If \(\mathbf{E}=20 e^{-5 y}\left(\cos 5 x \mathbf{a}_{x}-\sin 5 x \mathbf{a}_{y}\right)\), find \((a)|\mathbf{E}|\) at \(P(\pi / 6,0.1,2) ;(b)\) a unit vector in the direction of \(\mathbf{E}\) at \(P ;(c)\) the equation of the direction line passing through \(P\).

#tag_title# Step 2: Calculating the magnitude of E #tag_content# Next, we need to find the magnitude of the vector \(\mathbf{E}_{P}\). To do this, we first calculate the components of \(\mathbf{E}_{P}\): \(E_{Px} = 20 e^{-0.5} \cos\left(5 \frac{\pi}{6}\right)\) \(E_{Py} = -20 e^{-0.5} \sin\left(5 \frac{\pi}{6}\right)\) Now, let's find the magnitude of \(\mathbf{E}_{P}\): \(|\mathbf{E}_{P}| = \sqrt{E_{Px}^2 + E_{Py}^2} = \sqrt{(20 e^{-0.5} \cos\left(5 \frac{\pi}{6}\right))^2 + (-20 e^{-0.5} \sin\left(5 \frac{\pi}{6}\right))^2}\) #tag_title# Step 3: Finding the unit vector in the direction of E #tag_content# To find the unit vector in the direction of \(\mathbf{E}_{P}\), we need to divide each component by the magnitude of \(\mathbf{E}_{P}\): \(\mathbf{\hat E}_{P} = \frac{\mathbf{E}_{P}}{|\mathbf{E}_{P}|} = \frac{1}{|\mathbf{E}_{P}|}(E_{Px} \mathbf{a}_{x} + E_{Py} \mathbf{a}_{y})\) #tag_title# Step 4: Obtaining the direction line passing through point P #tag_content# Finally, let's find the equation of the direction line passing through point \(P\). We can use the following equation of the line, where \(r\) is a position vector: \(\mathbf{r} = \mathbf{P} + t\mathbf{\hat E}_{P}\) In coordinates, this is given by: \(x = P_x + t\hat E_{Px}\) \(y = P_y + t\hat E_{Py}\) \(z = P_z + t\hat E_{Pz}\) Substitute the values for point \(P\) and unit vector \(\mathbf{\hat E}_{P}\) to obtain the direction line passing through point \(P\).