# Need help with a vector question.

• August 28th 2009, 12:29 AM
toffeefan
Need help with a vector question.
Given the following vector:

a = i + 2j -2k

Find a possible set of values for y and z such that the unit vector shown below is perpendicular to vector a.

b = (y)j + (z)k

I've tried using both of these rules [ (i) |a x b| = |a||b|sin90 (ii)a.b = 0 ]to form an equation but i can't seem to get an answer.

I suspect it has something to do with matrices also, as this is part of a matrices question.
• August 28th 2009, 12:49 AM
red_dog
$\mathbf{a}\perp\mathbf{b}\Leftrightarrow \mathbf{a}\cdot\mathbf{b}=0$

$\mathbf{a}\cdot\mathbf{b}=2y-2z\Rightarrow y=z$
• August 28th 2009, 12:55 AM
toffeefan
Is that all? that was my answer too. but shouldn't y and z have some values instead of just y=z?
• August 28th 2009, 01:39 AM
red_dog
y (or z) can take any value.

For example:

$y=z=1$ or $y=z=-10$ etc.
• August 28th 2009, 02:16 AM
toffeefan
i see. Thanks!
• August 28th 2009, 03:10 AM
Plato
Quote:

Originally Posted by toffeefan
Given the following vector: a = i + 2j -2k

Find a possible set of values for y and z such that the unit vector shown below is perpendicular to vector a.
b = (y)j + (z)k

Did you notice that vector b should be a unit vector?
So $y^2+z^2=1$.
• August 28th 2009, 05:01 AM
toffeefan
Quote:

Originally Posted by Plato
Did you notice that vector b should be a unit vector?
So $y^2+z^2=1$.

ok. but i'm still not sure of how to go about solving x and y. what i've done is:

|a| = 3
|b| = 1 (since b is unit vector)

a x b = (2z + 2y)i - (z)j + (y)k

|a x b| = square root of (2z + 2y)^2 + z^2 + y^2

since y^2 + z^2 = 1,

|a x b| = square root of 1 + 4(1) + 6zy

Using |a x b| = |a||b|sin90,

square root of 1 + 4(1) + 6zy = 3
y = -z/3

How do i form the other eqn?