91Ó°ÊÓ

To find the cross product \(\vec{d_{1}} \times \vec{d_{2}}\) in two dimensions, use the determinant method. Treat the vectors as three-dimensional with the k-component as zero: \[\vec{d_{1}} = (4.0 \mathrm{~m}) \hat{\mathrm{i}} + (5.0 \mathrm{~m}) \hat{\mathrm{j}} + 0 \hat{\mathrm{k}}\]\[\vec{d_{2}} = (-3.0 \mathrm{~m}) \hat{\mathrm{i}} + (4.0 \mathrm{~m}) \hat{\mathrm{j}} + 0 \hat{\mathrm{k}}\] The cross product is calculated as: \[\vec{d_{1}} \times \vec{d_{2}} = \begin{vmatrix} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \ 4.0 & 5.0 & 0 \ -3.0 & 4.0 & 0 \end{vmatrix}\]Calculating the determinant, we only get the k-component: \[\vec{d_{1}} \times \vec{d_{2}} = (4.0 \cdot 4.0) - (5.0 \cdot -3.0) \hat{\mathrm{k}} = (16 + 15) \hat{\mathrm{k}} = 31 \hat{\mathrm{k}} \mathrm{~m^2}\]

The dot product \(\vec{d_{1}} \cdot \vec{d_{2}}\) is computed by multiplying corresponding components and summing them up:\[(4.0 \cdot -3.0) + (5.0 \cdot 4.0) = -12.0 + 20.0 = 8.0 \mathrm{~m^2}\]

First, find \(\vec{d_{1}} + \vec{d_{2}}\): \[\vec{d_{1}} + \vec{d_{2}} = (4.0 - 3.0) \hat{\mathrm{i}} + (5.0 + 4.0) \hat{\mathrm{j}} = (1.0 \mathrm{~m}) \hat{\mathrm{i}} + (9.0 \mathrm{~m}) \hat{\mathrm{j}}\]Then, calculate the dot product of this result and \(\vec{d_{2}}\): \[(1.0 \cdot -3.0) + (9.0 \cdot 4.0) = -3.0 + 36.0 = 33.0 \mathrm{~m^2}\]

The component of \(\vec{d_{1}}\) along \(\vec{d_{2}}\) is given by projecting \(\vec{d_{1}}\) onto \(\vec{d_{2}}\). Use the formula for projection: \[\text{Component} = \frac{\vec{d_{1}} \cdot \vec{d_{2}}}{\|\vec{d_{2}}\|}\]First, find the magnitude of \(\vec{d_{2}}\): \[\|\vec{d_{2}}\| = \sqrt{(-3.0)^2 + (4.0)^2} = \sqrt{9 + 16} = 5 \mathrm{~m}\]Therefore, compute the component: \[\text{Component} = \frac{8.0}{5} = 1.6 \mathrm{~m}\]