Vector help needed

[lasse.h.lysne]

can somebody please help me with this vector problem. I need to find the angle between a and u.
| a ⃗|=3
| b ⃗| = 2
a ⃗⋅ b ⃗=3
u ⃗= a ⃗+2b ⃗

find ∠ between a ⃗ and u ⃗

thanks

Lasse Norway.

 

[pka]

can somebody please help me with this vector problem. I need to find the angle between a and u.
| a ⃗|=3, | b ⃗| = 2, a ⃗⋅ b ⃗=3, u ⃗= a ⃗+2b ⃗

Here are some facts you need.
The angle between vectors \(\displaystyle v~\&~w\) is \(\displaystyle \arccos \left( {\dfrac{{v \cdot w}}{{\left\| v \right\|\left\| w \right\|}}} \right)\).

To do this from the given use \(\displaystyle \|u\|^2=u\cdot u\).
Here \(\displaystyle (a+2b)\cdot(a+2b)=a\cdot a+4a\cdot b+4b\cdot b\).

 

[lasse.h.lysne]

Here are some facts you need.
The angle between vectors \(\displaystyle v~\&~w\) is \(\displaystyle \arccos \left( {\dfrac{{v \cdot w}}{{\left\| v \right\|\left\| w \right\|}}} \right)\).

To do this from the given use \(\displaystyle \|u\|^2=u\cdot u\).
Here \(\displaystyle (a+2b)\cdot(a+2b)=a\cdot a+4a\cdot b+4b\cdot b\).

Thanks for the help, could I ask you to explain it for a simpler mind?

lasse

 

[Deleted member 4993]

Thanks for the help, could I ask you to explain it for a simpler mind?

lasse

Since you need to find angle between a^→ and u^→ :

You need to find the a^→ \(\displaystyle \cdot\) u^→

and find magnitude of u^→ and a^→
pka has shown you the way to find the magnitude of u^→

and then use all those information to find cos(Θ) (pka has given you the equation) and from that find Θ.

Cannot get simpler than that .....

 

[lasse.h.lysne]

Here are some facts you need.
The angle between vectors \(\displaystyle v~\&~w\) is \(\displaystyle \arccos \left( {\dfrac{{v \cdot w}}{{\left\| v \right\|\left\| w \right\|}}} \right)\).

To do this from the given use \(\displaystyle \|u\|^2=u\cdot u\).
Here \(\displaystyle (a+2b)\cdot(a+2b)=a\cdot a+4a\cdot b+4b\cdot b\).

is this right?? or am I far off??

(a ) ⃗ ⋅ (u ) ⃗=a ⃗ × a ⃗ × 2b ⃗
(a^2 ) ⃗+2a ⃗b ⃗
3^2 + 2×3
(a ) ⃗ ⋅ (u ) ⃗=15
|u ⃗|^2= (a^2 ) ⃗+4a ⃗b ⃗+ 4(b^2 ) ⃗
|u ⃗|^2= 3^2 + (4×3) + 4×22
|u ⃗|^2= 37
|u ⃗|= 6,08
cosθ= ((a ) ⃗ ⋅ (u ) ⃗) / (|a ⃗ |⋅|(u ) ⃗|)
cosθ= 15/(3⋅6,08)
cosθ= 0,82
∠ a ⃗ and u ⃗=34,677°

 

[Deleted member 4993]

is this right?? or am I far off??

(a ) ⃗ ⋅ (u ) ⃗=a ⃗ × a ⃗ × 2b ⃗ <<< This is incorrect and not needed

(a^2 ) +2a ⃗.b ⃗

3^2 + 2×3
(a ) ⃗ ⋅ (u ) ⃗=15
|u ⃗|^2= (a^2 ) ⃗+4a ⃗b ⃗+ 4(b^2 ) ⃗
|u ⃗|^2= 3^2 + (4×3) + 4×22
|u ⃗|^2= 37
|u ⃗|= 6,08
cosθ= ((a ) ⃗ ⋅ (u ) ⃗) / (|a ⃗ |⋅|(u ) ⃗|)
cosθ= 15/(3⋅6,08)
cosθ= 0,82
∠ a ⃗ and u ⃗=34,677°

Looks good to me - except for the corrections as noted.