Let A vector = iA cos$\theta$ j + A sin$\theta$ be any vector. Another vector B which is normal to A is
How I solved it:
I thought that there can be two possibilities in which A vector is perpendicular to two other vecotrs in which I choose either of them to be our B vector.if we make the figure look like image 1.
OD AND OE are iAcos $\theta$ and iBsin$\theta$. Then I thought for others and made the angles equal according to image 2.
For that answer should be Bcos $\theta$(- i ) and B sin $\theta$ j
Image 2
Image 1
My answer did not come right.Pls help where am I wrong and how should I approach it.
Please let me know where you find it understand my answer or the question.
Thank you
The given vector $\, \vec{OA}$ is $(A \cos \theta, A \sin \theta) \,$.
Unit vector in the direction of $\vec{A}$ is $ \,\hat{a} = (\cos \theta, \sin \theta)$.
Now unit vector orthogonal to $\vec{a}$ will be $(\cos (90^0 + \theta), \sin (90^0 + \theta)) = (-\sin \theta, \cos \theta)$. Please see the diagram. In your image $1$, please note that $\angle GOI = (90^0 - \theta) \, $ and not $\theta$.
So we have a unit vector $(-\sin \theta, \cos \theta) \,$ in the orthogonal direction of $\vec{OA}$. We will also have another orthogonal vector in the direction of $\vec{OC}$ and that is just opposite direction of $\vec{OB}$ and so it is $(\sin \theta, -\cos \theta)$.
Now if you want a vector of magnitude of $B$, just multiply the unit vector by the magnitude. So orthogonal vector $\vec{OB} = (-B \sin \theta, B \cos \theta)$.
As we know the dot product of two orthogonal vectors is zero, the easiest way to find orthogonal vector to a given vector in $2D$ space is to find the unit vector and then interchange values of $i$ and $j$ position with one of them being negative sign.
So if you have a unit vector $(x,y)$, the orthogonal vectors will be $(y,-x)$ and $(-y, x)$. You can then assign the required magnitude.