(a) Magnitude of $( \hat{i}+ \hat{j})=| \hat{i}+ \hat{j} |$ $=\sqrt{(1)^{2}+(1)^{2}}=\sqrt{2}$ Let the vector ( $\hat{i}+\hat{j}$ ) makes an angle θ with the direction of $\hat{i}$ , then $\cos \theta = \frac{ \widehat{ (i+\hat{j}) \cdot \hat{i} } }{ \widehat{ \left| i+\hat{j} \right| \left| \hat{i} \right| } }$ $= \frac{1}{(\sqrt{2})(1)} = \frac{1}{\sqrt{2}} = \cos 45^{\circ}$ or θ = 45° Magnitude of $( \hat{i}- \hat{j} ) = | \hat{i}- \hat{j} |$ $= \sqrt{(1)^{2}+(-1)^{2}} = \sqrt{2}$ Similarly, if θ is the angle which the vector ( $\hat{i}-\hat{j}$ ) makes with the direction of $\hat{i}$ , then$\cos \theta = \frac{ \widehat{ (i-\hat{j}) \cdot \hat{i} } }{ \widehat{ \left| i-\hat{j} \right| \left| \hat{i} \right| } }$ $= \frac{1}{\sqrt{2}} = \cos 45^{\circ}$ or θ = 45°Here, $\theta = -45^{\circ}$ with $\hat{i}$.