91Ó°ÊÓ

First, let's write the given vectors in their component form: $$2 \mathbf{j}= \langle 0, 2, 0\rangle$$ $$-5 \mathbf{i}= \langle-5,0,0\rangle$$ Now, both vectors are in their component form: \(\langle x, y, z\rangle\).

Next, we will find the cross product of the two component vectors using the determinant formula for cross product: $$\mathbf{A} \times \mathbf{B} = \left\langle\begin{array}{1} (A_yB_z - A_zB_y),\\ (A_zB_x - A_xB_z),\\ (A_xB_y - A_yB_x) \end{array}\right\rangle$$ Apply the formula to our vectors: $$2\mathbf{j} \times (-5)\mathbf{i}= \left\langle\begin{array}{1} (2 \cdot 0 - 0 \cdot 0),\\ (0 \cdot (-5) - 0 \cdot 0),\\ (0 \cdot 0 - 2 \cdot (-5)) \end{array}\right\rangle = \langle 0, 0, 10 \rangle$$ So, the cross product is given by the vector \(\langle0, 0, 10\rangle\).

Finally, let's sketch the two given vectors (\(2\mathbf{j}\) and \(-5\mathbf{i}\)) and their cross product (\(\langle0, 0, 10\rangle\)). Draw a x-y-z coordinate system on a graph paper with its origin O. Now, plot each vector as follows: 1. \(-5\mathbf{i}\): from the origin O, move 5 units in the negative x-direction. 2. \(2\mathbf{j}\): from the origin O, move 2 units in the positive y-direction. 3. \(\langle0, 0, 10\rangle\): from the origin O, move 10 units in the positive z-direction. Now, connect the origin O to each point, representing each vector. You should see that the vectors \(-5\mathbf{i}\) and \(2\mathbf{j}\) lie in the x-y plane (their z-components are equal to zero). In contrast, their cross product vector \(\langle0, 0, 10\rangle\) lies in the z-axis, perpendicular to both \(2\mathbf{j}\) and \(-5\mathbf{i}\) vectors.