# A unit vector in xy plane that makes an angle 45 degree with the vector i+j and an angle 60 degree with the vector 3i-4j is

Solution:

$\vec{p}=a\hat{i}+b \hat{j}\Rightarrow \left | \vec{p} \right |=1\Rightarrow a^{2}+b^{2}=1.$

$\Rightarrow$      $(a\hat{i}+b\hat{j})(\hat{i}+\hat{j})=\left | a\hat{i}+b\hat{j} \right |\left | \hat{i}+\hat{j} \right |\cos 45^{\circ}$

$a+b=\sqrt{a^{2}+b^{2}}\times \sqrt{2}\times \frac{1}{\sqrt{2}}$

$a+b=1$               $..........(1)$

$\Rightarrow$        $(a\hat{i}+b\hat{j})(3\hat{i}-4\hat{j})=\left | a\hat{i}+b\hat{j} \right |\left | 3\hat{i}-4\hat{j} \right |\cos 60^{\circ}$

$3a-4b=5\times \frac{1}{2}$            $.......(2)$

Sloving   $(1)$    and   $(2)$  We get

Hence ,       $a=\frac{13}{14}, b=\frac{1}{14}$

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