# plz check my work..

• Sep 9th 2007, 09:57 PM
harry
plz check my work..
Fine two vectors in opposite directions that are orthogonal to the vector u.
u= <((1/2)i,(-2/3)j

Solution:Two vectors are orthogonal if they are perpendicular.
There fore a vector v =(x,y) is orthogonal to u if

(x,y) . (1/2,-2/3) =1/2x-2/3y=0
now I have to find values of x and y that makes 0..?

I am stuck here...plz help? not getting right ans....
• Sep 9th 2007, 10:33 PM
CaptainBlack
Quote:

Originally Posted by harry
Fine two vectors in opposite directions that are orthogonal to the vector u.
u= <((1/2)i,(-2/3)j

Solution:Two vectors are orthogonal if they are perpendicular.
There fore a vector v =(x,y) is orthogonal to u if
(x,y) . (1/2,-2/3) =1/2x-2/3y=0
now I have to find values of x and y that makes 0..?

I am stuck here...plz help? not getting right ans....

(1/2)x=(2/3)y.

Choose any value you like for x, say x=1, then y=3/4, and so on.

RonL
• Sep 10th 2007, 02:30 AM
Soroban
Hello, Harry!

Quote:

Find two vectors in opposite directions that are orthogonal to the vector: . $\left\langle \frac{1}{2},\:-\frac{2}{3}\right\rangle$

Solution: Two vectors are orthogonal if they are perpendicular.

Therefore, a vector $\vec{v} \,=\,\langle x,y\rangle$ is orthogonal to $\vec{u}$ if:
. . $\langle x,\,y\rangle\cdot\left\langle\frac{1}{2},\,-\frac{2}{3}\right\rangle \;=\;\frac{1}{2}x - \frac{2}{3}y \;=\;0$

now I have to find values of x and y that makes 0 ? . . . . yes!

I am stuck here ...plz help? not getting right ans....
. . What is given as "the right answer" ?

You have: . $\frac{1}{2}x - \frac{2}{3}y \:=\:0\quad\Rightarrow\quad 3x \:=\:4y$

As CaptainBlack pointed out, use any pair of values that satisfy the equation.

The most obvious is: . $x=4,\:y=3\quad\Rightarrow\quad \vec{v} \:=\:\langle 4,\,3\rangle$
. . An opposite vector would be: . $-\vec{v} \;=\;\langle-4,\,-3\rangle$