Given: \(\mathbf{i}\) along x‑axis, \(\mathbf{j}\) along y‑axis.
(a) Vector \(\mathbf{i}+\mathbf{j}\)
\[ \mathbf{i} + \mathbf{j} = (1,0) + (0,1) = (1,1) \] Magnitude: \[ |\mathbf{i}+\mathbf{j}| = \sqrt{1^2 + 1^2} = \sqrt{2} \] Direction (angle \(\theta\) from +x‑axis): \[ \tan\theta = \frac{1}{1} = 1 \Rightarrow \theta = 45^\circ \] So \(\mathbf{i}+\mathbf{j}\) has magnitude \(\sqrt{2}\) and is directed at \(45^\circ\) above the +x‑axis.
(b) Vector \(\mathbf{i}-\mathbf{j}\)
\[ \mathbf{i} - \mathbf{j} = (1,0) - (0,1) = (1,-1) \] Magnitude: \[ |\mathbf{i}-\mathbf{j}| = \sqrt{1^2 + (-1)^2} = \sqrt{2} \] Direction (angle \(\phi\) from +x‑axis): \[ \tan\phi = \frac{-1}{1} = -1 \Rightarrow \phi = -45^\circ \] So \(\mathbf{i}-\mathbf{j}\) has magnitude \(\sqrt{2}\) and is directed at \(45^\circ\) below the +x‑axis.
First write the unit vectors along the directions of \(\mathbf{i}+\mathbf{j}\) and \(\mathbf{i}-\mathbf{j}\):
\[ \hat{u}_1 = \frac{\mathbf{i}+\mathbf{j}}{|\mathbf{i}+\mathbf{j}|} = \frac{\mathbf{i}+\mathbf{j}}{\sqrt{2}} \] \[ \hat{u}_2 = \frac{\mathbf{i}-\mathbf{j}}{|\mathbf{i}-\mathbf{j}|} = \frac{\mathbf{i}-\mathbf{j}}{\sqrt{2}} \]
(a) Component of \(\mathbf{A}\) along \(\mathbf{i}+\mathbf{j}\)
\[ \mathbf{A} = 2\mathbf{i} + 3\mathbf{j} \] Scalar component along \(\hat{u}_1\): \[ A_1 = \mathbf{A} \cdot \hat{u}_1 = (2\mathbf{i} + 3\mathbf{j}) \cdot \frac{\mathbf{i}+\mathbf{j}}{\sqrt{2}} \] \[ = \frac{1}{\sqrt{2}}[(2\mathbf{i}\cdot\mathbf{i}) + (2\mathbf{i}\cdot\mathbf{j}) + (3\mathbf{j}\cdot\mathbf{i}) + (3\mathbf{j}\cdot\mathbf{j})] \] \[ = \frac{1}{\sqrt{2}}[2(1) + 0 + 0 + 3(1)] = \frac{1}{\sqrt{2}}(5) = \frac{5}{\sqrt{2}} \] So the component along \(\mathbf{i}+\mathbf{j}\) is \(\dfrac{5}{\sqrt{2}}\), and the vector component is \[ \mathbf{A}_{\parallel (i+j)} = \frac{5}{\sqrt{2}}\,\hat{u}_1 = \frac{5}{\sqrt{2}}\cdot\frac{\mathbf{i}+\mathbf{j}}{\sqrt{2}} = \frac{5}{2}(\mathbf{i}+\mathbf{j}). \]
(b) Component of \(\mathbf{A}\) along \(\mathbf{i}-\mathbf{j}\)
Scalar component along \(\hat{u}_2\): \[ A_2 = \mathbf{A} \cdot \hat{u}_2 = (2\mathbf{i} + 3\mathbf{j}) \cdot \frac{\mathbf{i}-\mathbf{j}}{\sqrt{2}} \] \[ = \frac{1}{\sqrt{2}}[(2\mathbf{i}\cdot\mathbf{i}) + (2\mathbf{i}\cdot(-\mathbf{j})) + (3\mathbf{j}\cdot\mathbf{i}) + (3\mathbf{j}\cdot(-\mathbf{j}))] \] \[ = \frac{1}{\sqrt{2}}[2(1) + 0 + 0 - 3(1)] = \frac{1}{\sqrt{2}}(-1) = -\frac{1}{\sqrt{2}} \] So the component along \(\mathbf{i}-\mathbf{j}\) is \(-\dfrac{1}{\sqrt{2}}\). The negative sign indicates that \(\mathbf{A}\) has a component opposite to the direction of \(\mathbf{i}-\mathbf{j}\). The corresponding vector component is \[ \mathbf{A}_{\parallel (i-j)} = -\frac{1}{\sqrt{2}}\,\hat{u}_2 = -\frac{1}{\sqrt{2}}\cdot\frac{\mathbf{i}-\mathbf{j}}{\sqrt{2}} = -\frac{1}{2}(\mathbf{i}-\mathbf{j}). \]
Summary (scalar components):