# Problem Statement Given: \(\mathbf{F} = 16\mathbf{i} + 10\mathbf{j} + 14\mathbf{k}\) (in Newtons) **Task:** Determine the **magnitude of the projection** of \(\mathbf{F}\) along an axis (or pole) that is **perpendicular to the vector** \(\mathbf{a} = 3\mathbf{i} + 4\mathbf{j} - 12\mathbf{k}\). # Step 1: Find the Unit Vector Along the Perpendicular Axis Since we need the projection **along a direction perpendicular to \(\mathbf{a}\)**, let's call the unit vector along this axis \(\mathbf{n}\), where \(\mathbf{n} \cdot \mathbf{a} = \). But unless a specific perpendicular direction is given, the **maximum possible projection** of \(\mathbf{F}\) perpendicular to \(\mathbf{a}\) is the magnitude of the component of \(\mathbf{F}\) perpendicular to \(\mathbf{a}\). This is calculated as: \[ |\mathbf{F}_{\perp}| = |\mathbf{F} - \mathbf{F}_{\parallel}| \] where \[ \mathbf{F}_{\parallel} = \text{proj}_{\mathbf{a}} \mathbf{F} = \left( \frac{\mathbf{F} \cdot \mathbf{a}}{|\mathbf{a}|^2} \right) \mathbf{a} \] # Step 2: Compute the Magnitude of \(\mathbf{a}\) \[ |\mathbf{a}| = \sqrt{3^2 + 4^2 + (-12)^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13 \] # Step 3: Compute the Dot Product \(\mathbf{F} \cdot \mathbf{a}\) \[ \mathbf{F} \cdot \mathbf{a} = (16)(3) + (10)(4) + (14)(-12) = 48 + 40 - 168 = 88 - 168 = -80 \] # Step 4: Compute the Parallel Component \[ \mathbf{F}_{\parallel} = \left( \frac{-80}{13^2} \right)\mathbf{a} = \left( \frac{-80}{169} \right)\mathbf{a} \] \[ = \left( \frac{-80}{169} \right)(3\mathbf{i} + 4\mathbf{j} - 12\mathbf{k}) \] \[ = \left( \frac{-240}{169} \right)\mathbf{i} + \left( \frac{-320}{169} \right)\mathbf{j} + \left( \frac{960}{169} \right)\mathbf{k} \] # Step 5: Compute the Perpendicular Component \[ \mathbf{F}_{\perp} = \mathbf{F} - \mathbf{F}_{\parallel} \] Component-wise: - \(i\): \(16 - \frac{-240}{169} = 16 + \frac{240}{169} = \frac{16 \times 169 + 240}{169} = \frac{2704 + 240}{169} = \frac{2944}{169}\) - \(j\): \(10 - \frac{-320}{169} = 10 + \frac{320}{169} = \frac{169 + 320}{169} = \frac{201}{169}\) - \(k\): \(14 - \frac{960}{169} = \frac{14 \times 169 - 960}{169} = \frac{2366 - 960}{169} = \frac{1406}{169}\) So, \[ \mathbf{F}_{\perp} = \frac{2944}{169}\mathbf{i} + \frac{201}{169}\mathbf{j} + \frac{1406}{169}\mathbf{k} \] # Step 6: Find the Magnitude of the Perpendicular Component \[ |\mathbf{F}_{\perp}| = \sqrt{\left(\frac{2944}{169}\right)^2 + \left(\frac{201}{169}\right)^2 + \left(\frac{1406}{169}\right)^2} \] \[ = \frac{1}{169}\sqrt{2944^2 + 201^2 + 1406^2} \] Calculate each square: - \(2944^2 = 8,667,136\) - \(201^2 = 4,040,100\) - \(1406^2 = 1,977,636\) Sum: \(8,667,136 + 4,040,100 + 1,977,636 = 14,684,872\) Take the square root: \[ \sqrt{14,684,872} \approx 3,832.16 \] Now divide by 169: \[ |\mathbf{F}_{\perp}| = \frac{3,832.16}{169} \approx 22.68 \text{ N} \] # Final Answer **The magnitude of the projection of \(\mathbf{F}\) along any axis perpendicular to \(\mathbf{a}\) is:** \[ \boxed{22.7\ \text{N}} \] (Rounded to one decimal place) --- **Summary Table** | Step | Value | |------|--------------------------| | \(|\mathbf{a}|\) | \(13\) | | \(\mathbf{F} \cdot \mathbf{a}\) | \(-80\) | | \(\mathbf{F}_{\parallel}\) | See above | | \(\mathbf{F}_{\perp}\) | See above | | Magnitude | \(\approx 22.7\) N | --- **Note:** If you need the projection along a *specific* perpendicular vector, you would need to specify which one, as there are infinitely many perpendicular directions to a given vector in 3D space. --- **No image included as this is a purely vector calculation.**