Question

given u = [1,2,3] and v=[2,-2,-6] use dot product to find a unit vector perpendicular to both

Answer 1

\[
\mathbf{w}\cdot \mathbf{u} = 0 \\
\mathbf{w}\cdot \mathbf{v} = 0
\]First find \(\mathbf{w}\). Then make it a unit vector.

Answer 2

Consider letting \(\mathbf{w} = [w_1,w_2,w_3]\).
You'll get a system of equations.

Answer 3

cos u could easily find it by cross product (: but then my teacher asks us to do by dot

Answer 4

Well get it.

Answer 5

and you could use any point for w1, w2, w3? how could you know which points to match with both vectors

Answer 6

Do you know how to solve a system of equations?

Answer 7

You have two equations and three variables.... that means you have one free variable. Just set the free variable to 0 to make your life easy.

Answer 8

or set it to 1

Answer 9

okay

Answer 10

Do you have to write out the equations for you?

Answer 11

I never solve this by dot, only with cross (:

Answer 12

\[\begin{array}{rrl}
\mathbf{w}\cdot \mathbf{u} = 0 & \implies
\begin{bmatrix}
w_1 \\ w_2 \\ w_3
\end{bmatrix}
\cdot
\begin{bmatrix}
1 \\ 2 \\ 3
\end{bmatrix}=0
&\implies
w_1+2w_2+3w_3=0
\\
\mathbf{w}\cdot \mathbf{v} = 0 &\implies
\begin{bmatrix}
w_1 \\ w_2 \\ w_3
\end{bmatrix}
\cdot
\begin{bmatrix}
2 \\ -2 \\ -6
\end{bmatrix}=0

&\implies
2w_1-2w_2-6w_3=0
\end{array}
\]

Answer 13

Now, we let \(w_3=1\) since it is a free variable.
That gives us: \[
\begin{array}{rcl}
w_1+2w_2+3w_3&=&0\\
2w_1-2w_2-6w_3&=&0 \\
w_3 &=& 1
\end{array}
\]

Answer 14

@pinky_cute1995 Do you remember how to solve systems of equations?

Answer 15

so w1 = -1 ?

Answer 16

You should solve for \(w_1\) and \(w_2\)

Do you know how to do substitution or elimination?

Answer 17

okayyy

Answer 18

I got it

Answer 19

the vector is [1,-2,1]

Answer 20

An axle has two wheels of radii 0.95 m and 0.15 m attached to it. A 30 N force is applied horizontally to the edge of the larger wheel and a similar weight hangs from the edge of the smaller wheel.
a. the net torque acting on the axle
b. the maximum weight that the 30N force can pull up

Answer 21

You have to now make the vector into a unit vector.

Answer 22

I know how to do the unit vector hehe

Answer 23

@pinky_cute1995 mention me in the other question if you want me to respond to it.

Answer 24

okayyyy

Answer 25

Divide it by the magnitude...

Answer 26

\[
||\mathbf{w}|| = \sqrt{\mathbf{w}\cdot \mathbf{w}} = \sqrt{1^2+(-2)^2+1^2} = \sqrt{6}
\]

Answer 27

it's okay

Answer 28

So the unit vector is \[
\left[\frac{1}{\sqrt{6}},\frac{-2}{\sqrt{6}},\frac{1}{\sqrt{6}}\right]
\]

Answer 29

I know how to do that (:

Answer 30

omggg

Answer 31

You said you didn't...

Answer 32

read it again ^^

Answer 33

Okay

Answer 34

flutter

Answer 35

FUUCK

Answer 36

anyway thank you again

Answer 37

you are my angel hahaha

Answer 38

For the physics one

Answer 39

did you draw the picture?

Answer 40

which one ?

Answer 41

the plane one or the axle

Answer 42

the two wheels thing

Answer 43

yeah

Answer 44

I try

Answer 45

but then it kinda weird :[

Answer 46

Hmmm, well to get the net torque we need the value of the weight

Answer 47

Ok we could answer in terms of the weight.

Answer 48

If the mass is \(m\) then the net torque is going to be: \[
30 \times 0.95 - m\times9.8\times 0.15
\]

Answer 49

so what is toque ?

Answer 50

Torque is the force vector cross product with the radius vector.

Answer 51

The radius is vector is just the vector from the axis of symmetry to the point where the force is applied.

Answer 52

In this case since the radius and vector are perpendicular, the magnitude of the cross product is the the product of the magnitudes...

Answer 53

But you should read up on it.

Answer 54

okay