δ are real constants and a,b,c,d are vectors.

Hence, <αa+βb, γc+δd> = α<a, γc+δd> + β<b, γc+δd>

= α(<a, γc> + <a, δd>) + β (<b, γc> + <b, δd>)

= α(γ<a, c> + δ<a, d>) + β (γ<b, c> + δ<b, d>)

=αγ<a, c> + δα<a, d>) + βγ<b, c> + δβ<b, d>)

Here,

<w₁+ 2w₂, 3w₁-w₂> = <w₁, 3w₁-w₂> +  <2w₂, 3w₁-w₂>

=1<w₁, 3w₁> + 1<w₁, -w₂> +  2<w₂, 3w₁> + 2 <w₂, -w₂>

=1(3)<w₁, w₁> + 1(-1)<w₁, w₂> +  2(3)<w₂, w₁> + 2(-1) <w₂, w₂>

Hence,

<w₁+ 2w₂, 3w₁-w₂>=

=3<w₁, w₁> -1<w₁, w₂> +  6<w₂, w₁> - 2<w₂, w₂>

From previous solutions;

<w₁, w₁> = ll w₁ ll²

<w₁, w₂> = <w₂, w₁>

<w₁+ 2w₂, 3w₁-w₂>

=3 ll w₁ ll² - <w₁, w₂> +  6<w₂, w₁> - 2 ll w₂ ll²

 =3 ll w₁ ll² - <w₁, w₂> +  6< w_1,w₂> - 2 ll w₂ ll²

=3 ll w₁ ll² +  5< w_1,w₂> - 2 ll w₂ ll²

From (1) , (2) , (3) we have

 ll w₁ ll = $\sqrt 5$  , ll w₂ ll = 4, and < w_1,w₂> = 2 $\sqrt 5$ 

Therefore,

<w₁+ 2w₂, 3w₁-w₂> = 3($\sqrt 5$)² + 5(2$\sqrt 5$)- 2(4)²

  = 3(5) + 10$\sqrt 5$ - 2 (16)

= 15+ 10$\sqrt 5$ - 32

=-17+10$\sqrt 5$

Hence, <w₁+ 2w₂, 3w₁-w₂> = -17+10$\sqrt 5$