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Unit Vector Q

Gmart

Doing my head in :confused:

Find a unit vector that

makes an angle of 60 degrees with vector (3i + 4j)

Cos 60 is a half, but then you get two unknowns?? Is this a projection q?

Reply 1

BabyMaths

Original post by Gmart

Doing my head in :confused:

Find a unit vector that

makes an angle of 60 degrees with vector (3i + 4j)

Cos 60 is a half, but then you get two unknowns?? Is this a projection q?

One way to do it is to find cos and sin of

\arctan\left(\frac{4}{3}\right) \pm 60

.

Why?

(edited 10 years ago)

Reply 2

Gmart

Original post by BabyMaths

One way to do it is to find cos and sin of

\arctan\left(\frac{4}{3}\right) \pm 60

.

Why?

I'm sorry, but I have no idea what that means, or why it might give me an answer.

arctan 4/3 is 53 degrees

So if I add 60, then we have 113 degrees

the sin of that is 0.92

and the cos is -0.4

How is this helpful?

Maybe the I can use the dot product rule, but in two dimensions that gives me two unknowns

Reply 3

Doctor_Einstein

Original post by Gmart

Do you know the following rule:

tan(theta) = opposite/adjacent?

We can apply that to vector given to calculate the angle of the vector given.

Doing so gives,

tan(theta) = j-component/x-component = 4/3

So,

tan(theta) = 4/3

To find theta, we take the inverse tan of both sides, otherwise known as arctan, to give:

theta = inversetan(4/3)

Now this gives us the angle of the 3i + 4j vector with respect to the x-axis, in the anti-clockwise direction.

Now a vector that has an angle 60 degrees more or less than our initial vector therefore has an angle of + or - 60 from inversetan(4/3).

The next step is then to calculate the components of such a vector.

We use the rule:

x-component = hypotenuse*cos(theta)

y-component = hypotenuse*sin(theta)

The above rules are derived from the standard trigonometric equations.

For a unit vector, hypotenuse = 1, therefore

x-component = cos(arctan(4/3) +- 60)
y-component = sin(arctan(4/3) +- 60)

I'll leave computing the actual numbers to you.

Remember to express your final answer in terms of i and j. That is, icomponent*i + jcomponent*j.

(edited 10 years ago)

Reply 4

Doctor_Einstein

Original post by Gmart

Alternatively you could use the dot product rule.

Yes there are 2 unknowns, but also 2 equations.

This can be done as follows:

Let the unit vector have components ai + bj, where a and b are unknowns.

Since it is a unit vector, the magnitude must be 1, so therefore:

a^2 + b^2 = 1

Now there is a rule for dot product of two vectors that:

A.B= |A||B|cos(angle between)

Now |A| = 1 (unit vector) and |B| = sqrt(3^2 + 4^2) = 5.

Therefore

(ai + bj).(3i + 4j) = 5cos(60)

Therefore,

3a + 4b = 5cos(60)

So we have two equations and two unknowns, which can be solved simultanously. That is:

(1) a^2 + b^2 = 1
(2) 3a + 4b = 5cos(60)

I'll leave solving this up to you.

Reply 5

james22

This can be done more neatly, let a be the vector you have been given and b be the one you need to find, so that |b|=1.

Then a.b=|a||b|cos(60)

a.b=5/2*

You also have that |b|=1 so you have 2 equations in 2 unknowns. You can simply find the general vector satisfying * then normalise it to avoid solving anyting simultaneous.

Reply 6

BabyMaths

Original post by Gmart

It's helpful because it allows you to pretty much just write down the answer to this question (as you did :smile:

) and any similar questions.

Did you check the angle between -0.3907 i + 0.9205 j and 3i+4j?

Did you check the magnitude of -0.3907 i + 0.9205 j?

Not a necessary step obviously because