position vectors

**Amanda-UK** · August 17th 2007, 05:08 AM

how do i tackle these :
1A)
Consider the following position vectors, v=i-j and w=j-2k

find i) the angle between the above vectors
ii) a vector which perpendicular on both vectors

b) Let z=2-3j and s=1+2j find

2+z
Z^2-Z and 1+s

c) solve the following equation:

z^3 - z^2 +z = 0

amanda xx

**topsquark** · August 17th 2007, 05:33 AM

Originally Posted by Amanda-UK

how do i tackle these :
1A)
Consider the following position vectors, v=i-j and w=j-2k

find i) the angle between the above vectors
ii) a vector which perpendicular on both vectors

i) $v \cdot w = |v| |w| cos(\theta)$

ii) The vector $v \times w$ is perpendicular to both v and w.

-Dan

**topsquark** · August 17th 2007, 05:37 AM

Originally Posted by Amanda-UK

b) Let z=2-3j and s=1+2j find

2+z
Z^2-Z and 1+s

$\frac{2 + z}{z^2 - z} = \frac{2 + (2 - 3j)}{(2 - 3j)^2 - (2 - 3j)}$

$= \frac{4 - 3j}{4 - 12j + 9j^2 - 2 + 3j}$

$= \frac{4 - 3j}{2 - 9j - 9} = \frac{4 - 3j}{-7 - 9j}$

Of course, you need to simplify this expression...

(Do you really need someone to do 1 + s for you??)

-Dan

**Amanda-UK** · August 17th 2007, 05:37 AM

Originally Posted by topsquark

i) $v \cdot w = |v| |w| cos(\theta)$

ii) The vector $v \times w$ is perpendicular to both v and w.

-Dan

how would i then go on from there to get the answer? x

**topsquark** · August 17th 2007, 05:38 AM

Originally Posted by Amanda-UK

c) solve the following equation:

z^3 - z^2 +z = 0

Factor:
$z^3 - z^2 + z = 0$

$z(z^2 - z + 1) = 0$

I leave the rest to you.

-Dan

**topsquark** · August 17th 2007, 05:42 AM

Originally Posted by topsquark

i) $v \cdot w = |v| |w| cos(\theta)$

ii) The vector $v \times w$ is perpendicular to both v and w.

-Dan

Originally Posted by Amanda-UK

how would i then go on from there to get the answer? x

i) This is a dot product:
$v \cdot w = v_x w_x + v_y w_y + v_z w_z$

$|v| = \sqrt{v_x^2 + v_y^2 + v_z^2}$

ii) $v \times w = \text{det} \left [ \begin{matrix} i & j & k \\ v_x & v_y & v_z \\ w_x & w_y & w_z \end{matrix} \right ]$

-Dan

**Amanda-UK** · August 17th 2007, 05:45 AM

Originally Posted by topsquark

$\frac{2 + z}{z^2 - z} = \frac{2 + (2 - 3j)}{(2 - 3j)^2 - (2 - 3j)}$

$= \frac{4 - 3j}{4 - 6j + 9j^2 - 2 + 3j}$

$= \frac{4 - 3j}{2 - 3j - 9} = \frac{4 - 3j}{-11 - 3j}$

Of course, you need to simplify this expression...

(Do you really need someone to do 1 + s for you??)

-Dan

sorry i wrote in eqution wrong ( not to sure how to write maths language on here: should be this:

Let z = 2-3j and s= 1+2j find:
.
Z^2 - Z (there should be a line above the last Z but not lettin me do it)

and

2+z
1+s

Let z = 2-3j and s= 1+2j find:

**Amanda-UK** · August 17th 2007, 05:55 AM

Originally Posted by topsquark

i) This is a dot product:
$v \cdot w = v_x w_x + v_y w_y + v_z w_z$

$|v| = \sqrt{v_x^2 + v_y^2 + v_z^2}$

ii) $v \times w = \text{det} \left [ \begin{matrix} i & j & k \\ v_x & v_y & v_z \\ w_x & w_y & w_z \end{matrix} \right ]$

-Dan

would you be able to show me this method more please x

**topsquark** · August 17th 2007, 07:34 AM

Originally Posted by Amanda-UK

sorry i wrote in eqution wrong ( not to sure how to write maths language on here: should be this:

Let z = 2-3j and s= 1+2j find:
.
Z^2 - Z (there should be a line above the last Z but not lettin me do it)

and

2+z
1+s

Let z = 2-3j and s= 1+2j find:

Just as well, I goofed in the denominator when I posted anyway.

Okay, so the first one is
$z^2 - \bar{z} = (2 - 3j)^2 - (2 + 3j)$

$= (4 - 12j + 9j^2) - (2 + 3j)$

$= 4 - 12j - 9 - 2 - 3j$

$= -7 - 15j$

The second one is now $\frac{2 + z}{1 + s}$ ?

$= \frac{2 + (2 - 3j)}{1 + (1 + 2j)}$

$= \frac{2 + 2 - 3j}{1 + 1 + 2j}$

$= \frac{4 - 3j}{2 + 2j}$

Because this is so simple up to here I presume you need help simplifying it. $j = \sqrt{-1}$ , so we have a radical in the denominator that we wish to get rid of. Now, recall the identity:
$(a + b)(a - b) = a^2 - b^2$

Thus if we have $2 + 2\sqrt{-1}$ we can multiply it by $2 - 2\sqrt{-1}$ to get rid of the radical. But what we do to the denominator we must do to the numerator...
$= \frac{4 - 3j}{2 + 2j} \cdot \frac{2 - 2j}{2 - 2j}$

$= \frac{(4 - 3j)(2 - 2j)}{(2 + 2j)(2 - 2j)}$

$= \frac{8 - 14j + 6j^2}{2^2 - (2j)^2}$

$= \frac{8 - 14j - 6}{4 - 4j^2}$

$= \frac{2 - 14j}{4 + 4}$

$= \frac{2 - 14j}{8}$ <-- Now cancel the common factor of 2

$= \frac{1 - 7j}{4}$

-Dan

**topsquark** · August 17th 2007, 07:38 AM

Originally Posted by topsquark

i) This is a dot product:
$v \cdot w = v_x w_x + v_y w_y + v_z w_z$

$|v| = \sqrt{v_x^2 + v_y^2 + v_z^2}$

ii) $v \times w = \text{det} \left [ \begin{matrix} i & j & k \\ v_x & v_y & v_z \\ w_x & w_y & w_z \end{matrix} \right ]$

-Dan

Originally Posted by Amanda-UK

would you be able to show me this method more please x

If you don't know what I'm talking about here then you must be using another method in your class. Can you tell me what you've done on this?

-Dan

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