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Q. A unit vector in $x y$ - plane makes an angle of $45^{\circ}$ with the vector $\hat{ i }+\hat{ j }$ and an angle of $60^{\circ}$ with the vector $3 \hat{ i }-4 \hat{ j }$ is

Vector Algebra

Solution:

Let the required unit vector be $r=a \hat{i}+b \hat{j}$.
Then, $|r| =1$
$ \Rightarrow a^2+b^2 =1 ...$(i)
Since, $r$ makes an angle of $45^{\circ}$ with $\hat{i}+\hat{j}$ and an angle of $60^{\circ}$ with $3 \hat{i}-4 \hat{j}$, therefore,
$ \cos \frac{\pi}{4}=\frac{r \cdot(\hat{i}+\hat{j})}{|r||\hat{i}+\hat{j}|}$
and $ \cos \frac{\pi}{3}=\frac{r \cdot(3 \hat{i}-4 \hat{j})}{|r||3 \hat{i}-4 \hat{j}|} $
$\Rightarrow \frac{1}{\sqrt{2}} =\frac{a+b}{\sqrt{2}} $
and $ \frac{1}{2}=\frac{3 a-4 b}{5}$
$ \Rightarrow a+b =1 $
and $ a-4 b =\frac{5}{2} $
$ \Rightarrow a =\frac{13}{14}, b=\frac{1}{14} $
$\therefore r =\frac{13}{14} \hat{i}+\frac{1}{14} \hat{j}$