Question

Find the angle between the line \[ \mathbf{r} = ( \hat{i} + 2\hat{j} + \hat{k} ) + \lambda( \hat{i} + \hat{j} + \hat{k} ) \] and the plane \[ \mathbf{r} \cdot (2\hat{i} + \hat{j} + \hat{k}) = 8. \]

Hint:

To find the angle between a line and a plane, use the formula: \[ \sin \theta = \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}| |\mathbf{n}|}. \]

Answer 1

Step 1: Identify the Direction and Normal Vectors 
- The direction vector of the line is \( \mathbf{d} = (1, 1, 1) \). 
- The normal vector of the plane is \( \mathbf{n} = (2, 1, 1) \). 
Step 2: Apply the Formula for the Angle Between Line and Plane 
The angle \( \theta \) between the line and the plane can be calculated using the angle between the direction vector of the line and the normal vector of the plane: \[ \sin \theta = \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}| |\mathbf{n}|}. \] \[ = \frac{| (1,1,1) \cdot (2,1,1) |}{\sqrt{1^2 + 1^2 + 1^2} \times \sqrt{2^2 + 1^2 + 1^2}}. \] \[ = \frac{| 2 + 1 + 1 |}{\sqrt{3} \times \sqrt{6}}. \] \[ = \frac{4}{\sqrt{18}} = \frac{2\sqrt{2}}{3}. \] \[ \theta = \sin^{-1}\left( \frac{2\sqrt{2}}{3} \right). \]