1. ## Vectors!

Hello!

If a = i - j + 5k and b = 2i - j - 3k find the vector c in the direction of a such that IcI = IbI -( magnitude)

I would appreaciate a explanation to how to tackle this Q, so I can do it myself.

Thanks!

2. Originally Posted by classicstrings
Hello!

If a = i - j + 5k and b = 2i - j - 3k find the vector c in the direction of a such that IcI = IbI -( magnitude)

I would appreaciate a explanation to how to tackle this Q, so I can do it myself.

Thanks!
You have the equation
|c|=|b|
you are given c is in direction of a
c=ka, where k is any real no.

Malay

3. I still dont understand Malay.

So Magnitude b = root(14)

Then what?

4. Originally Posted by classicstrings
Hello!

If a = i - j + 5k and b = 2i - j - 3k find the vector c in the direction of a such that IcI = IbI -( magnitude)

I would appreaciate a explanation to how to tackle this Q, so I can do it myself.

Thanks!
You seek $\displaystyle \bold{c}=|\bold{b}|\ \hat{\bold{a}}=\frac{|\bold{b}|}{|\bold{a}|}\ \bold{a}$

RonL

5. Hi Capt Black,

Using your formula i got [root(14)/3root(3)]*(i - j + 5k)

But, I still do not understand why it is that.

$\displaystyle \bold{c}=|\bold{b}|\ \hat{\bold{a}}=\frac{|\bold{b}|}{|\bold{a}|}\ \bold{a}$

How do you get that you have to do that:

$\displaystyle \bold{c}=|\bold{b}|\ \hat{\bold{a}}$

Thanks so much.

6. Originally Posted by classicstrings
Hi Capt Black,

Using your formula i got [root(14)/3root(3)]*(i - j + 5k)

But, I still do not understand why it is that.

$\displaystyle \bold{c}=|\bold{b}|\ \hat{\bold{a}}=\frac{|\bold{b}|}{|\bold{a}|}\ \bold{a}$

How do you get that you have to do that:

$\displaystyle \bold{c}=|\bold{b}|\ \hat{\bold{a}}$

Thanks so much.
Could you understand my equation?
If yes, the aim is to determine k

Keep Smiling
Malay

7. I'll go over it again tommorrow as it's 1:16 am here, and I need to get to school tommorrow! Thanks!

8. Originally Posted by classicstrings
Hi Capt Black,

Using your formula i got [root(14)/3root(3)]*(i - j + 5k)

But, I still do not understand why it is that.

$\displaystyle \bold{c}=|\bold{b}|\ \hat{\bold{a}}=\frac{|\bold{b}|}{|\bold{a}|}\ \bold{a}$

How do you get that you have to do that:

$\displaystyle \bold{c}=|\bold{b}|\ \hat{\bold{a}}$

Thanks so much.
$\displaystyle \bold{c}=|\bold{b}|\ \hat{\bold{a}}$

is a vector in the direction of $\displaystyle \bold{a}$ (that is it is a scalar times
a unit vector pointing in the required direction. Also as $\displaystyle \hat{\bold{a}}$
is a unit vector:

$\displaystyle ||\bold{b}|\ \hat{\bold{a}}|=|\bold{b}||\hat{\bold{a}}|=|\bold{ b}|$

RonL

9. Hello, classicstrings!

Let me give it a try . . .

If $\displaystyle \vec a \,= \,i - j + 5k$ and $\displaystyle \vec b \,= \,2i - j - 3k$
find the vector $\displaystyle \vec c$ in the direction of $\displaystyle \vec a$ such that $\displaystyle |\vec c| = |\vec b|$

We want $\displaystyle \vec c$ to have the same magnitude as $\displaystyle \vec b$.

Very well, let's find the magnitude (length) of $\displaystyle \vec b:$
. . $\displaystyle |b|\:=\:\sqrt{2^2 + (-1)^2 + (-3)^2} \:= \:\sqrt{14}$

So now we know the length of $\displaystyle \vec c$.

We want $\displaystyle \vec c$ to have the same direction at $\displaystyle \vec a$.
. . If $\displaystyle \vec a$ has the right magnitude $\displaystyle (\sqrt{14})$, we're done!

What is the magnitude of $\displaystyle \vec a$ ?
. . $\displaystyle |\vec a| \:= \:\sqrt{1^2 + (-1)^2 + 5&2} \:= \:\sqrt{27} \;= \;3\sqrt{3}$ . . . it's the wrong length

Can we change its magnitude to$\displaystyle \sqrt{14}$ ? . . . Yes!

If we divide $\displaystyle \vec a$ by $\displaystyle 3\sqrt{3}$, we have: .$\displaystyle \vec u \:=\:\frac{\vec a}{3\sqrt{3}}$

. . This is a unit vector in the direction of $\displaystyle \vec a$ (exactly one unit long).

To make it$\displaystyle \sqrt{14}$ units long, multiply it by$\displaystyle \sqrt{14}$, and we have: .$\displaystyle \sqrt{14}\cdot\frac{\vec a}{3\sqrt{3}}$

So we have: .$\displaystyle \vec c \;\;=\; \;\frac{\sqrt{14}}{3\sqrt{3}}\cdot\vec a \quad\Rightarrow$ . $\displaystyle \boxed{\vec c\;= \;\frac{\sqrt{42}}{9}(i - j + 5k)}$

This is the vector with the direction of $\displaystyle \vec a$ and the length of $\displaystyle \vec b$.

10. Originally Posted by classicstrings
Hello!

If a = i - j + 5k and b = 2i - j - 3k find the vector c in the direction of a such that IcI = IbI -( magnitude)

I would appreaciate a explanation to how to tackle this Q, so I can do it myself.

Thanks!
I would like to tell you something about vectors.
A vector has a direction and magnitude.These two properties are independent of each other. A vector is completely determined by its direction and magnitude.
There are two ways of representating a vector:
1)Using Cartesian co-ordinates
2)Using Polar co-ordinates
(These can be interchanged from one form to another, use the form you are comfortable with)
If two vectors have the same direction, then one can be expressed as a scalar times another.
In your question, c has a given magnitude(equal to b), and a definite direction(same as a). So, your vector c is unique.