Question

the vector OA where O is origin is given by OA= 2icap-2jcap now its rotated by 45 degree anticlockwise

Answer 1

Answer:

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Answer 2

Given vector is,

$\longrightarrow \vec{\sf{OA}}=\sf{2\hat i-2\hat j}$

First let us write the vector in the form $\sf{r\left(\cos\theta\,\hat i+\sin\theta\,\hat j\right).}$

$\longrightarrow \vec{\sf{OA}}=\sf{2\left(\hat i-\hat j\right)}$

$\longrightarrow \vec{\sf{OA}}=\sf{2\sqrt2\left(\dfrac{1}{\sqrt2}\ \hat i-\dfrac{1}{\sqrt2}\ \hat j\right)}$

$\longrightarrow \vec{\sf{OA}}=\sf{2\sqrt2\left(\cos45^o\ \hat i-\sin45^o\ \hat j\right)}$

$\longrightarrow \vec{\sf{OA}}=\sf{2\sqrt2\left(\cos(-45^o)\ \hat i+\sin(-45^o)\ \hat j\right)}$

Now it is in the form $\sf{r\left(\cos\theta\,\hat i+\sin\theta\,\hat j\right).}$

This means our vector has magnitude $\sf{2\sqrt2}$ and is at an angle $\sf{45^o}$ below positive x axis, i.e., in 4th quadrant.

If we rotate this vector by $\sf{45^o}$ anticlockwise about O, the angle gets added by this much angle. But the magnitude remains unchanged.

Hence the new vector is,

$\longrightarrow \vec{\sf{OA'}}=\sf{2\sqrt2\left(\cos\left(-45^o+45^o\right)\ \hat i+\sin\left(-45^o+45^o\right)\ \hat j\right)}$

$\longrightarrow \vec{\sf{OA'}}=\sf{2\sqrt2\left(\cos0^o\ \hat i+\sin0^o\ \hat j\right)}$

$\longrightarrow\underline{\underline{\vec{\sf{OA'}}=\sf{2\sqrt2\ \hat i}}}$