Answer:
\[|\hat{i}+\hat{j}|=\sqrt{{{1}^{2}}+{{1}^{2}}}=\sqrt{2}\] If the vector \[\hat{i}+\hat{j}\]makes angle \[\beta \] with \[x\]-axis, then \[\tan \beta =\frac{\text{coeff}\text{.of \hat{i}}}{\text{coeff}\text{.of \hat{j}}}=\frac{1}{1}=1\] \[\therefore \] \[\beta ={{45}^{\circ }}\]
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